Engineering Mathematics I

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Linear Systems & Eigenvalue Problems Solving Systems of Equations Problem: Solve the system of equations for $x, y, z$: $$ \begin{cases} x + y + z = 1 \\ x + 2y + 4z = \lambda \\ x + 4y + 10z = \lambda^2 \end{cases} $$ Solution: We form the augmented matrix and use Gaussian elimination: $$ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 4 & | & \lambda \\ 1 & 4 & 10 & | & \lambda^2 \end{pmatrix} $$ $R_2 \to R_2 - R_1$: $R_3 \to R_3 - R_1$: $$ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & \lambda - 1 \\ 0 & 3 & 9 & | & \lambda^2 - 1 \end{pmatrix} $$ $R_3 \to R_3 - 3R_2$: $$ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & \lambda - 1 \\ 0 & 0 & 0 & | & \lambda^2 - 1 - 3(\lambda - 1) \end{pmatrix} $$ Simplify the last element: $\lambda^2 - 1 - 3\lambda + 3 = \lambda^2 - 3\lambda + 2 = (\lambda - 1)(\lambda - 2)$. The augmented matrix becomes: $$ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & \lambda - 1 \\ 0 & 0 & 0 & | & (\lambda - 1)(\lambda - 2) \end{pmatrix} $$ Now, we analyze the cases based on the value of $\lambda$: Case 1: $\lambda = 1$ The last row becomes $0 = 0$. The system has infinitely many solutions. The matrix is: $$ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} $$ From the second row: $y + 3z = 0 \implies y = -3z$. From the first row: $x + y + z = 1 \implies x - 3z + z = 1 \implies x - 2z = 1 \implies x = 1 + 2z$. Let $z = t$ (a free parameter). The solutions are $(x, y, z) = (1 + 2t, -3t, t)$ for any real number $t$. Case 2: $\lambda = 2$ The last row becomes $0 = 0$. The system has infinitely many solutions. The matrix is: $$ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & 2 - 1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} $$ From the second row: $y + 3z = 1 \implies y = 1 - 3z$. From the first row: $x + y + z = 1 \implies x + (1 - 3z) + z = 1 \implies x + 1 - 2z = 1 \implies x = 2z$. Let $z = t$ (a free parameter). The solutions are $(x, y, z) = (2t, 1 - 3t, t)$ for any real number $t$. Case 3: $\lambda \ne 1$ and $\lambda \ne 2$ The last row implies $0 = (\lambda - 1)(\lambda - 2)$, which is a contradiction. Therefore, the system has no solution. Eigenvalues and Eigenvectors Problem: Given matrix $A = \begin{pmatrix} 4 & -1 & -1 \\ -1 & 4 & -1 \\ -1 & -1 & 4 \end{pmatrix}$. Find the eigenvalues of $A$. Find the eigenvectors of $A$. If $B = I - \frac{1}{4}A$, find the eigenvalues of $B$. Solution: Eigenvalues of A: We solve the characteristic equation $\det(A - \lambda I) = 0$: $$ \det \begin{pmatrix} 4-\lambda & -1 & -1 \\ -1 & 4-\lambda & -1 \\ -1 & -1 & 4-\lambda \end{pmatrix} = 0 $$ Expand the determinant: $(4-\lambda)[(4-\lambda)^2 - (-1)(-1)] - (-1)[(-1)(4-\lambda) - (-1)(-1)] + (-1)[(-1)(-1) - (-1)(4-\lambda)] = 0$ $(4-\lambda)[(4-\lambda)^2 - 1] + [-(4-\lambda) - 1] - [1 + (4-\lambda)] = 0$ $(4-\lambda)[(4-\lambda-1)(4-\lambda+1)] - [5-\lambda] - [5-\lambda] = 0$ $(4-\lambda)(3-\lambda)(5-\lambda) - 2(5-\lambda) = 0$ $(5-\lambda)[(4-\lambda)(3-\lambda) - 2] = 0$ $(5-\lambda)[\lambda^2 - 7\lambda + 12 - 2] = 0$ $(5-\lambda)[\lambda^2 - 7\lambda + 10] = 0$ $(5-\lambda)(\lambda - 2)(\lambda - 5) = 0$ This gives the eigenvalues: $\lambda_1 = 5$ (with multiplicity 2) and $\lambda_2 = 2$. Eigenvectors of A: For $\lambda = 5$: Solve $(A - 5I)\vec{v} = \vec{0}$: $$ \begin{pmatrix} 4-5 & -1 & -1 \\ -1 & 4-5 & -1 \\ -1 & -1 & 4-5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ All rows are identical: $-x - y - z = 0 \implies x + y + z = 0$. This is a plane passing through the origin. We need two linearly independent vectors satisfying this condition. Let $x=1, y=0 \implies z=-1$. So $\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$. Let $x=0, y=1 \implies z=-1$. So $\vec{v}_2 = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$. These are two linearly independent eigenvectors for $\lambda = 5$. For $\lambda = 2$: Solve $(A - 2I)\vec{v} = \vec{0}$: $$ \begin{pmatrix} 4-2 & -1 & -1 \\ -1 & 4-2 & -1 \\ -1 & -1 & 4-2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ Augmented matrix: $$ \begin{pmatrix} 2 & -1 & -1 & | & 0 \\ -1 & 2 & -1 & | & 0 \\ -1 & -1 & 2 & | & 0 \end{pmatrix} $$ $R_1 \leftrightarrow R_2$: $$ \begin{pmatrix} -1 & 2 & -1 & | & 0 \\ 2 & -1 & -1 & | & 0 \\ -1 & -1 & 2 & | & 0 \end{pmatrix} $$ $R_2 \to R_2 + 2R_1$: $R_3 \to R_3 - R_1$: $$ \begin{pmatrix} -1 & 2 & -1 & | & 0 \\ 0 & 3 & -3 & | & 0 \\ 0 & -3 & 3 & | & 0 \end{pmatrix} $$ $R_2 \to \frac{1}{3}R_2$: $R_3 \to R_3 + R_2$: $$ \begin{pmatrix} -1 & 2 & -1 & | & 0 \\ 0 & 1 & -1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} $$ From the second row: $y - z = 0 \implies y = z$. From the first row: $-x + 2y - z = 0 \implies -x + 2z - z = 0 \implies -x + z = 0 \implies x = z$. So, $x = y = z$. Let $z=1$. Then $\vec{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. Eigenvalues of B: Given $B = I - \frac{1}{4}A$. If $\vec{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\vec{v} = \lambda\vec{v}$. Now consider $B\vec{v}$: $B\vec{v} = (I - \frac{1}{4}A)\vec{v} = I\vec{v} - \frac{1}{4}A\vec{v} = \vec{v} - \frac{1}{4}(\lambda\vec{v}) = (1 - \frac{1}{4}\lambda)\vec{v}$. So, the eigenvalues of $B$ are $\mu = 1 - \frac{1}{4}\lambda$. For $\lambda_1 = 5$: $\mu_1 = 1 - \frac{1}{4}(5) = 1 - \frac{5}{4} = -\frac{1}{4}$. For $\lambda_2 = 2$: $\mu_2 = 1 - \frac{1}{4}(2) = 1 - \frac{1}{2} = \frac{1}{2}$. The eigenvalues of $B$ are $-\frac{1}{4}$ (with multiplicity 2) and $\frac{1}{2}$. Differential Equations & Series Expansion Higher-Order Differential Equations Problem: If $y = a(x + \sqrt{x^2-1})^n + b(x - \sqrt{x^2-1})^n$, prove: $(x^2-1)y'' + xy' - n^2y = 0$ $(x^2-1)y_{n+2} + (2n+1)xy_{n+1} + (n^2-m^2)y_n = 0$ (This looks like a typo from a Legendre equation, usually it's $y_{n+2}$ and $y_n$ for $n$-th derivative, not $n+2$-th and $n$-th derivative of $y$. Assuming it's a recurrence relation for derivatives.) Let's re-interpret the second part as a recurrence relation for the $k$-th derivatives, usually denoted $y^{(k)}$. If the problem implies a relationship for $y^{(n+2)}$ and $y^{(n+1)}$ then the original equation would be related to Legendre's. However, the given equation is not Legendre's. Let's assume the problem meant for the first part: Prove $(x^2-1)y'' + xy' - n^2y = 0$. The second part looks like a standard recurrence relation derived from Legendre's ODE, but the first ODE is not exactly Legendre's. Let's focus on the first part given the expression for $y$. Solution for part 1: Let $u = (x + \sqrt{x^2-1})^n$ and $v = (x - \sqrt{x^2-1})^n$. Then $y = au + bv$. First, find $u'$. Let $w = x + \sqrt{x^2-1}$. $w' = 1 + \frac{1}{2\sqrt{x^2-1}} (2x) = 1 + \frac{x}{\sqrt{x^2-1}} = \frac{\sqrt{x^2-1} + x}{\sqrt{x^2-1}} = \frac{w}{\sqrt{x^2-1}}$. So, $u' = n w^{n-1} w' = n (x + \sqrt{x^2-1})^{n-1} \frac{x + \sqrt{x^2-1}}{\sqrt{x^2-1}} = n \frac{(x + \sqrt{x^2-1})^n}{\sqrt{x^2-1}} = \frac{n u}{\sqrt{x^2-1}}$. Similarly, for $v$, let $z = x - \sqrt{x^2-1}$. $z' = 1 - \frac{x}{\sqrt{x^2-1}} = \frac{\sqrt{x^2-1} - x}{\sqrt{x^2-1}} = \frac{z}{\sqrt{x^2-1}}$. So, $v' = n z^{n-1} z' = n (x - \sqrt{x^2-1})^{n-1} \frac{x - \sqrt{x^2-1}}{\sqrt{x^2-1}} = \frac{n v}{\sqrt{x^2-1}}$. Now, $y' = au' + bv' = \frac{n}{\sqrt{x^2-1}}(au + bv) = \frac{ny}{\sqrt{x^2-1}}$. This implies $\sqrt{x^2-1} y' = ny$. Square both sides: $(x^2-1)(y')^2 = n^2 y^2$. Differentiate both sides with respect to $x$: $2x(y')^2 + (x^2-1) 2y'y'' = n^2 2y y'$ Divide by $2y'$ (assuming $y' \ne 0$): $x y' + (x^2-1) y'' = n^2 y$ Rearranging, we get: $(x^2-1)y'' + xy' - n^2y = 0$. This proves the first part. Regarding part 2: $(x^2-1)y_{n+2} + (2n+1)xy_{n+1} + (n^2-m^2)y_n = 0$. This expression is characteristic of a recurrence relation for derivatives of solutions to Legendre's differential equation, typically written as $(1-x^2)y'' - 2xy' + n(n+1)y = 0$. The given ODE is $(x^2-1)y'' + xy' - n^2y = 0$. This is not Legendre's equation. If the question intended to ask about the $k$-th derivative using Leibniz's rule for the first ODE, it would be a much more complex derivation for a general $n$. Given the phrasing $y_{n+2}$ and $y_{n+1}$, it often implies the $(n+2)$-th and $(n+1)$-th derivatives of $y$. However, if the first part is correct, the second part as stated doesn't directly follow as a general identity from the given $y$ for arbitrary $n$. It might be a specific case or a misremembered formula. Without further clarification on the notation $y_{n+2}$, etc., interpreting them as higher derivatives of $y$ leads to a very complex derivation that is unlikely for a cheatsheet. If it were a recurrence for polynomial solutions, that would be a different context. Taylor Series Expansion Problem: Expand $f(x, y) = e^x \cos y$ at $(1, \pi/4)$ up to 3 terms (i.e., up to first-order terms and the second-order terms). Solution: The Taylor series expansion for a function $f(x,y)$ around a point $(a,b)$ is given by: $$ f(x,y) = f(a,b) + (x-a)f_x(a,b) + (y-b)f_y(a,b) + \frac{1}{2!