Multiple Integrals Practice

Cheatsheet Content

### Q1. Evaluate the following double integrals: #### (i) $\int_0^1 \int_0^x e^{x/y} dy dx$ **Solution:** This integral is challenging to evaluate directly in the given order. Let's re-examine the limits from the OCR: $\int_0^1 \int_0^x e^{x/y} dy dx$. This order is incorrect based on typical integral setups where the inner integral depends on the outer variable. Assuming the OCR meant: $\int_0^1 \int_y^1 e^{x/y} dx dy$ (integrating with respect to x first) or $\int_0^1 \int_0^y e^{x/y} dx dy$. Given the problem structure, it's highly probable it should be integrated $dy dx$ or $dx dy$. Let's assume the limits are $\int_0^1 \int_0^x e^{x/y} dy dx$. Let's try to change the order of integration if the limits were $\int_0^1 \int_0^x \dots dy dx$. The region of integration is $0 \le x \le 1$ and $0 \le y \le x$. This region is a triangle with vertices (0,0), (1,0), (1,1). Changing the order to $dx dy$: $0 \le y \le 1$ and $y \le x \le 1$. So the integral becomes: $$ I = \int_0^1 \int_y^1 e^{x/y} dx dy $$ Let $u = x/y$, then $du = (1/y) dx$. $$ I = \int_0^1 \left[ y e^{x/y} \right]_y^1 dy $$ $$ I = \int_0^1 (y e^{1/y} - y e^{y/y}) dy $$ $$ I = \int_0^1 (y e^{1/y} - y e) dy $$ This integral is still complex due to $y e^{1/y}$. Let's re-evaluate the OCR interpretation. If it was $\int_0^1 \int_y^1 e^{x/y} dx dy$, then the region is $0 \le y \le 1$ and $y \le x \le 1$. If the original was $\int_0^1 \int_0^x e^{x/y} dy dx$, then $y$ is the inner variable. The integral $\int e^{x/y} dy$ is not elementary. Given the context of multiple integrals, it's possible that the expression is intended to be simpler. Let's assume a common typo and that $x/y$ was meant to be $x y$ or $x^2$. However, we must work with the provided text. If the OCR for (i) was $\int_0^1 \int_0^x e^{x/y} dx dy$, then: $$ \int_0^1 \left[ y e^{x/y} \right]_0^x dy = \int_0^1 (y e^{x/y}|_0^x) dy $$ This is not correct. Let's reconsider the OCR: $\int_0^1 \int_0^x e^{x/y} dy dx$. The inner integral is $\int_0^x e^{x/y} dy$. Let $u = 1/y$. Then $du = -1/y^2 dy$, so $dy = -1/u^2 du$. When $y=0$, $u \to \infty$. When $y=x$, $u = 1/x$. $$ \int_{\infty}^{1/x} e^{xu} (-1/u^2) du = \int_{1/x}^{\infty} e^{xu}/u^2 du $$ This is a non-elementary integral. There might be a typo in the question itself. Assuming it's a standard problem, let's consider the possibility that it was $\int_0^1 \int_0^x e^{xy} dy dx$. Then $\int_0^x e^{xy} dy = [e^{xy}/x]_0^x = (e^{x^2} - 1)/x$. So $\int_0^1 (e^{x^2} - 1)/x dx$, which is also not elementary. Let's assume the question meant $\int_0^1 \int_y^1 e^{x/y} dx dy$. $$ \int_0^1 \left[ y e^{x/y} \right]_y^1 dy = \int_0^1 (y e^{1/y} - y e) dy $$ This is still problematic. Given the typical level of such problems, there might be a simpler interpretation or a common class of problems this belongs to. If it was $\int_0^1 \int_x^1 e^{y/x} dy dx$: $$ \int_0^1 \left[ x e^{y/x} \right]_x^1 dx = \int_0^1 (x e^{1/x} - x e) dx $$ Still not simple. Let's assume the OCR was correct and the problem is intended to be changed to polar coordinates, or a different substitution. If the integrand was $e^{x^2+y^2}$, then polar coordinates would be used. Given the difficulty, and without further clarification, it's hard to provide a definitive elementary solution for $\int_0^1 \int_0^x e^{x/y} dy dx$. This might be a question testing integral properties or a non-elementary integral. #### (ii) $\int_0^1 \int_0^{\sqrt{1+x^2}} \frac{1}{1+x^2+y^2} dy dx$ **Solution:** The region of integration is $0 \le x \le 1$ and $0 \le y \le \sqrt{1+x^2}$. This integral looks like it could benefit from a trigonometric substitution for the inner integral. Let $y = \sqrt{1+x^2} \tan\theta$. Then $dy = \sqrt{1+x^2} \sec^2\theta d\theta$. When $y=0$, $\theta=0$. When $y=\sqrt{1+x^2}$, $\tan\theta=1$, so $\theta=\pi/4$. The inner integral becomes: $$ \int_0^{\pi/4} \frac{1}{1+x^2 + (1+x^2)\tan^2\theta} \sqrt{1+x^2} \sec^2\theta d\theta $$ $$ = \int_0^{\pi/4} \frac{1}{(1+x^2)(1+\tan^2\theta)} \sqrt{1+x^2} \sec^2\theta d\theta $$ $$ = \int_0^{\pi/4} \frac{1}{(1+x^2)\sec^2\theta} \sqrt{1+x^2} \sec^2\theta d\theta $$ $$ = \int_0^{\pi/4} \frac{1}{\sqrt{1+x^2}} d\theta $$ $$ = \frac{1}{\sqrt{1+x^2}} [\theta]_0^{\pi/4} = \frac{\pi/4}{\sqrt{1+x^2}} $$ Now, substitute this back into the outer integral: $$ I = \int_0^1 \frac{\pi/4}{\sqrt{1+x^2}} dx $$ $$ I = \frac{\pi}{4} \int_0^1 \frac{1}{\sqrt{1+x^2}} dx $$ This is a standard integral: $\int \frac{1}{\sqrt{a^2+x^2}} dx = \ln|x + \sqrt{a^2+x^2}|$. $$ I = \frac{\pi}{4} \left[ \ln|x + \sqrt{1+x^2}| \right]_0^1 $$ $$ I = \frac{\pi}{4} \left( \ln|1 + \sqrt{1+1}| - \ln|0 + \sqrt{1+0}| \right) $$ $$ I = \frac{\pi}{4} \left( \ln(1 + \sqrt{2}) - \ln(1) \right) $$ $$ I = \frac{\pi}{4} \ln(1 + \sqrt{2}) $$ #### (iii) $\int_0^1 \int_0^{2\sqrt{x}} (x^2+y^2) dy dx$ **Solution:** The region of integration is $0 \le x \le 1$ and $0 \le y \le 2\sqrt{x}$. First, evaluate the inner integral with respect to $y$: $$ \int_0^{2\sqrt{x}} (x^2+y^2) dy = \left[ x^2y + \frac{y^3}{3} \right]_0^{2\sqrt{x}} $$ $$ = x^2(2\sqrt{x}) + \frac{(2\sqrt{x})^3}{3} - (0) $$ $$ = 2x^{5/2} + \frac{8x^{3/2}}{3} $$ Now, integrate this expression with respect to $x$ from 0 to 1: $$ I = \int_0^1 \left( 2x^{5/2} + \frac{8}{3}x^{3/2} \right) dx $$ $$ = \left[ 2 \frac{x^{7/2}}{7/2} + \frac{8}{3} \frac{x^{5/2}}{5/2} \right]_0^1 $$ $$ = \left[ \frac{4}{7}x^{7/2} + \frac{16}{15}x^{5/2} \right]_0^1 $$ $$ = \left( \frac{4}{7}(1)^{7/2} + \frac{16}{15}(1)^{5/2} \right) - (0) $$ $$ = \frac{4}{7} + \frac{16}{15} $$ To combine these fractions, find a common denominator (105): $$ = \frac{4 \times 15}{7 \times 15} + \frac{16 \times 7}{15 \times 7} = \frac{60}{105} + \frac{112}{105} $$ $$ = \frac{172}{105} $$ #### (iv) $\iint_D (1+x+y) dx dy$: D is the region bounded by $x=-y$, $x=\sqrt{y}$, $y=0$, $y=2$. **Solution:** The region D is defined by $x=-y$, $x=\sqrt{y}$, $y=0$, $y=2$. We can integrate with respect to $x$ first, then $y$. For a given $y$, $x$ ranges from $-y$ to $\sqrt{y}$. The outer integral for $y$ ranges from $0$ to $2$. $$ I = \int_0^2 \int_{-y}^{\sqrt{y}} (1+x+y) dx dy $$ First, evaluate the inner integral with respect to $x$: $$ \int_{-y}^{\sqrt{y}} (1+x+y) dx = \left[ x + \frac{x^2}{2} + yx \right]_{-y}^{\sqrt{y}} $$ $$ = \left( \sqrt{y} + \frac{(\sqrt{y})^2}{2} + y\sqrt{y} \right) - \left( (-y) + \frac{(-y)^2}{2} + y(-y) \right) $$ $$ = \left( y^{1/2} + \frac{y}{2} + y^{3/2} \right) - \left( -y + \frac{y^2}{2} - y^2 \right) $$ $$ = y^{1/2} + \frac{y}{2} + y^{3/2} + y - \frac{y^2}{2} + y^2 $$ $$ = y^{3/2} + \frac{3}{2}y + y^{1/2} + \frac{y^2}{2} $$ Now, integrate this expression with respect to $y$ from 0 to 2: $$ I = \int_0^2 \left( y^{3/2} + \frac{3}{2}y + y^{1/2} + \frac{y^2}{2} \right) dy $$ $$ = \left[ \frac{y^{5/2}}{5/2} + \frac{3}{2} \frac{y^2}{2} + \frac{y^{3/2}}{3/2} + \frac{1}{2} \frac{y^3}{3} \right]_0^2 $$ $$ = \left[ \frac{2}{5}y^{5/2} + \frac{3}{4}y^2 + \frac{2}{3}y^{3/2} + \frac{1}{6}y^3 \right]_0^2 $$ Substitute $y=2$: $$ = \frac{2}{5}(2)^{5/2} + \frac{3}{4}(2)^2 + \frac{2}{3}(2)^{3/2} + \frac{1}{6}(2)^3 $$ $$ = \frac{2}{5}(4\sqrt{2}) + \frac{3}{4}(4) + \frac{2}{3}(2\sqrt{2}) + \frac{1}{6}(8) $$ $$ = \frac{8\sqrt{2}}{5} + 3 + \frac{4\sqrt{2}}{3} + \frac{4}{3} $$ Combine terms: $$ = \left( \frac{8\sqrt{2}}{5} + \frac{4\sqrt{2}}{3} \right) + \left( 3 + \frac{4}{3} \right) $$ $$ = \frac{24\sqrt{2} + 20\sqrt{2}}{15} + \frac{9+4}{3} $$ $$ = \frac{44\sqrt{2}}{15} + \frac{13}{3} $$ #### (v) $\iint_D xy(x+y) dx dy$: D is the area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. **Solution:** The integral is $\iint_D (x^2y + xy^2) dx dy$ over the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Consider the symmetry of the integrand and the region. The integrand $f(x,y) = x^2y + xy^2$. The region D is symmetric with respect to both the x-axis and the y-axis. Let's check the parity of the terms in the integrand: - For $x^2y$: If we replace $y$ with $-y$, $x^2(-y) = -x^2y$. This term is odd with respect to $y$. - For $xy^2$: If we replace $x$ with $-x$, $(-x)y^2 = -xy^2$. This term is odd with respect to $x$. For a region symmetric about the x-axis, if $f(x,y)$ is odd with respect to $y$ (i.e., $f(x,-y) = -f(x,y)$), then $\iint_D f(x,y) dx dy = 0$. For a region symmetric about the y-axis, if $f(x,y)$ is odd with respect to $x$ (i.e., $f(-x,y) = -f(x,y)$), then $\iint_D f(x,y) dx dy = 0$. Let's apply this: $$ \iint_D (x^2y + xy^2) dx dy = \iint_D x^2y \, dx dy + \iint_D xy^2 \, dx dy $$ For $\iint_D x^2y \, dx dy$: The integrand $x^2y$ is odd with respect to $y$. Since the ellipse is symmetric about the x-axis, this integral is 0. For $\iint_D xy^2 \, dx dy$: The integrand $xy^2$ is odd with respect to $x$. Since the ellipse is symmetric about the y-axis, this integral is 0. Therefore, the total integral is $0 + 0 = 0$. **Answer:** $0$ ### Q2. Evaluate the following integrals (Polar Coordinates): #### (i) $\iint_D r \sin\theta \, dr d\theta$ over the cardioid $r = a(1-\cos\theta)$ above the initial line. **Solution:** "Above the initial line" means $y \ge 0$, which corresponds to $0 \le \theta \le \pi$. The region D is described by $0 \le \theta \le \pi$ and $0 \le r \le a(1-\cos\theta)$. The integral is: $$ I = \int_0^\pi \int_0^{a(1-\cos\theta)} r \sin\theta \, r dr d\theta $$ Note: The problem statement gives $\iint r \sin\theta dr d\theta$. In polar coordinates, the area element is $dA = r dr d\theta$. So the integrand $r \sin\theta$ is already part of the function $f(r,\theta)$ and the $r$ from the Jacobian is missing. Assuming the problem wants to integrate $f(r,\theta) = r \sin\theta$ with the Jacobian $r$, the integral should be $\iint (r \sin\theta) r dr d\theta = \iint r^2 \sin\theta dr d\theta$. If it meant $f(r,\theta) = \sin\theta$, then it would be $\iint \sin\theta r dr d\theta$. Given the wording, it's safer to assume the integrand is $r \sin\theta$ and the $r$ from $dA$ is applied. Let's proceed with $\iint r^2 \sin\theta dr d\theta$: $$ I = \int_0^\pi \int_0^{a(1-\cos\theta)} r^2 \sin\theta \, dr d\theta $$ First, integrate with respect to $r$: $$ \int_0^{a(1-\cos\theta)} r^2 \sin\theta \, dr = \sin\theta \left[ \frac{r^3}{3} \right]_0^{a(1-\cos\theta)} $$ $$ = \sin\theta \frac{(a(1-\cos\theta))^3}{3} = \frac{a^3}{3} (1-\cos\theta)^3 \sin\theta $$ Now, integrate this with respect to $\theta$ from $0$ to $\pi$: $$ I = \int_0^\pi \frac{a^3}{3} (1-\cos\theta)^3 \sin\theta \, d\theta $$ Let $u = 1-\cos\theta$. Then $du = \sin\theta \, d\theta$. When $\theta=0$, $u = 1-\cos(0) = 1-1=0$. When $\theta=\pi$, $u = 1-\cos(\pi) = 1-(-1)=2$. $$ I = \frac{a^3}{3} \int_0^2 u^3 \, du $$ $$ = \frac{a^3}{3} \left[ \frac{u^4}{4} \right]_0^2 $$ $$ = \frac{a^3}{3} \left( \frac{2^4}{4} - 0 \right) = \frac{a^3}{3} \frac{16}{4} = \frac{a^3}{3} \times 4 = \frac{4a^3}{3} $$ #### (ii) $\iint_D r^3 \, dr d\theta$ over the area included between the circles $r = 2 \sin\theta$ and $r = 4 \sin\theta$. **Solution:** The region is between two circles. Both circles pass through the origin. $r = 2 \sin\theta$ is a circle with diameter 2, centered at $(0,1)$ in Cartesian coordinates. $r = 4 \sin\theta$ is a circle with diameter 4, centered at $(0,2)$ in Cartesian coordinates. Both circles are traced out as $\theta$ goes from $0$ to $\pi$. So, the limits for $\theta$ are $0$ to $\pi$. For a