Stokes Integrals Partials 8B7802

1. **Problem Statement:** (a)(i) State Stokes' theorem without proof. (a)(ii) Find $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F}(x,y,z) = xyz \mathbf{i} + 2xz \mathbf{j} + e^{xy} \mathbf{k}$ where $C$ is the circle $x^2 + y^2 = 16$, $z=5$. (b) Evaluate $\iint_R 2e^{\frac{x+y}{x-y}} dA$ where $R$ is trapezoid with vertices $(1,0)$, $(2,0)$, $(0,-2)$, $(0,-1)$. (c) Given $z=f(x,y)$ with continuous second derivatives and $x=r^2+s^2$, $y=2rs$, find $\frac{\partial^2 z}{\partial s^2}$. --- 2. **(a)(i) Stokes' Theorem Statement:** Stokes' theorem states: $$\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$$ where $C$ is a positively oriented, piecewise-smooth, simple closed curve bounding the surface $S$, and $\mathbf{n}$ is the unit normal to $S$. --- 3. **(a)(ii) Calculate $\int_C \mathbf{F} \cdot d\mathbf{r}$ using Stokes' theorem:** - $C$ is the circle $x^2 + y^2 = 16$, $z=5$. - Choose $S$ as the disk $x^2 + y^2 \leq 16$, $z=5$. - Compute curl $\nabla \times \mathbf{F}$: $$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xyz & 2xz & e^{xy} \end{vmatrix}$$ Calculate components: - $\frac{\partial}{\partial y}(e^{xy}) = x e^{xy}$ - $\frac{\partial}{\partial z}(2xz) = 2x$ - $\frac{\partial}{\partial x}(e^{xy}) = y e^{xy}$ - $\frac{\partial}{\partial z}(xyz) = xy$ - $\frac{\partial}{\partial x}(2xz) = 2z$ - $\frac{\partial}{\partial y}(xyz) = xz$ So, $$\nabla \times \mathbf{F} = \left( x e^{xy} - 2x \right) \mathbf{i} - \left( y e^{xy} - xy \right) \mathbf{j} + \left( 2z - xz \right) \mathbf{k}$$ Simplify: $$\nabla \times \mathbf{F} = (x e^{xy} - 2x) \mathbf{i} - (y e^{xy} - xy) \mathbf{j} + (2z - xz) \mathbf{k}$$ At $z=5$: $$\nabla \times \mathbf{F} = (x e^{xy} - 2x) \mathbf{i} - (y e^{xy} - xy) \mathbf{j} + (10 - 5x) \mathbf{k}$$ --- 4. **Surface integral:** Since $S$ is the disk in plane $z=5$, the unit normal vector $\mathbf{n} = \mathbf{k}$. Therefore, $$\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \iint_D (10 - 5x) \, dA$$ where $D$ is the disk $x^2 + y^2 \leq 16$ in $xy$-plane. --- 5. **Evaluate integral:** $$\iint_D (10 - 5x) \, dA = \iint_D 10 \, dA - 5 \iint_D x \, dA$$ - $\iint_D 10 \, dA = 10 \times$ area of disk $= 10 \times \pi \times 16 = 160\pi$ - $\iint_D x \, dA = 0$ by symmetry (disk symmetric about $y$-axis) So, $$\int_C \mathbf{F} \cdot d\mathbf{r} = 160\pi$$ --- 6. **(b) Evaluate $\iint_R 2e^{\frac{x+y}{x-y}} dA$ over trapezoid $R$ with vertices $(1,0)$, $(2,0)$, $(0,-2)$, $(0,-1)$:** - Use substitution: $$u = x + y, \quad v = x - y$$ - Then, $$x = \frac{u+v}{2}, \quad y = \frac{u - v}{2}$$ - Jacobian determinant: $$J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}$$ - Absolute value $|J| = \frac{1}{2}$ - The integrand becomes: $$2 e^{\frac{x+y}{x-y}} = 2 e^{\frac{u}{v}}$$ - Find new region $S$ in $uv$-plane: Vertices: - $(1,0)$: $u=1+0=1$, $v=1-0=1$ - $(2,0)$: $u=2+0=2$, $v=2-0=2$ - $(0,-2)$: $u=0+(-2)=-2$, $v=0-(-2)=2$ - $(0,-1)$: $u=0+(-1)=-1$, $v=0-(-1)=1$ So $S$ is trapezoid with vertices $(1,1)$, $(2,2)$, $(-2,2)$, $(-1,1)$. - $v$ ranges from 1 to 2. - For fixed $v$, $u$ ranges from $-v$ to $v$. --- 7. **Integral in $uv$ coordinates:** $$\iint_R 2 e^{\frac{u}{v}} dA = \int_{v=1}^2 \int_{u=-v}^v 2 e^{\frac{u}{v}} \times \frac{1}{2} du dv = \int_1^2 \int_{-v}^v e^{\frac{u}{v}} du dv$$ --- 8. **Evaluate inner integral:** $$\int_{-v}^v e^{\frac{u}{v}} du$$ Substitute $w = \frac{u}{v}$, so $du = v dw$. Limits: when $u=-v$, $w=-1$; when $u=v$, $w=1$. So, $$\int_{-v}^v e^{\frac{u}{v}} du = \int_{-1}^1 e^w v dw = v \int_{-1}^1 e^w dw = v \left[ e^w \right]_{-1}^1 = v (e - e^{-1})$$ --- 9. **Evaluate outer integral:** $$\int_1^2 v (e - e^{-1}) dv = (e - e^{-1}) \int_1^2 v dv = (e - e^{-1}) \left[ \frac{v^2}{2} \right]_1^2 = (e - e^{-1}) \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{3}{2} (e - e^{-1})$$ --- 10. **Answer for (b):** $$\iint_R 2 e^{\frac{x+y}{x-y}} dA = \frac{3}{2} (e - e^{-1})$$ --- 11. **(c) Find $\frac{\partial^2 z}{\partial s^2}$ given $z=f(x,y)$, $x=r^2 + s^2$, $y=2rs$:** - Use chain rule: $$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} = f_x (2s) + f_y (2r)$$ - Differentiate again w.r.t. $s$: $$\frac{\partial^2 z}{\partial s^2} = \frac{\partial}{\partial s} (2s f_x + 2r f_y) = 2 f_x + 2s \frac{\partial f_x}{\partial s} + 2r \frac{\partial f_y}{\partial s}$$ - Compute $\frac{\partial f_x}{\partial s}$ and $\frac{\partial f_y}{\partial s}$ using chain rule: $$\frac{\partial f_x}{\partial s} = f_{xx} \frac{\partial x}{\partial s} + f_{xy} \frac{\partial y}{\partial s} = f_{xx} (2s) + f_{xy} (2r)$$ $$\frac{\partial f_y}{\partial s} = f_{yx} \frac{\partial x}{\partial s} + f_{yy} \frac{\partial y}{\partial s} = f_{yx} (2s) + f_{yy} (2r)$$ - Substitute back: $$\frac{\partial^2 z}{\partial s^2} = 2 f_x + 2s [f_{xx} (2s) + f_{xy} (2r)] + 2r [f_{yx} (2s) + f_{yy} (2r)]$$ - Simplify: $$= 2 f_x + 4 s^2 f_{xx} + 4 r s f_{xy} + 4 r s f_{yx} + 4 r^2 f_{yy}$$ - Since $f_{xy} = f_{yx}$ (continuous second derivatives), $$\frac{\partial^2 z}{\partial s^2} = 2 f_x + 4 s^2 f_{xx} + 8 r s f_{xy} + 4 r^2 f_{yy}$$ --- **Final answers:** (a)(i) Stokes' theorem: $$\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$$ (a)(ii) $$\int_C \mathbf{F} \cdot d\mathbf{r} = 160 \pi$$ (b) $$\iint_R 2 e^{\frac{x+y}{x-y}} dA = \frac{3}{2} (e - e^{-1})$$ (c) $$\frac{\partial^2 z}{\partial s^2} = 2 f_x + 4 s^2 f_{xx} + 8 r s f_{xy} + 4 r^2 f_{yy}$$