} \left[ (x-a)^2 f_{xx}(a,b) + 2(x-a)(y-b) f_{xy}(a,b) + (y-b)^2 f_{yy}(a,b) \right] + \dots $$ Here, $(a,b) = (1, \pi/4)$. First, evaluate the function and its partial derivatives at $(1, \pi/4)$: $f(x, y) = e^x \cos y$ $f(1, \pi/4) = e^1 \cos(\pi/4) = e \frac{\sqrt{2}}{2} = \frac{e\sqrt{2}}{2}$. Partial derivatives: $f_x = \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y$ $f_y = \frac{\partial}{\partial y}(e^x \cos y) = -e^x \sin y$ Second partial derivatives: $f_{xx} = \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y$ $f_{yy} = \frac{\partial}{\partial y}(-e^x \sin y) = -e^x \cos y$ $f_{xy} = \frac{\partial}{\partial y}(e^x \cos y) = -e^x \sin y$ Now, evaluate these at $(1, \pi/4)$: $f_x(1, \pi/4) = e^1 \cos(\pi/4) = \frac{e\sqrt{2}}{2}$ $f_y(1, \pi/4) = -e^1 \sin(\pi/4) = -\frac{e\sqrt{2}}{2}$ $f_{xx}(1, \pi/4) = e^1 \cos(\pi/4) = \frac{e\sqrt{2}}{2}$ $f_{yy}(1, \pi/4) = -e^1 \cos(\pi/4) = -\frac{e\sqrt{2}}{2}$ $f_{xy}(1, \pi/4) = -e^1 \sin(\pi/4) = -\frac{e\sqrt{2}}{2}$ Substitute these values into the Taylor series formula: $$ f(x,y) \approx \frac{e\sqrt{2}}{2} + (x-1)\left(\frac{e\sqrt{2}}{2}\right) + (y-\frac{\pi}{4})\left(-\frac{e\sqrt{2}}{2}\right) + \frac{1}{2} \left[ (x-1)^2\left(\frac{e\sqrt{2}}{2}\right) + 2(x-1)(y-\frac{\pi}{4})\left(-\frac{e\sqrt{2}}{2}\right) + (y-\frac{\pi}{4})^2\left(-\frac{e\sqrt{2}}{2}\right) \right] $$ Factor out $\frac{e\sqrt{2}}{2}$: $$ f(x,y) \approx \frac{e\sqrt{2}}{2} \left[ 1 + (x-1) - (y-\frac{\pi}{4}) + \frac{1}{2} \left( (x-1)^2 - 2(x-1)(y-\frac{\pi}{4}) - (y-\frac{\pi}{4})^2 \right) \right] $$ Extreme Values of Functions Problem: Examine the extreme values of $f(x, y) = x^2 - xy + y^2 + 3x - 2y + 1$. Solution: To find extreme values, we first find critical points by setting the first partial derivatives to zero. $f_x = \frac{\partial}{\partial x}(x^2 - xy + y^2 + 3x - 2y + 1) = 2x - y + 3$ $f_y = \frac{\partial}{\partial y}(x^2 - xy + y^2 + 3x - 2y + 1) = -x + 2y - 2$ Set $f_x = 0$ and $f_y = 0$: 1) $2x - y + 3 = 0 \implies y = 2x + 3$ 2) $-x + 2y - 2 = 0$ Substitute (1) into (2): $-x + 2(2x + 3) - 2 = 0$ $-x + 4x + 6 - 2 = 0$ $3x + 4 = 0 \implies x = -\frac{4}{3}$ Now find $y$: $y = 2(-\frac{4}{3}) + 3 = -\frac{8}{3} + \frac{9}{3} = \frac{1}{3}$. So, the only critical point is $(-\frac{4}{3}, \frac{1}{3})$. Next, we use the Second Derivative Test. We need the second partial derivatives: $f_{xx} = \frac{\partial}{\partial x}(2x - y + 3) = 2$ $f_{yy} = \frac{\partial}{\partial y}(-x + 2y - 2) = 2$ $f_{xy} = \frac{\partial}{\partial y}(2x - y + 3) = -1$ Calculate the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$: $D = (2)(2) - (-1)^2 = 4 - 1 = 3$. At the critical point $(-\frac{4}{3}, \frac{1}{3})$, $D = 3$. Since $D > 0$ and $f_{xx} = 2 > 0$, the function has a local minimum at $(-\frac{4}{3}, \frac{1}{3})$. The value of the minimum is: $f(-\frac{4}{3}, \frac{1}{3}) = (-\frac{4}{3})^2 - (-\frac{4}{3})(\frac{1}{3}) + (\frac{1}{3})^2 + 3(-\frac{4}{3}) - 2(\frac{1}{3}) + 1$ $= \frac{16}{9} + \frac{4}{9} + \frac{1}{9} - 4 - \frac{2}{3} + 1$ $= \frac{21}{9} - 3 - \frac{2}{3} = \frac{7}{3} - 3 - \frac{2}{3} = \frac{5}{3} - 3 = \frac{5 - 9}{3} = -\frac{4}{3}$. Since this is the only critical point and $D > 0$, this local minimum is also a global minimum. Multivariable Calculus - Integration Volume Calculation Problem: Find the volume of the solid cut from the first octant by the surface $z = 4 - x^2 - y$. Solution: The first octant implies $x \ge 0, y \ge 0, z \ge 0$. The surface is $z = 4 - x^2 - y$. For $z \ge 0$, we must have $4 - x^2 - y \ge 0$, which means $y \le 4 - x^2$. The region of integration $R$ in the $xy$-plane is defined by: $x \ge 0$ $y \ge 0$ $y \le 4 - x^2$ To find the limits for $x$, we set $y=0$ in $y = 4 - x^2$, which gives $0 = 4 - x^2 \implies x^2 = 4 \implies x = 2$ (since $x \ge 0$). So, the region $R$ is $0 \le x \le 2$ and $0 \le y \le 4 - x^2$. The volume $V$ is given by the double integral of $z$ over $R$: $$ V = \iint_R (4 - x^2 - y) \, dA = \int_{0}^{2} \int_{0}^{4-x^2} (4 - x^2 - y) \, dy \, dx $$ First, integrate with respect to $y$: $$ \int_{0}^{4-x^2} (4 - x^2 - y) \, dy = \left[ (4 - x^2)y - \frac{1}{2}y^2 \right]_{0}^{4-x^2} $$ $$ = (4 - x^2)(4 - x^2) - \frac{1}{2}(4 - x^2)^2 - 0 $$ $$ = (4 - x^2)^2 - \frac{1}{2}(4 - x^2)^2 = \frac{1}{2}(4 - x^2)^2 $$ Now, integrate this expression with respect to $x$: $$ V = \int_{0}^{2} \frac{1}{2}(4 - x^2)^2 \, dx = \frac{1}{2} \int_{0}^{2} (16 - 8x^2 + x^4) \, dx $$ $$ = \frac{1}{2} \left[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \right]_{0}^{2} $$ $$ = \frac{1}{2} \left[ 16(2) - \frac{8}{3}(2)^3 + \frac{1}{5}(2)^5 - (0) \right] $$ $$ = \frac{1}{2} \left[ 32 - \frac{8}{3}(8) + \frac{1}{5}(32) \right] $$ $$ = \frac{1}{2} \left[ 32 - \frac{64}{3} + \frac{32}{5} \right] $$ $$ = 16 - \frac{32}{3} + \frac{16}{5} $$ To combine these, find a common denominator (15): $$ = \frac{16 \times 15}{15} - \frac{32 \times 5}{15} + \frac{16 \times 3}{15} $$ $$ = \frac{240 - 160 + 48}{15} = \frac{80 + 48}{15} = \frac{128}{15} $$ The volume of the solid is $\frac{128}{15}$. Arc Length of a Curve Problem: Find the length of the curve $y = \frac{4\sqrt{2}}{3}x^{3/2} - 1$, for $0 \le x \le 1$. Solution: The arc length $L$ of a curve $y=f(x)$ from $a$ to $b$ is given by the formula: $$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ First, find the derivative $\frac{dy}{dx}$: $y = \frac{4\sqrt{2}}{3}x^{3/2} - 1$ $\frac{dy}{dx} = \frac{4\sqrt{2}}{3} \cdot \frac{3}{2}x^{3/2 - 1} - 0 = 