given $\theta$, $r$ ranges from $2\sin\theta$ to $4\sin\theta$. The integral is $\iint r^3 \, dr d\theta$. With the Jacobian $r$, the integrand becomes $r^3 \times r = r^4$. $$ I = \int_0^\pi \int_{2\sin\theta}^{4\sin\theta} r^4 \, dr d\theta $$ First, integrate with respect to $r$: $$ \int_{2\sin\theta}^{4\sin\theta} r^4 \, dr = \left[ \frac{r^5}{5} \right]_{2\sin\theta}^{4\sin\theta} $$ $$ = \frac{(4\sin\theta)^5}{5} - \frac{(2\sin\theta)^5}{5} $$ $$ = \frac{1024\sin^5\theta}{5} - \frac{32\sin^5\theta}{5} $$ $$ = \frac{992}{5} \sin^5\theta $$ Now, integrate this with respect to $\theta$ from $0$ to $\pi$: $$ I = \int_0^\pi \frac{992}{5} \sin^5\theta \, d\theta $$ Since $\sin^5\theta$ is symmetric about $\theta = \pi/2$ (i.e., $\sin^5(\pi-\theta) = \sin^5\theta$), we can write: $$ I = \frac{992}{5} \times 2 \int_0^{\pi/2} \sin^5\theta \, d\theta $$ Using Wallis' Integral formula for $\int_0^{\pi/2} \sin^n x dx$: $$ \int_0^{\pi/2} \sin^5\theta \, d\theta = \frac{(5-1)(5-3)}{5 \times 3 \times 1} \times 1 = \frac{4 \times 2}{5 \times 3 \times 1} = \frac{8}{15} $$ $$ I = \frac{992}{5} \times 2 \times \frac{8}{15} $$ $$ I = \frac{992 \times 16}{75} = \frac{15872}{75} $$ #### (iii) $\iint_D r^2 \sin\theta \, dr d\theta$ over the semi-circle $r = 2a \cos\theta$ above the initial line. **Solution:** The semi-circle $r = 2a \cos\theta$ above the initial line ($y \ge 0$) corresponds to $0 \le \theta \le \pi/2$. (Note: $r=2a\cos\theta$ traces a circle for $0 \le \theta \le \pi$. For $y \ge 0$, $\sin\theta \ge 0$, which is $0 \le \theta \le \pi$. However, $r$ must be positive, so $\cos\theta \ge 0$, which means $0 \le \theta \le \pi/2$. This describes the right half of the circle. The problem specifies "semi-circle above the initial line", which implies the upper half of the circle. A common interpretation for $r=2a\cos\theta$ is a circle centered at $(a,0)$ with radius $a$. The upper half of this circle is indeed $0 \le \theta \le \pi/2$.) The region D is described by $0 \le \theta \le \pi/2$ and $0 \le r \le 2a \cos\theta$. The integral is $\iint r^2 \sin\theta \, dr d\theta$. With the Jacobian $r$, the integrand becomes $r^2 \sin\theta \times r = r^3 \sin\theta$. $$ I = \int_0^{\pi/2} \int_0^{2a\cos\theta} r^3 \sin\theta \, dr d\theta $$ First, integrate with respect to $r$: $$ \int_0^{2a\cos\theta} r^3 \sin\theta \, dr = \sin\theta \left[ \frac{r^4}{4} \right]_0^{2a\cos\theta} $$ $$ = \sin\theta \frac{(2a\cos\theta)^4}{4} = \sin\theta \frac{16a^4\cos^4\theta}{4} = 4a^4 \cos^4\theta \sin\theta $$ Now, integrate this with respect to $\theta$ from $0$ to $\pi/2$: $$ I = \int_0^{\pi/2} 4a^4 \cos^4\theta \sin\theta \, d\theta $$ Let $u = \cos\theta$. Then $du = -\sin\theta \, d\theta$. When $\theta=0$, $u = \cos(0) = 1$. When $\theta=\pi/2$, $u = \cos(\pi/2) = 0$. $$ I = 4a^4 \int_1^0 u^4 (-du) $$ $$ = 4a^4 \int_0^1 u^4 \, du $$ $$ = 4a^4 \left[ \frac{u^5}{5} \right]_0^1 $$ $$ = 4a^4 \left( \frac{1^5}{5} - 0 \right) = \frac{4a^4}{5} $$ ### Q3. Evaluate the following integrals by changing the order of integration: #### (i) $\int_0^\infty \int_x^\infty \frac{e^{-y}}{y} dy dx$ **Solution:** The region of integration is $0 \le x ### Q4. Find the area lying inside the cardioid $r = a(1 + \cos\theta)$ and outside the circle $r = a$. **Solution:** The area in polar coordinates is given by $\frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta$. We need to find the area of the cardioid minus the area of the circle that is inside the cardioid. First, find the intersection points of $r = a(1+\cos\theta)$ and $r=a$: $a = a(1+\cos\theta) \Rightarrow 1 = 1+\cos\theta \Rightarrow \cos\theta = 0$. So $\theta = \pi/2$ and $\theta = -\pi/2$ (or $3\pi/2$). The cardioid is traced from $\theta=0$ to $\theta=2\pi$. The circle is centered at the origin. The region "inside the cardioid and outside the circle" means $a \le r \le a(1+\cos\theta)$. This condition $a \le a(1+\cos\theta)$ implies $1 \le 1+\cos\theta \Rightarrow \cos\theta \ge 0$. So $\theta$ must be in the range $[-\pi/2, \pi/2]$. The area $A$ is given by: $$ A = \frac{1}{2} \int_{-\pi/2}^{\pi/2} \left[ (a(1+\cos\theta))^2 - a^2 \right] d\theta $$ Due to symmetry, we can integrate from $0$ to $\pi/2$ and multiply by 2: $$ A = \int_0^{\pi/2} \left[ a^2(1+2\cos\theta+\cos^2\theta) - a^2 \right] d\theta $$ $$ A = \int_0^{\pi/2} \left[ a^2+2a^2\cos\theta+a^2\cos^2\theta - a^2 \right] d\theta $$ $$ A = \int_0^{\pi/2} \left[ 2a^2\cos\theta+a^2\cos^2\theta \right] d\theta $$ $$ A = a^2 \int_0^{\pi/2} (2\cos\theta+\cos^2\theta) d\theta $$ Use the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$: $$ A = a^2 \int_0^{\pi/2} \left( 2\cos\theta + \frac{1+\cos(2\theta)}{2} \right) d\theta $$ $$ A = a^2 \left[ 2\sin\theta + \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_0^{\pi/2} $$ $$ A = a^2 \left( \left( 2\sin(\pi/2) + \frac{\pi/2}{2} + \frac{\sin(\pi)}{4} \right) - \left( 2\sin(0) + 0 + \frac{\sin(0)}{4} \right) \right) $$ $$ A = a^2 \left( 2(1) + \frac{\pi}{4} + 0 - 0 \right) $$ $$ A = a^2 \left( 2 + \frac{\pi}{4} \right) = 2a^2 + \frac{\pi a^2}{4} $$ ### Q5. Find the area lying between the parabola $y = x^2$ and the line $x - y + z = 0$. **Solution:** This question seems to have a typo, as "z" is usually a 3D coordinate. Assuming the line is in the xy-plane, it should be $x-y+C=0$ or $x-y=0$ (i.e. $y=x$). Let's assume the line is $y=x$. We need to find the area between $y=x^2$ and $y=x$. First, find the intersection points: $x^2 = x \Rightarrow x^2-x=0 \Rightarrow x(x-1)=0$. So $x=0$ and $x=1$. The corresponding y-values are $y=0^2=0$ and $y=1^2=1$. The intersection points are $(0,0)$ and $(1,1)$. Between $x=0$ and $x=1$, the line $y=x$ is above the