Multiple Integrals Cheatsheet

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Standard Integral Formulas $\int dx = ax + C$ $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ $\int e^{ax} dx = \frac{1}{a} e^{ax} + C$ $\int \sin x dx = -\cos x + C$ $\int \cos x dx = \sin x + C$ $\int \sec^2 x dx = \tan x + C$ $\int \csc^2 x dx = -\cot x + C$ $\int \sec x \tan x dx = \sec x + C$ $\int \csc x \cot x dx = -\csc x + C$ $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{\sqrt{x^2-a^2}} dx = \cosh^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{x\sqrt{x^2-a^2}} dx = \frac{1}{a} \sec^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{x\sqrt{a^2-x^2}} dx = -\frac{1}{a} \text{sech}^{-1}\left(\frac{x}{a}\right) + C$ Double Integrals Evaluation Example 1 Evaluate $\int_2^3 \int_0^1 xy(1+x+y) \, dy \, dx$ The integral becomes: $$ \int_2^3 \int_0^1 (xy + x^2y + xy^2) \, dy \, dx $$ $$ \int_2^3 \left[ \frac{xy^2}{2} + \frac{x^2y^2}{2} + \frac{xy^3}{3} \right]_0^1 \, dx $$ $$ \int_2^3 \left( \frac{x}{2} + \frac{x^2}{2} + \frac{x}{3} \right) \, dx $$ $$ \int_2^3 \left( \frac{5x}{6} + \frac{x^2}{2} \right) \, dx $$ $$ \left[ \frac{5x^2}{12} + \frac{x^3}{6} \right]_2^3 = \left( \frac{5(9)}{12} + \frac{27}{6} \right) - \left( \frac{5(4)}{12} + \frac{8}{6} \right) $$ $$ = \left( \frac{45}{12} + \frac{54}{12} \right) - \left( \frac{20}{12} + \frac{16}{12} \right) = \frac{99}{12} - \frac{36}{12} = \frac{63}{12} = \frac{21}{4} $$ Evaluation Example 2 Evaluate $\int_0^a \int_0^{\sqrt{a^2-y^2}} (x^2+y^2) \, dx \, dy$ by changing to polar coordinates. Region of integration: $x^2+y^2=a^2$ in the first quadrant. Limits for polar coordinates: $r$ from $0$ to $a$, $\theta$ from $0$ to $\pi/2$. The integral becomes: $$ \int_0^{\pi/2} \int_0^a (r^2) r \, dr \, d\theta = \int_0^{\pi/2} \int_0^a r^3 \, dr \, d\theta $$ $$ \int_0^{\pi/2} \left[ \frac{r^4}{4} \right]_0^a \, d\theta = \int_0^{\pi/2} \frac{a^4}{4} \, d\theta $$ $$ \left[ \frac{a^4}{4} \theta \right]_0^{\pi/2} = \frac{a^4}{4} \left( \frac{\pi}{2} \right) = \frac{\pi a^4}{8} $$ Change of Order of Integration Example Evaluate $\int_0^2 \int_{y/2}^1 \frac{y^2}{\sqrt{x^2+y^2}} \, dx \, dy$. Region: $x=y/2$ to $x=1$, $y=0$ to $y=2$. This is a triangle with vertices $(0,0), (1,0), (1,2)$. Changing order: $y$ from $0$ to $2x$, $x$ from $0$ to $1$. The integral becomes: $$ \int_0^1 \int_0^{2x} \frac{y^2}{\sqrt{x^2+y^2}} \, dy \, dx $$ This integral is complex. Let's consider a simpler change of order problem. If we had $\int_0^1 \int_0^{y} (x^2+y^2) \, dx \, dy$ and changed order: Region: $x=0$ to $x=y$, $y=0$ to $y=1$. This is a triangle with vertices $(0,0), (0,1), (1,1)$. Changing order: $y=x$ to $y=1$, $x=0$ to $x=1$. $$ \int_0^1 \int_x^1 (x^2+y^2) \, dy \, dx $$ $$ \int_0^1 \left[ x^2y + \frac{y^3}{3} \right]_x^1 \, dx = \int_0^1 \left( x^2 + \frac{1}{3} - x^3 - \frac{x^3}{3} \right) \, dx $$ $$ \int_0^1 \left( x^2 + \frac{1}{3} - \frac{4x^3}{3} \right) \, dx = \left[ \frac{x^3}{3} + \frac{x}{3} - \frac{x^4}{3} \right]_0^1 = \frac{1}{3} + \frac{1}{3} - \frac{1}{3} = \frac{1}{3} $$ Triple Integrals Evaluation Example 1 Evaluate $\int_0^1 \int_0^x \int_0^{x+y} xyz \, dz \, dy \, dx$ The integral becomes: $$ \int_0^1 \int_0^x \left[ \frac{xyz^2}{2} \right]_0^{x+y} \, dy \, dx = \int_0^1 \int_0^x \frac{xy(x+y)^2}{2} \, dy \, dx $$ $$ \frac{1}{2} \int_0^1 \int_0^x xy(x^2+2xy+y^2) \, dy \, dx = \frac{1}{2} \int_0^1 \int_0^x (x^3y+2x^2y^2+xy^3) \, dy \, dx $$ $$ \frac{1}{2} \int_0^1 \left[ \frac{x^3y^2}{2} + \frac{2x^2y^3}{3} + \frac{xy^4}{4} \right]_0^x \, dx $$ $$ \frac{1}{2} \int_0^1 \left( \frac{x^5}{2} + \frac{2x^5}{3} + \frac{x^5}{4} \right) \, dx = \frac{1}{2} \int_0^1 \left( \frac{6x^5+8x^5+3x^5}{12} \right) \, dx $$ $$ \frac{1}{2} \int_0^1 \frac{17x^5}{12} \, dx = \frac{17}{24} \left[ \frac{x^6}{6} \right]_0^1 = \frac{17}{24} \cdot \frac{1}{6} = \frac{17}{144} $$ Evaluation Example 2 Evaluate $\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} z \, dz \, dy \, dx$. This is for a sphere in the first octant. The integral becomes: $$ \int_0^1 \int_0^{\sqrt{1-x^2}} \left[ \frac{z^2}{2} \right]_0^{\sqrt{1-x^2-y^2}} \, dy \, dx = \int_0^1 \int_0^{\sqrt{1-x^2}} \frac{1-x^2-y^2}{2} \, dy \, dx $$ $$ \frac{1}{2} \int_0^1 \left[ (1-x^2)y - \frac{y^3}{3} \right]_0^{\sqrt{1-x^2}} \, dx $$ Let $1-x^2 = R^2$. Then $y$ goes from $0$ to $R$. $$ \frac{1}{2} \int_0^1 \left[ R^2 \cdot R - \frac{R^3}{3} \right] \, dx = \frac{1}{2} \int_0^1 \left( R^3 - \frac{R^3}{3} \right) \, dx = \frac{1}{2} \int_0^1 \frac{2R^3}{3} \, dx $$ $$ \frac{1}{3} \int_0^1 (1-x^2)^{3/2} \, dx $$ To solve $\int_0^1 (1-x^2)^{3/2} \, dx$, let $x = \sin \theta$, $dx = \cos \theta \, d\theta$. Limits: $0 \to 0$, $1 \to \pi/2$. $$ \frac{1}{3} \int_0^{\pi/2} (\cos^2 \theta)^{3/2} \cos \theta \, d\theta = \frac{1}{3} \int_0^{\pi/2} \cos^4 \theta \, d\theta $$ Using Wallis' formula for $\int_0^{\pi/2} \cos^n x \, dx = \frac{(n-1)!!}{n!!} \frac{\pi}{2}$ if $n$ is even: $$ \frac{1}{3} \cdot \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{1}{3} \cdot \frac{3}{8} \cdot \frac{\pi}{2} = \frac{\pi}{16} $$ Change of Variables General Formula For $x = f(u,v)$ and $y = g(u,v)$, the Jacobian $J = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}$. Then $dx \, dy = |J| \, du \, dv$. Polar Coordinates $x = r \cos \theta$, $y = r \sin \theta$. $J = \det \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix} = r \cos^2 \theta + r \sin^2 \theta = r$. So, $dx \, dy = r \, dr \, d\theta$. Spherical Coordinates $x = \rho \sin \phi \cos \theta$, $y = \rho \sin \phi \sin \theta$, $z = \rho \cos \phi$. The Jacobian is $J = \rho^2 \sin \phi$. So, $dx \, dy \, dz = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$. Cylindrical Coordinates $x = r \cos \theta$, $y = r \sin \theta$, $z = z$. The Jacobian is $J = r$. So, $dx \, dy \, dz = r \, dr \, d\theta \, dz$. Volume of Ellipsoid The volume of the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ is given by $V = \iiint_R \, dx \, dy \, dz$. Consider the first octant where $x,y,z \ge 0$. $z$ from $0$ to $c \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$. $y$ from $0$ to $b \sqrt{1 - \frac{x^2}{a^2}}$. $x$ from $0$ to $a$. The total volume is $8$ times the volume in the first octant. $V = 8 \int_0^a \int_0^{b\sqrt{1-x^2/a^2}} \int_0^{c\sqrt{1-x^2/a^2-y^2/b^2}} \, dz \, dy \, dx$ Using the transformation $x=a\rho\sin\phi\cos\theta$, $y=b\rho\sin\phi\sin\theta$, $z=c\rho\cos\phi$, the Jacobian is $abc\rho^2\sin\phi$. The integral becomes: $V = \int_0^{2\pi} \int_0^{\pi} \int_0^1 (abc\rho^2\sin\phi) \, d\rho \, d\phi \, d\theta$ $V = abc \left( \int_0^1 \rho^2 \, d\rho \right) \left( \int_0^{\pi} \sin\phi \, d\phi \right) \left( \int_0^{2\pi} \, d\theta \right)$ $V = abc \left[ \frac{\rho^3}{3} \right]_0^1 \left[ -\cos\phi \right]_0^{\pi} \left[ \theta \right]_0^{2\pi}$ $V = abc \left( \frac{1}{3} \right) (1 - (-1)) (2\pi) = abc \left( \frac{1}{3} \right) (2) (2\pi) = \frac{4}{3}\pi abc$ Volume of Sphere The volume of the sphere $x^2+y^2+z^2=a^2$ is found using spherical coordinates. Limits: $\rho$ from $0$ to $a$, $\phi$ from $0$ to $\pi$, $\theta$ from $0$ to $2\pi$. $V = \int_0^{2\pi} \int_0^{\pi} \int_0^a \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ $V = \left( \int_0^a \rho^2 \, d\rho \right) \left( \int_0^{\pi} \sin\phi \, d\phi \right) \left( \int_0^{2\pi} \, d\theta \right)$ $V = \left[ \frac{\rho^3}{3} \right]_0^a \left[ -\cos\phi \right]_0^{\pi} \left[ \theta \right]_0^{2\pi}$ $V = \left( \frac{a^3}{3} \right) (1 - (-1)) (2\pi) = \frac{a^3}{3} \cdot 2 \cdot 2\pi = \frac{4}{3}\pi a^3$