2\sqrt{2}x^{1/2} = 2\sqrt{2x}$. Next, square the derivative: $\left(\frac{dy}{dx}\right)^2 = (2\sqrt{2x})^2 = 4(2x) = 8x$. Now, substitute this into the arc length formula with limits $a=0$ and $b=1$: $$ L = \int_{0}^{1} \sqrt{1 + 8x} \, dx $$ To evaluate this integral, use a substitution. Let $u = 1 + 8x$. Then $du = 8 \, dx \implies dx = \frac{1}{8}du$. When $x=0$, $u = 1 + 8(0) = 1$. When $x=1$, $u = 1 + 8(1) = 9$. $$ L = \int_{1}^{9} \sqrt{u} \, \frac{1}{8}du = \frac{1}{8} \int_{1}^{9} u^{1/2} \, du $$ $$ = \frac{1}{8} \left[ \frac{u^{3/2}}{3/2} \right]_{1}^{9} = \frac{1}{8} \left[ \frac{2}{3}u^{3/2} \right]_{1}^{9} $$ $$ = \frac{1}{12} \left[ u^{3/2} \right]_{1}^{9} = \frac{1}{12} (9^{3/2} - 1^{3/2}) $$ $$ = \frac{1}{12} ((\sqrt{9})^3 - 1) = \frac{1}{12} (3^3 - 1) $$ $$ = \frac{1}{12} (27 - 1) = \frac{26}{12} = \frac{13}{6} $$ The length of the curve is $\frac{13}{6}$. Line Integrals Problem: Evaluate $\int_C (x - y + z - 2) \, ds$ where $C$: Line segment from $(0, 1, 1)$ to $(1, 0, 1)$. Solution: First, we need to parametrize the line segment $C$. A line segment from point $P_0$ to $P_1$ can be parametrized as $\vec{r}(t) = (1-t)P_0 + tP_1$ for $0 \le t \le 1$. Here, $P_0 = (0, 1, 1)$ and $P_1 = (1, 0, 1)$. $\vec{r}(t) = (1-t)(0, 1, 1) + t(1, 0, 1)$ $\vec{r}(t) = (0, 1-t, 1-t) + (t, 0, t)$ $\vec{r}(t) = (t, 1-t, 1)$. So, $x(t) = t$, $y(t) = 1-t$, $z(t) = 1$. Next, we need to find $ds$. First, find $\vec{r}'(t)$: $\vec{r}'(t) = \frac{d}{dt}(t, 1-t, 1) = (1, -1, 0)$. Then, calculate $||\vec{r}'(t)|| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2}$. So, $ds = ||\vec{r}'(t)|| \, dt = \sqrt{2} \, dt$. Now, substitute $x(t), y(t), z(t)$ and $ds$ into the integral: $$ \int_C (x - y + z - 2) \, ds = \int_{0}^{1} (t - (1-t) + 1 - 2) \sqrt{2} \, dt $$ Simplify the integrand: $t - 1 + t + 1 - 2 = 2t - 2$. So the integral becomes: $$ \int_{0}^{1} (2t - 2) \sqrt{2} \, dt = \sqrt{2} \int_{0}^{1} (2t - 2) \, dt $$ Evaluate the integral: $$ \sqrt{2} \left[ t^2 - 2t \right]_{0}^{1} $$ $$ = \sqrt{2} \left[ (1^2 - 2(1)) - (0^2 - 2(0)) \right] $$ $$ = \sqrt{2} \left[ (1 - 2) - 0 \right] $$ $$ = \sqrt{2} (-1) = -\sqrt{2} $$ The value of the line integral is $-\sqrt{2}$. Vector Calculus - Flux and Divergence Theorem Outward Flux Problem: Find the outward flux of the field $\vec{F} = xz\mathbf{i} + yz\mathbf{j} + \mathbf{k}$ across the surface of the upper cap cut from the solid sphere $x^2 + y^2 + z^2 \le 25$ by the plane $z = 3$. Solution: The surface $S$ is the upper part of the sphere $x^2 + y^2 + z^2 = 25$ for $z \ge 3$. This is an open surface (a cap). We can use the Divergence Theorem, which states that $\iint_S \vec{F} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \vec{F} \, dV$ for a *closed* surface $S$ enclosing a volume $V$. To use the Divergence Theorem, we must close the surface. The boundary of the cap $S$ is a circle at $z=3$. Let's call this disk $S_1$. The equation of the sphere is $x^2 + y^2 + z^2 = 25$. When $z=3$, $x^2 + y^2 + 3^2 = 25 \implies x^2 + y^2 = 16$. So $S_1$ is the disk $x^2+y^2 \le 16$ in the plane $z=3$. Let $S_{total}$ be the closed surface consisting of the cap $S$ and the disk $S_1$. The outward normal for $S_1$ points downwards, i.e., $\mathbf{n}_{S_1} = -\mathbf{k}$. By the Divergence Theorem: $$ \iint_{S_{total}} \vec{F} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \vec{F} \, dV $$ where $V$ is the volume of the spherical segment (the region $x^2+y^2+z^2 \le 25$ with $3 \le z \le 5$). First, calculate the divergence of $\vec{F}$: $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(1) = z + z + 0 = 2z$. So, the triple integral is $\iiint_V 2z \, dV$. It is simplest to evaluate this integral using cylindrical coordinates. $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. $dV = r \, dz \, dr \, d\theta$. The limits for $z$ are from $3$ to $\sqrt{25 - x^2 - y^2} = \sqrt{25 - r^2}$. The limits for $r$ are from $0$ to $4$ (since $x^2+y^2=16$ at $z=3$). The limits for $\theta$ are from $0$ to $2\pi$. $$ \iiint_V 2z \, dV = \int_{0}^{2\pi} \int_{0}^{4} \int_{3}^{\sqrt{25-r^2}} 2z \cdot r \, dz \, dr \, d\theta $$ Integrate with respect to $z$: $$ \int_{3}^{\sqrt{25-r^2}} 2zr \, dz = r \left[ z^2 \right]_{3}^{\sqrt{25-r^2}} = r((\sqrt{25-r^2})^2 - 3^2) = r(25 - r^2 - 9) = r(16 - r^2) $$ Now integrate with respect to $r$: $$ \int_{0}^{4} r(16 - r^2) \, dr = \int_{0}^{4} (16r - r^3) \, dr = \left[ 8r^2 - \frac{1}{4}r^4 \right]_{0}^{4} $$ $$ = (8(4^2) - \frac{1}{4}(4^4)) - 0 = 8(16) - \frac{1}{4}(256) = 128 - 64 = 64 $$ Finally, integrate with respect to $\theta$: $$ \int_{0}^{2\pi} 64 \, d\theta = 64 [\theta]_{0}^{2\pi} = 64(2\pi) = 128\pi $$ So, $\iint_{S_{total}} \vec{F} \cdot \mathbf{n} \, dS = 128\pi$. Now we need to subtract the flux through the bottom disk $S_1$. For $S_1$, $z=3$ and $\mathbf{n}_{S_1} = -\mathbf{k}$. $\vec{F} \cdot \mathbf{n}_{S_1} = (xz\mathbf{i} + yz\mathbf{j} + \mathbf{k}) \cdot (-\mathbf{k}) = -1$. The integral over $S_1$ is $\iint_{S_1} (-1) \, dA = - \text{Area}(S_1)$. $S_1$ is a disk of radius 4 (since $x^2+y^2=16$). Area$(S_1) = \pi (4^2) = 16\pi$. So, $\iint_{S_1} \vec{F} \cdot \mathbf{n}_{S_1} \, dS = -16\pi$. The outward flux across the upper cap $S$ is: $$ \iint_S \vec{F} \cdot \mathbf{n} \, dS = \iint_{S_{total}} \vec{F} \cdot \mathbf{n} \, dS - \iint_{S_1} \vec{F} \cdot \mathbf{n}_{S_1} \, dS $$ $$ = 128\pi - (-16\pi) = 128\pi + 16\pi = 144\pi $$ The outward flux is $144\pi$. Divergence Theorem with Spherical Shell (Ellipsoid) Problem: Let $\mathbf{n}$ be the outer unit normal of the spherical shell $S: 4x^2 + 4y^2 + 36z^2 = 36, z \ge 0$. Let $\vec{F} = y\mathbf{i} + x^2\mathbf{j} + (x^2 + y^2)^{3/2}\mathbf{k}$. Find the value of $\iint_S \vec{F} \cdot \mathbf{n} \, dS$. Solution: The surface $S$ is the upper half ($z \ge 0$) of the