parabola $y=x^2$ (e.g., at $x=0.5$, $y=0.5$ for the line, $y=0.25$ for the parabola). The area $A$ is given by: $$ A = \int_0^1 (x - x^2) dx $$ $$ A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 $$ $$ A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - (0) $$ $$ A = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6} $$ If the line was $x-y+C=0$, the value would depend on $C$. Given $x-y+z=0$, it is most likely a typo and meant $y=x$. ### Q6. Evaluate $\iint_R (x + y)^2 dx dy$, where R is the parallelogram in the $xy$-plane with vertices $(1, 0), (3, 1), (2, 2), (0, 1)$ using the transformation $u = x + y$ and $v = x - 2y$. **Solution:** The transformation is $u = x+y$ and $v = x-2y$. We need to find $x$ and $y$ in terms of $u$ and $v$. From $u=x+y$, $y=u-x$. Substitute into the second equation: $v = x - 2(u-x) = x - 2u + 2x = 3x - 2u$. So $3x = v+2u \Rightarrow x = \frac{1}{3}(v+2u)$. Then $y = u - x = u - \frac{1}{3}(v+2u) = \frac{3u - v - 2u}{3} = \frac{1}{3}(u-v)$. So, $x = \frac{1}{3}(2u+v)$ and $y = \frac{1}{3}(u-v)$. Next, we need to find the Jacobian of the transformation: $J = \left| \frac{\partial(x,y)}{\partial(u,v)} \right|$. $$ J = \left| \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \right| = \left| \begin{vmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{vmatrix} \right| $$ $$ J = \left| (2/3)(-1/3) - (1/3)(1/3) \right| = \left| -2/9 - 1/9 \right| = \left| -3/9 \right| = \left| -1/3 \right| = 1/3 $$ So $dx dy = |J| du dv = \frac{1}{3} du dv$. Now, transform the vertices of the parallelogram from $(x,y)$ to $(u,v)$: 1. $(x,y)=(1,0)$: $u=1+0=1$, $v=1-2(0)=1$. Point $(1,1)$. 2. $(x,y)=(3,1)$: $u=3+1=4$, $v=3-2(1)=1$. Point $(4,1)$. 3. $(x,y)=(2,2)$: $u=2+2=4$, $v=2-2(2)=-2$. Point $(4,-2)$. 4. $(x,y)=(0,1)$: $u=0+1=1$, $v=0-2(1)=-2$. Point $(1,-2)$. In the $uv$-plane, the vertices are $(1,1), (4,1), (4,-2), (1,-2)$. This is a rectangle with sides parallel to the axes. The limits for $u$ are from $1$ to $4$. The limits for $v$ are from $-2$ to $1$. The integrand is $(x+y)^2 = u^2$. So the integral becomes: $$ I = \iint_{R'} u^2 \left( \frac{1}{3} \right) du dv $$ $$ I = \frac{1}{3} \int_1^4 \int_{-2}^1 u^2 \, dv du $$ First, integrate with respect to $v$: $$ \int_{-2}^1 u^2 \, dv = u^2 [v]_{-2}^1 = u^2 (1 - (-2)) = 3u^2 $$ Now, integrate with respect to $u$: $$ I = \frac{1}{3} \int_1^4 3u^2 \, du = \int_1^4 u^2 \, du $$ $$ = \left[ \frac{u^3}{3} \right]_1^4 $$ $$ = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 $$ ### Q7. Evaluate $\iint_D [xy(1-x - y)]^{1/2} dx dy$ over the region D bounded by $x = 0, y = 0, x + y = 1$ by changing the variables. **Solution:** The region D is a triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$. This region is suitable for the transformation $u=x$, $v=y$, $w=1-x-y$. However, a more common transformation for such a triangular region is $u=x$, $v=y/(1-x)$ or $u=x+y$, $v=y$. Let's use the transformation $u=x+y$ and $v=y$. (Or $u=x$, $v=y$, and $s=1-x-y$ but that is for triple integrals). Let $u = x+y$ and $v = y$. Then $x = u-v$ and $y=v$. The Jacobian $J = \left| \frac{\partial(x,y)}{\partial(u,v)} \right|$: $$ J = \left| \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \right| = \left| \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} \right| $$ $$ J = |(1)(1) - (-1)(0)| = |1| = 1 $$ So $dx dy = du dv$. Now, transform the region D: - $x=0$: $u-v=0 \Rightarrow u=v$. - $y=0$: $v=0$. - $x+y=1$: $u=1$. The new region $D'$ in the $uv$-plane is bounded by $v=0$, $u=1$, and $u=v$. This is a triangle with vertices $(0,0)$, $(1,0)$, $(1,1)$. The limits are $0 \le u \le 1$ and $0 \le v \le u$. (Or $0 \le v \le 1$ and $v \le u \le 1$). The integrand is $[xy(1-x-y)]^{1/2}$. Substitute $x=u-v$, $y=v$, and $1-x-y = 1-(u-v)-v = 1-u$. So the integrand becomes $[(u-v)v(1-u)]^{1/2}$. The integral is: $$ I = \int_0^1 \int_0^u [(u-v)v(1-u)]^{1/2} \, dv du $$ $$ I = \int_0^1 (1-u)^{1/2} \int_0^u [v(u-v)]^{1/2} \, dv du $$ Consider the inner integral: $\int_0^u [v(u-v)]^{1/2} \, dv$. Let $v = u \sin^2\phi$. Then $dv = 2u \sin\phi \cos\phi \, d\phi$. When $v=0$, $\phi=0$. When $v=u$, $\sin^2\phi=1 \Rightarrow \phi=\pi/2$. $$ \int_0^{\pi/2} [u\sin^2\phi (u-u\sin^2\phi)]^{1/2} (2u \sin\phi \cos\phi) \, d\phi $$ $$ = \int_0^{\pi/2} [u^2\sin^2\phi (1-\sin^2\phi)]^{1/2} (2u \sin\phi \cos\phi) \, d\phi $$ $$ = \int_0^{\pi/2} [u^2\sin^2\phi \cos^2\phi]^{1/2} (2u \sin\phi \cos\phi) \, d\phi $$ $$ = \int_0^{\pi/2} |u\sin\phi \cos\phi| (2u \sin\phi \cos\phi) \, d\phi $$ Since $u \ge 0$ and $\sin\phi, \cos\phi \ge 0$ for $0 \le \phi \le \pi/2$: $$ = \int_0^{\pi/2} u\sin\phi \cos\phi (2u \sin\phi \cos\phi) \, d\phi $$ $$ = 2u^2 \int_0^{\pi/2} \sin^2\phi \cos^2\phi \, d\phi $$ Using $\sin\phi\cos\phi = \frac{1}{2}\sin(2\phi)$: $$ = 2u^2 \int_0^{\pi/2} \left( \frac{1}{2}\sin(2\phi) \right)^2 \, d\phi $$ $$ = 2u^2 \int_0^{\pi/2} \frac{1}{4}\sin^2(2\phi) \, d\phi $$ $$ = \frac{u^2}{2} \int_0^{\pi/2} \sin^2(2\phi) \, d\phi $$ Using $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$: $$ = \frac{u^2}{2} \int_0^{\pi/2} \frac{1-\cos(4\phi)}{2} \, d\phi $$ $$ = \frac{u^2}{4} \left[ \phi - \frac{\sin(4\phi)}{4} \right]_0^{\pi/2} $$ $$ = \frac{u^2}{4} \left( \left( \frac{\pi}{2} - \frac{\sin(2\pi)}{4} \right) - (0) \right) $$ $$ = \frac{u^2}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi u^2}{8} $$ Now, substitute this back into the outer integral: $$ I = \int_0^1 (1-u)^{1/2} \frac{\pi u^2}{8} \, du $$ $$ = \frac{\pi}{8} \int_0^1 u^2 (1-u)^{1/2} \, du $$ This is a Beta function integral: $B(a,b) = \int_0^1 t^{a-1}(1-t)^{b-1} dt = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. Here $a-1=2 \Rightarrow a=3$. And $b-1=1/2 \Rightarrow b=3/2$. $$ I = \frac{\pi}{8} B(3, 3/2) = \frac{\pi}{8} \frac{\Gamma(3)\Gamma(3/2)}{\Gamma(3+3/2)} = \frac{\pi}{8} \frac{\Gamma(3)\Gamma(3/2)}{\Gamma(9/2)} $$ We know $\Gamma(3) = 2! = 2$. $\Gamma(3/2) = \frac{1}{2}\Gamma(1/2) = \frac{\sqrt{\pi}}{2}$. $\Gamma(9/2) = \frac{7}{2}\Gamma(7/2) = \frac{7}{2}\frac{5}{2}\Gamma(5/2) = \frac{7}{2}\frac{5}{2}\frac{3}{2}\Gamma(3/2) = \frac{7}{2}\frac{5}{2}\frac{3}{2}\frac{\sqrt{\pi}}{2} = \frac{105\sqrt{\pi}}{16}$. $$ I = \frac{\pi}{8} \frac{2 \times (\sqrt{\pi}/2)}{105\sqrt{\pi}/16} = \frac{\pi}{8} \frac{\sqrt{\pi}}{105\sqrt{\pi}/16} $$ $$ = \frac{\pi}{8} \frac{16}{105} = \frac{2\pi}{105} $$ ### Q8. Show that $\int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dy dx \ne \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dx dy$. **Solution:** This problem demonstrates that Fubini's Theorem (which allows swapping order of integration) does not apply when the integrand is not absolutely integrable over the region. Let $f(x,y) = \frac{x-y}{(x+y)^3}$. We need to evaluate both iterated integrals. First integral: $I_1 = \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dy dx$. Inner integral: $\int_0^1 \frac{x-y}{(x+y)^3} dy$. Let $u = x+y$, then $du = dy$. And $y=u-x$. $$ \int_x^{x+1} \frac{x-(u-x)}{u^3} du = \int_x^{x+1} \frac{2x-u}{u^3} du $$ $$ = \int_x^{x+1} \left( \frac{2x}{u^3} - \frac{1}{u^2} \right) du $$ $$ = \left[ -\frac{x}{u^2} + \frac{1}{u} \right]_x^{x+1} $$ $$ = \left( -\frac{x}{(x+1)^2} + \frac{1}{x+1} \right) - \left( -\frac{x}{x^2} + \frac{1}{x} \right) $$ $$ = \left( \frac{-x + (x+1)}{(x+1)^2} \right) - \left( -\frac{1}{x} + \frac{1}{x} \right) $$ $$ = \frac{1}{(x+1)^2} - 0 = \frac{1}{(x+1)^2} $$ Now, integrate this with respect to $x$: $$ I_1 = \int_0^1 \frac{1}{(x+1)^2} dx $$ $$ = \left[ -\frac{1}{x+1} \right]_0^1 $$ $$ = \left( -\frac{1}{1+1} \right) - \left( -\frac{1}{0+1} \right) $$ $$ = -\frac{1}{2} - (-1) = 1 - \frac{1}{2} = \frac{1}{2} $$ Second integral: $I_2 = \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dx dy$. Inner integral: $\int_0^1 \frac{x-y}{(x+y)^3} dx$. Let $u = x+y$, then $du = dx$. And $x=u-y$. $$ \int_y^{y+1} \frac{(u-y)-y}{u^3} du = \int_y^{y+1} \frac{u-2y}{u^3} du $$ $$ = \int_y^{y+1} \left( \frac{1}{u^2} - \frac{2y}{u^3} \right) du $$ $$ = \left[ -\frac{1}{u} + \frac{y}{u^2} \right]_y^{y+1} $$ $$ = \left( -\frac{1}{y+1} + \frac{y}{(y+1)^2} \right) - \left( -\frac{1}{y} + \frac{y}{y^2} \right) $$ $$ = \left( \frac{-(y+1) + y}{(y+1)^2} \right) - \left( -\frac{1}{y} + \frac{1}{y} \right) $$ $$ = \frac{-1}{(y+1)^2} - 0 = -\frac{1}{(y+1)^2} $$ Now, integrate this with respect to $y$: $$ I_2 = \int_0^1 -\frac{1}{(y+1)^2} dy $$ $$ = -\left[ -\frac{1}{y+1} \right]_0^1 $$ $$ = -\left( \left( -\frac{1}{2} \right) - \left( -1 \right) \right) $$ $$ = -\left( -\frac{1}{2} + 1 \right) = -\left( \frac{1}{2} \right) = -\frac{1}{2} $$ Since $I_1 = 1/2$ and $I_2 = -1/2$, they are not equal. This shows that $\int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dy dx \ne \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dx dy$. ### Q9. Evaluate $\iint_D x^2y^2 dx dy$ over the circle $x^2 + y^2 = 1$. **Solution:** The region D is the unit circle centered at the origin. It is convenient to use polar coordinates. $x = r\cos\theta$, $y = r\sin\theta$. $dx dy = r dr d\theta$. The circle $x^2+y^2=1$ becomes $r^2=1 \Rightarrow r=1$. The limits for $r$ are $0 \le r \le 1$. The limits for $\theta$ are $0 \le \theta \le 2\pi$. The integrand $x^2y^2 = (r\cos\theta)^2(r\sin\theta)^2 = r^2\cos^2\theta r^2\sin^2\theta = r^4\cos^2\theta\sin^2\theta$. The integral becomes: $$ I = \int_0^{2\pi} \int_0^1 r^4\cos^2\theta\sin^2\theta \, r dr d\theta $$ $$ I = \int_0^{2\pi} \int_0^1 r^5\cos^2\theta\sin^2\theta \, dr d\theta $$ Separate the integrals: $$ I = \left( \int_0^1 r^5 \, dr \right) \left( \int_0^{2\pi} \cos^2\theta\sin^2\theta \, d\theta \right) $$ First part: $$ \int_0^1 r^5 \, dr = \left[ \frac{r^6}{6} \right]_0^1 = \frac{1}{6} $$ Second part: Use $\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$: $$ \int_0^{2\pi} \left( \frac{1}{2}\sin(2\theta) \right)^2 \, d\theta = \int_0^{2\pi} \frac{1}{4}\sin^2(2\theta) \, d\theta $$ Use $\sin^2\phi = \frac{1-\cos(2\phi)}{2}$: $$ = \frac{1}{4} \int_0^{2\pi} \frac{1-\cos(4\theta)}{2} \, d\theta $$ $$ = \frac{1}{8} \int_0^{2\pi} (1-\cos(4\theta)) \, d\theta $$ $$ = \frac{1}{8} \left[ \theta - \frac{\sin(4\theta)}{4} \right]_0^{2\pi} $$ $$ = \frac{1}{8} \left( (2\pi - \frac{\sin(8\pi)}{4}) - (0 - \frac{\sin(0)}{4}) \right) $$ $$ = \frac{1}{8} (2\pi - 0 - 0 + 0) = \frac{2\pi}{8} = \frac{\pi}{4} $$ Combine the parts: $$ I = \frac{1}{6} \times \frac{\pi}{4} = \frac{\pi}{24} $$ ### Q10. Use double integral to find the volume of the following solids: #### (i) $x^2 + y^2 + z^2 = r^2$ **Solution:** This equation describes a sphere of radius $r$. The volume of a solid can be found by $\iint_R z \, dA$, where $R$ is the projection of the solid onto the $xy$-plane. For a sphere $x^2+y^2+z^2=r^2$, the upper hemisphere is $z = \sqrt{r^2-x^2-y^2}$. The projection onto the $xy$-plane is the disk $x^2+y^2 \le r^2$. The total volume of the sphere is twice the volume of the upper hemisphere. $$ V = 2 \iint_D \sqrt{r^2-x^2-y^2} \, dx dy $$ where $D$ is the disk $x^2+y^2 \le r^2$. It is best to use polar coordinates: $x = \rho\cos\phi$, $y = \rho\sin\phi$, $dx dy = \rho d\rho d\phi$. Note: I'm using $\rho$ for the radial coordinate to avoid confusion with the given $r$ for the radius of the sphere. $$ V = 2 \int_0^{2\pi} \int_0^r \sqrt{r^2-\rho^2} \, \rho d\rho d\phi $$ First, integrate with respect to $\rho$: Let $u = r^2-\rho^2$. Then $du = -2\rho d\rho \Rightarrow \rho d\rho = -\frac{1}{2} du$. When $\rho=0$, $u=r^2$. When $\rho=r$, $u=0$. $$ \int_{r^2}^0 \sqrt{u} \left( -\frac{1}{2} \right) du = -\frac{1}{2} \int_{r^2}^0 u^{1/2} du $$ $$ = \frac{1}{2} \int_0^{r^2} u^{1/2} du = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_0^{r^2} $$ $$ = \frac{1}{2} \frac{2}{3} [u^{3/2}]_0^{r^2} = \frac{1}{3} ((r^2)^{3/2} - 0) = \frac{1}{3} r^3 $$ Now, integrate with respect to $\phi$: $$ V = 2 \int_0^{2\pi} \frac{1}{3} r^3 \, d\phi $$ $$ V = \frac{2}{3} r^3 [\phi]_0^{2\pi} = \frac{2}{3} r^3 (2\pi - 0) = \frac{4}{3}\pi r^3 $$ This is the standard formula for the volume of a sphere. #### (ii) Cylinder $x^2 + y^2 = 4$ and the planes $y + z = 4$ and $z = 0$. **Solution:** The base of the solid is the disk $x^2+y^2 \le 4$ in the $xy$-plane (since $z=0$ is the lower bound). The upper surface is the plane $y+z=4$, which means $z = 4-y$. So we need to integrate $z=4-y$ over the disk $x^2+y^2 \le 4$. $$ V = \iint_D (4-y) \, dx dy $$ where $D$ is the disk $x^2+y^2 \le 4$. It is best to use polar coordinates: $x = r\cos\theta$, $y = r\sin\theta$, $dx dy = r dr d\theta$. The disk $x^2+y^2 \le 4$ becomes $0 \le r \le 2$ and $0 \le \theta \le 2\pi$. $$ V = \int_0^{2\pi} \int_0^2 (4 - r\sin\theta) r \, dr d\theta $$ $$ V = \int_0^{2\pi} \int_0^2 (4r - r^2\sin\theta) \, dr d\theta $$ First, integrate with respect to $r$: $$ \int_0^2 (4r - r^2\sin\theta) \, dr = \left[ 2r^2 - \frac{r^3}{3}\sin\theta \right]_0^2 $$ $$ = \left( 2(2^2) - \frac{2^3}{3}\sin\theta \right) - (0) $$ $$ = 8 - \frac{8}{3}\sin\theta $$ Now, integrate with respect to $\theta$: $$ V = \int_0^{2\pi} \left( 8 - \frac{8}{3}\sin\theta \right) d\theta $$ $$ = \left[ 8\theta + \frac{8}{3}\cos\theta \right]_0^{2\pi} $$ $$ = \left( 8(2\pi) + \frac{8}{3}\cos(2\pi) \right) - \left( 8(0) + \frac{8}{3}\cos(0) \right) $$ $$ = \left( 16\pi + \frac{8}{3}(1) \right) - \left( 0 + \frac{8}{3}(1) \right) $$ $$ = 16\pi + \frac{8}{3} - \frac{8}{3} = 16\pi $$ ### Q11. Change into polar coordinates and evaluate $\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} dy dx$. **Solution:** The region of integration is the entire first quadrant ($x \ge 0, y \ge 0$). In polar coordinates: $x^2+y^2 = r^2$. $dy dx = r dr d\theta$. For the first quadrant, $0 \le r ### Q12. Evaluate the following integrals: #### (i) $\int_0^{\log 2} \int_0^z \int_0^{x+\log y} e^{x+y+z} dx dy dz$ **Solution:** The OCR for the innermost limit is $x+\log y$. This is unusual for the variable of integration. It's more likely to be $\log y$ or $x$. Let's assume it was $\int_0^z \int_0^y e^{x+y+z} dx dy dz$ or similar. Given the exact OCR, it's $\int_0^{\log 2} \int_0^z \int_0^{x+\log y} e^{x+y+z} dx dy dz$. The innermost integral is $\int_0^{x+\log y} e^{x+y+z} dx$. This implies $x$ is the integration variable for this part. This is a typo. The limit $x+\log y$ can't be for $x$. It should be $x$ or $y$ or $z$. Let's assume the order of integration is $dx dy dz$. So the innermost integral is with respect to $x$. The limit $x+\log y$ cannot be an upper limit for $x$. Typical forms are $\int_A^B \int_{g_1(z)}^{g_2(z)} \int_{h_1(y,z)}^{h_2(y,z)} f(x,y,z) dx dy dz$. Thus, $x+\log y$ must be an upper limit for $x$, which is impossible as it contains $x$. Let's assume the question meant $\int_0^{\log 2} \int_0^z \int_0^{y} e^{x+y+z} dx dy dz$. This is a common setup. Inner integral: $\int_0^y e^{x+y+z} dx = e^{y+z} \int_0^y e^x dx = e^{y+z} [e^x]_0^y = e^{y+z} (e^y - 1) = e^{2y+z} - e^{y+z}$. Middle integral: $\int_0^z (e^{2y+z} - e^{y+z}) dy = \left[ \frac{1}{2}e^{2y+z} - e^{y+z} \right]_0^z$. $$ = \left( \frac{1}{2}e^{2z+z} - e^{z+z} \right) - \left( \frac{1}{2}e^{z} - e^{z} \right) $$ $$ = \frac{1}{2}e^{3z} - e^{2z} - \left( -\frac{1}{2}e^z \right) = \frac{1}{2}e^{3z} - e^{2z} + \frac{1}{2}e^z $$ Outer integral: $\int_0^{\log 2} \left( \frac{1}{2}e^{3z} - e^{2z} + \frac{1}{2}e^z \right) dz$. $$ = \left[ \frac{1}{2}\frac{e^{3z}}{3} - \frac{e^{2z}}{2} + \frac{1}{2}e^z \right]_0^{\log 2} $$ $$ = \left[ \frac{1}{6}e^{3z} - \frac{1}{2}e^{2z} + \frac{1}{2}e^z \right]_0^{\log 2} $$ Substitute $z=\log 2$: $$ = \left( \frac{1}{6}e^{3\log 2} - \frac{1}{2}e^{2\log 2} + \frac{1}{2}e^{\log 2} \right) - \left( \frac{1}{6}e^0 - \frac{1}{2}e^0 + \frac{1}{2}e^0 \right) $$ $$ = \left( \frac{1}{6}e^{\log 2^3} - \frac{1}{2}e^{\log 2^2} + \frac{1}{2}e^{\log 2} \right) - \left( \frac{1}{6} - \frac{1}{2} + \frac{1}{2} \right) $$ $$ = \left( \frac{1}{6}(8) - \frac{1}{2}(4) + \frac{1}{2}(2) \right) - \left( \frac{1}{6} \right) $$ $$ = \left( \frac{8}{6} - 2 + 1 \right) - \frac{1}{6} = \left( \frac{4}{3} - 1 \right) - \frac{1}{6} $$ $$ = \frac{1}{3} - \frac{1}{6} = \frac{2-1}{6} = \frac{1}{6} $$ #### (ii) $\int_0^{\pi/2} \int_0^{a\cos\theta} \int_0^{\sqrt{a^2-r^2}} r \, dz dr d\theta$ **Solution:** The OCR for the upper limit of the second integral is $a \cos\theta$. For the innermost integral, it's $\sqrt{a^2-r^2}$. The integrand is $r$. This integral seems to be in cylindrical coordinates. Inner integral: $\int_0^{\sqrt{a^2-r^2}} r \, dz = r [z]_0^{\sqrt{a^2-r^2}} = r\sqrt{a^2-r^2}$. Middle integral: $\int_0^{a\cos\theta} r\sqrt{a^2-r^2} \, dr$. Let $u = a^2-r^2$. Then $du = -2r dr \Rightarrow r dr = -\frac{1}{2} du$. When $r=0$, $u=a^2$. When $r=a\cos\theta$, $u = a^2-(a\cos\theta)^2 = a^2(1-\cos^2\theta) = a^2\sin^2\theta$. $$ \int_{a^2}^{a^2\sin^2\theta} \sqrt{u} \left( -\frac{1}{2} \right) du = -\frac{1}{2} \int_{a^2}^{a^2\sin^2\theta} u^{1/2} du $$ $$ = \frac{1}{2} \int_{a^2\sin^2\theta}^{a^2} u^{1/2} du = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{a^2\sin^2\theta}^{a^2} $$ $$ = \frac{1}{3} [u^{3/2}]_{a^2\sin^2\theta}^{a^2} = \frac{1}{3} \left( (a^2)^{3/2} - (a^2\sin^2\theta)^{3/2} \right) $$ $$ = \frac{1}{3} \left( a^3 - a^3\sin^3\theta \right) = \frac{a^3}{3} (1-\sin^3\theta) $$ Outer integral: $\int_0^{\pi/2} \frac{a^3}{3} (1-\sin^3\theta) \, d\theta$. $$ = \frac{a^3}{3} \int_0^{\pi/2} (1-\sin^3\theta) \, d\theta $$ $$ = \frac{a^3}{3} \left[ \theta - \int_0^{\pi/2} \sin^3\theta \, d\theta \right]_0^{\pi/2} $$ Using Wallis' formula for $\int_0^{\pi/2} \sin^n x dx$: $$ \int_0^{\pi/2} \sin^3\theta \, d\theta = \frac{(3-1)}{3 \times 1} \times 1 = \frac{2}{3} $$ $$ = \frac{a^3}{3} \left[ \theta \right]_0^{\pi/2} - \frac{a^3}{3} \frac{2}{3} $$ $$ = \frac{a^3}{3} \left( \frac{\pi}{2} - 0 \right) - \frac{2a^3}{9} $$ $$ = \frac{\pi a^3}{6} - \frac{2a^3}{9} = a^3 \left( \frac{\pi}{6} - \frac{2}{9} \right) $$ $$ = a^3 \left( \frac{3\pi - 4}{18} \right) $$ ### Q13. Evaluate $\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \frac{dz dy dx}{\sqrt{1-x^2-y^2-z^2}}$ and $\iiint_D xyzdxdydz$ over the positive octant of the sphere $x^2 + y^2 + z^2 = b^2$ by changing to the spherical coordinates. **Solution:** #### Part 1: $\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \frac{1}{\sqrt{1-x^2-y^2-z^2}} dz dy dx$ The region of integration is the positive octant of the unit sphere ($x \ge 0, y \ge 0, z \ge 0, x^2+y^2+z^2 \le 1$). We change to spherical coordinates: $x = \rho\sin\phi\cos\theta$ $y = \rho\sin\phi\sin\theta$ $z = \rho\cos\phi$ $dV = \rho^2\sin\phi \, d\rho d\phi d\theta$. The sphere $x^2+y^2+z^2=1$ becomes $\rho=1$. For the positive octant: $0 \le \rho \le 1$ $0 \le \phi \le \pi/2$ (from z-axis to xy-plane) $0 \le \theta \le \pi/2$ (first quadrant in xy-plane) The integrand is $\frac{1}{\sqrt{1-(x^2+y^2+z^2)}} = \frac{1}{\sqrt{1-\rho^2}}$. The integral becomes: $$ I = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 \frac{1}{\sqrt{1-\rho^2}} \rho^2\sin\phi \, d\rho d\phi d\theta $$ Separate the integrals: $$ I = \left( \int_0^1 \frac{\rho^2}{\sqrt{1-\rho^2}} d\rho \right) \left( \int_0^{\pi/2} \sin\phi \, d\phi \right) \left( \int_0^{\pi/2} 1 \, d\theta \right) $$ Third part: $\int_0^{\pi/2} 1 \, d\theta = [\theta]_0^{\pi/2} = \pi/2$. Second part: $\int_0^{\pi/2} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi/2} = (-\cos(\pi/2)) - (-\cos(0)) = 0 - (-1) = 1$. First part: $\int_0^1 \frac{\rho^2}{\sqrt{1-\rho^2}} d\rho$. Let $\rho = \sin t$. Then $d\rho = \cos t \, dt$. When $\rho=0$, $t=0$. When $\rho=1$, $t=\pi/2$. $$ \int_0^{\pi/2} \frac{\sin^2 t}{\sqrt{1-\sin^2 t}} \cos t \, dt = \int_0^{\pi/2} \frac{\sin^2 t}{\cos t} \cos t \, dt $$ $$ = \int_0^{\pi/2} \sin^2 t \, dt $$ Using Wallis' formula (or $\sin^2 t = \frac{1-\cos(2t)}{2}$): $$ = \frac{1}{2} \int_0^{\pi/2} (1-\cos(2t)) dt = \frac{1}{2} \left[ t - \frac{\sin(2t)}{2} \right]_0^{\pi/2} $$ $$ = \frac{1}{2} \left( (\pi/2 - 0) - (0 - 0) \right) = \frac{\pi}{4} $$ Combine all parts: $$ I = \frac{\pi}{4} \times 1 \times \frac{\pi}{2} = \frac{\pi^2}{8} $$ #### Part 2: $\iiint_D xyzdxdydz$ over the positive octant of the sphere $x^2 + y^2 + z^2 = b^2$. The region D is the positive octant of a sphere of radius $b$. In spherical coordinates: $x = \rho\sin\phi\cos\theta$ $y = \rho\sin\phi\sin\theta$ $z = \rho\cos\phi$ $dV = \rho^2\sin\phi \, d\rho d\phi d\theta$. For the positive octant: $0 \le \rho \le b$ $0 \le \phi \le \pi/2$ $0 \le \theta \le \pi/2$ The integrand $xyz = (\rho\sin\phi\cos\theta)(\rho\sin\phi\sin\theta)(\rho\cos\phi) = \rho^3\sin^2\phi\cos\phi\sin\theta\cos\theta$. The integral becomes: $$ I = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^b (\rho^3\sin^2\phi\cos\phi\sin\theta\cos\theta) \rho^2\sin\phi \, d\rho d\phi d\theta $$ $$ I = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^b \rho^5\sin^3\phi\cos\phi\sin\theta\cos\theta \, d\rho d\phi d\theta $$ Separate the integrals: $$ I = \left( \int_0^b \rho^5 \, d\rho \right) \left( \int_0^{\pi/2} \sin^3\phi\cos\phi \, d\phi \right) \left( \int_0^{\pi/2} \sin\theta\cos\theta \, d\theta \right) $$ First part: $\int_0^b \rho^5 \, d\rho = \left[ \frac{\rho^6}{6} \right]_0^b = \frac{b^6}{6}$. Second part: $\int_0^{\pi/2} \sin^3\phi\cos\phi \, d\phi$. Let $u=\sin\phi$. Then $du=\cos\phi \, d\phi$. When $\phi=0$, $u=0$. When $\phi=\pi/2$, $u=1$. $$ \int_0^1 u^3 \, du = \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{4} $$ Third part: $\int_0^{\pi/2} \sin\theta\cos\theta \, d\theta$. Let $w=\sin\theta$. Then $dw=\cos\theta \, d\theta$. When $\theta=0$, $w=0$. When $\theta=\pi/2$, $w=1$. $$ \int_0^1 w \, dw = \left[ \frac{w^2}{2} \right]_0^1 = \frac{1}{2} $$ Combine all parts: $$ I = \frac{b^6}{6} \times \frac{1}{4} \times \frac{1}{2} = \frac{b^6}{48} $$ ### Q14. Evaluate $\iint_D \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} dx dy$ over the positive quadrant of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. **Solution:** The region D is the portion of the ellipse in the first quadrant ($x \ge 0, y \ge 0$). We use generalized polar coordinates (or elliptic coordinates). Let $x = a r \cos\theta$ and $y = b r \sin\theta$. The Jacobian is $J = \left| \frac{\partial(x,y)}{\partial(r,\theta)} \right|$. $$ J = \left| \begin{vmatrix} a\cos\theta & -ar\sin\theta \\ b\sin\theta & br\cos\theta \end{vmatrix} \right| $$ $$ J = |(a\cos\theta)(br\cos\theta) - (-ar\sin\theta)(b\sin\theta)| $$ $$ J = |abr\cos^2\theta + abr\sin^2\theta| = |abr(\cos^2\theta+\sin^2\theta)| = abr $$ So $dx dy = abr \, dr d\theta$. The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ becomes $\frac{(ar\cos\theta)^2}{a^2} + \frac{(br\sin\theta)^2}{b^2} = 1$. $r^2\cos^2\theta + r^2\sin^2\theta = 1 \Rightarrow r^2=1 \Rightarrow r=1$. For the positive quadrant ($x \ge 0, y \ge 0$): $0 \le r \le 1$ $0 \le \theta \le \pi/2$ The integrand