ellipsoid given by $4x^2 + 4y^2 + 36z^2 = 36$. Divide by 36: $\frac{x^2}{9} + \frac{y^2}{9} + \frac{z^2}{1} = 1$. This is an ellipsoid with semi-axes $a=3, b=3, c=1$. The surface $S$ is the upper cap of this ellipsoid. It is an open surface. To use the Divergence Theorem, we need to close the surface. We add the bottom disk $S_0$ in the $xy$-plane ($z=0$). The boundary of $S$ is where $z=0$: $\frac{x^2}{9} + \frac{y^2}{9} = 1 \implies x^2 + y^2 = 9$. So $S_0$ is the disk $x^2+y^2 \le 9$ in the plane $z=0$. Let $S_{total} = S \cup S_0$ be the closed surface. The outward normal for $S_0$ is $\mathbf{n}_{S_0} = -\mathbf{k}$. By the Divergence Theorem: $$ \iint_{S_{total}} \vec{F} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \vec{F} \, dV $$ where $V$ is the volume enclosed by $S_{total}$ (the upper half of the ellipsoid). First, calculate the divergence of $\vec{F}$: $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(x^2) + \frac{\partial}{\partial z}((x^2 + y^2)^{3/2})$ $\nabla \cdot \vec{F} = 0 + 0 + 0 = 0$. Since the divergence is zero, the triple integral is zero: $$ \iiint_V \nabla \cdot \vec{F} \, dV = \iiint_V 0 \, dV = 0 $$ Therefore, $\iint_{S_{total}} \vec{F} \cdot \mathbf{n} \, dS = 0$. We can write this as: $$ \iint_S \vec{F} \cdot \mathbf{n} \, dS + \iint_{S_0} \vec{F} \cdot \mathbf{n}_{S_0} \, dS_0 = 0 $$ So, $\iint_S \vec{F} \cdot \mathbf{n} \, dS = - \iint_{S_0} \vec{F} \cdot \mathbf{n}_{S_0} \, dS_0$. Now, calculate the flux through $S_0$: On $S_0$, $z=0$. The outward normal for $S_0$ is $\mathbf{n}_{S_0} = -\mathbf{k}$. $\vec{F}$ on $S_0$ is $\vec{F}(x,y,0) = y\mathbf{i} + x^2\mathbf{j} + (x^2+y^2)^{3/2}\mathbf{k}$. $\vec{F} \cdot \mathbf{n}_{S_0} = (y\mathbf{i} + x^2\mathbf{j} + (x^2+y^2)^{3/2}\mathbf{k}) \cdot (-\mathbf{k}) = -(x^2+y^2)^{3/2}$. The integral over $S_0$ is: $$ \iint_{S_0} \vec{F} \cdot \mathbf{n}_{S_0} \, dS_0 = \iint_{S_0} -(x^2+y^2)^{3/2} \, dA $$ The region $S_0$ is the disk $x^2+y^2 \le 9$. This is a circle of radius 3 centered at the origin. Convert to polar coordinates: $x^2+y^2 = r^2$, $dA = r \, dr \, d\theta$. $$ \iint_{S_0} -(x^2+y^2)^{3/2} \, dA = \int_{0}^{2\pi} \int_{0}^{3} -(r^2)^{3/2} r \, dr \, d\theta $$ $$ = \int_{0}^{2\pi} \int_{0}^{3} -r^3 \cdot r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{3} -r^4 \, dr \, d\theta $$ Integrate with respect to $r$: $$ \int_{0}^{3} -r^4 \, dr = \left[ -\frac{1}{5}r^5 \right]_{0}^{3} = -\frac{1}{5}(3^5 - 0^5) = -\frac{243}{5} $$ Now integrate with respect to $\theta$: $$ \int_{0}^{2\pi} -\frac{243}{5} \, d\theta = -\frac{243}{5} [\theta]_{0}^{2\pi} = -\frac{243}{5}(2\pi) = -\frac{486\pi}{5} $$ So, $\iint_{S_0} \vec{F} \cdot \mathbf{n}_{S_0} \, dS_0 = -\frac{486\pi}{5}$. Finally, the outward flux across $S$ is: $$ \iint_S \vec{F} \cdot \mathbf{n} \, dS = - \left( -\frac{486\pi}{5} \right) = \frac{486\pi}{5} $$ The value of the integral is $\frac{486\pi}{5}$.