is $\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} = \sqrt{1 - r^2\cos^2\theta - r^2\sin^2\theta} = \sqrt{1-r^2}$. The integral becomes: $$ I = \int_0^{\pi/2} \int_0^1 \sqrt{1-r^2} (abr) \, dr d\theta $$ $$ I = ab \int_0^{\pi/2} \int_0^1 r\sqrt{1-r^2} \, dr d\theta $$ Separate the integrals: $$ I = ab \left( \int_0^1 r\sqrt{1-r^2} \, dr \right) \left( \int_0^{\pi/2} 1 \, d\theta \right) $$ Second part: $\int_0^{\pi/2} 1 \, d\theta = [\theta]_0^{\pi/2} = \pi/2$. First part: $\int_0^1 r\sqrt{1-r^2} \, dr$. Let $u = 1-r^2$. Then $du = -2r dr \Rightarrow r dr = -\frac{1}{2} du$. When $r=0$, $u=1$. When $r=1$, $u=0$. $$ \int_1^0 \sqrt{u} \left( -\frac{1}{2} \right) du = -\frac{1}{2} \int_1^0 u^{1/2} du $$ $$ = \frac{1}{2} \int_0^1 u^{1/2} du = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_0^1 $$ $$ = \frac{1}{2} \frac{2}{3} [u^{3/2}]_0^1 = \frac{1}{3} (1^{3/2} - 0) = \frac{1}{3} $$ Combine all parts: $$ I = ab \times \frac{1}{3} \times \frac{\pi}{2} = \frac{\pi ab}{6} $$ ### Q15. Find volume (using double and triple integrals) of the tetrahedron bounded by the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ and the coordinate axes. **Solution:** #### Using Triple Integral: The tetrahedron is bounded by $x=0, y=0, z=0$ and the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. From the plane equation, $z = c\left(1 - \frac{x}{a} - \frac{y}{b}\right)$. The limits of integration are: $0 \le x \le a$ $0 \le y \le b\left(1 - \frac{x}{a}\right)$ (from $y=0$ to the line $\frac{x}{a} + \frac{y}{b} = 1$ in the $xy$-plane) $0 \le z \le c\left(1 - \frac{x}{a} - \frac{y}{b}\right)$ The volume $V$ is given by: $$ V = \int_0^a \int_0^{b(1-x/a)} \int_0^{c(1-x/a-y/b)} dz dy dx $$ Inner integral: $\int_0^{c(1-x/a-y/b)} dz = c\left(1 - \frac{x}{a} - \frac{y}{b}\right)$. Middle integral: $\int_0^{b(1-x/a)} c\left(1 - \frac{x}{a} - \frac{y}{b}\right) dy$. Let $K = 1 - \frac{x}{a}$. $$ = c \int_0^{bK} \left( K - \frac{y}{b} \right) dy $$ $$ = c \left[ Ky - \frac{y^2}{2b} \right]_0^{bK} $$ $$ = c \left( K(bK) - \frac{(bK)^2}{2b} \right) - 0 $$ $$ = c \left( bK^2 - \frac{b^2K^2}{2b} \right) = c \left( bK^2 - \frac{bK^2}{2} \right) = c \frac{bK^2}{2} = \frac{bc}{2} K^2 $$ Substitute $K = 1 - \frac{x}{a}$: $$ = \frac{bc}{2} \left( 1 - \frac{x}{a} \right)^2 $$ Outer integral: $\int_0^a \frac{bc}{2} \left( 1 - \frac{x}{a} \right)^2 dx$. Let $u = 1 - \frac{x}{a}$. Then $du = -\frac{1}{a} dx \Rightarrow dx = -a \, du$. When $x=0$, $u=1$. When $x=a$, $u=0$. $$ = \frac{bc}{2} \int_1^0 u^2 (-a) du = -\frac{abc}{2} \int_1^0 u^2 du $$ $$ = \frac{abc}{2} \int_0^1 u^2 du = \frac{abc}{2} \left[ \frac{u^3}{3} \right]_0^1 $$ $$ = \frac{abc}{2} \left( \frac{1}{3} - 0 \right) = \frac{abc}{6} $$ #### Using Double Integral: The volume is given by $\iint_R z(x,y) dA$, where $R$ is the region in the $xy$-plane bounded by the intercepts. The plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. So $z = c\left(1 - \frac{x}{a} - \frac{y}{b}\right)$. The region $R$ in the $xy$-plane is bounded by $x=0, y=0$ and the line $\frac{x}{a} + \frac{y}{b} = 1$. The limits for $x$ are from $0$ to $a$. The limits for $y$ are from $0$ to $b(1-x/a)$. $$ V = \int_0^a \int_0^{b(1-x/a)} c\left(1 - \frac{x}{a} - \frac{y}{b}\right) dy dx $$ This is the same integral we evaluated in the triple integral method's middle and outer steps. From the previous calculation, this evaluates to $\frac{abc}{6}$. ### Q16. Find volume of the cylinder of radius $r$ and height $h$ units. **Solution:** A cylinder with radius $r$ and height $h$ can be described as the region $x^2+y^2 \le r^2$ for $0 \le z \le h$. We can use a triple integral in Cartesian coordinates: $$ V = \iiint_D dV = \int_{-r}^r \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \int_0^h dz dy dx $$ Inner integral: $\int_0^h dz = h$. Middle integral: $\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} h \, dy = h [y]_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} = h (\sqrt{r^2-x^2} - (-\sqrt{r^2-x^2})) = 2h\sqrt{r^2-x^2}$. Outer integral: $\int_{-r}^r 2h\sqrt{r^2-x^2} \, dx$. Let $x = r\sin\theta$. Then $dx = r\cos\theta \, d\theta$. When $x=-r$, $\theta=-\pi/2$. When $x=r$, $\theta=\pi/2$. $\sqrt{r^2-x^2} = \sqrt{r^2-r^2\sin^2\theta} = \sqrt{r^2(1-\sin^2\theta)} = \sqrt{r^2\cos^2\theta} = r|\cos\theta|$. For $-\pi/2 \le \theta \le \pi/2$, $\cos\theta \ge 0$, so $|\cos\theta|=\cos\theta$. $$ V = \int_{-\pi/2}^{\pi/2} 2h (r\cos\theta) (r\cos\theta) \, d\theta $$ $$ = 2hr^2 \int_{-\pi/2}^{\pi/2} \cos^2\theta \, d\theta $$ Using $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$: $$ = 2hr^2 \int_{-\pi/2}^{\pi/2} \frac{1+\cos(2\theta)}{2} \, d\theta $$ $$ = hr^2 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{-\pi/2}^{\pi/2} $$ $$ = hr^2 \left( \left( \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right) - \left( -\frac{\pi}{2} + \frac{\sin(-\pi)}{2} \right) \right) $$ $$ = hr^2 \left( \left( \frac{\pi}{2} + 0 \right) - \left( -\frac{\pi}{2} + 0 \right) \right) $$ $$ = hr^2 \left( \frac{\pi}{2} + \frac{\pi}{2} \right) = hr^2 (\pi) = \pi r^2 h $$ Alternatively, using cylindrical coordinates: $x = \rho\cos\theta$, $y = \rho\sin\theta$, $z=z$. $dV = \rho \, d\rho d\theta dz$. For the cylinder $x^2+y^2 \le r^2$ and $0 \le z \le h$: $0 \le \rho \le r$ $0 \le \theta \le 2\pi$ $0 \le z \le h$ $$ V = \int_0^{2\pi} \int_0^r \int_0^h \rho \, dz d\rho d\theta $$ Inner integral: $\int_0^h \rho \, dz = \rho [z]_0^h = \rho h$. Middle integral: $\int_0^r \rho h \, d\rho = h \left[ \frac{\rho^2}{2} \right]_0^r = h \frac{r^2}{2}$. Outer integral: $\int_0^{2\pi} h \frac{r^2}{2} \, d\theta = \frac{hr^2}{2} [\theta]_0^{2\pi} = \frac{hr^2}{2} (2\pi) = \pi r^2 h $.