Stokes Theorem PDF

# Stoke's Theorem: **Relation between Line & Surface Integrals** $ \int_C \textbf{F}. d\textbf{r} = \iint_S curl \textbf{F}. \textbf{n} dS$ ## Example 1: Verify Stoke's theorem for $\textbf{F} = (x+y^2) \textbf{i} - 2xy \textbf{j}$ taken around the rectangle bounded by the lines $x=a$, $x=-a$, $...

# Stoke's Theorem: **Relation between Line & Surface Integrals** $ \int_C \textbf{F}. d\textbf{r} = \iint_S curl \textbf{F}. \textbf{n} dS$ ## Example 1: Verify Stoke's theorem for $\textbf{F} = (x+y^2) \textbf{i} - 2xy \textbf{j}$ taken around the rectangle bounded by the lines $x=a$, $x=-a$, $y=0$, $y=b$. ### Solution: **Step 1:** calculate $\int_C \textbf{F}. d\textbf{r}$ $\int_C \textbf{F}. d\textbf{r} = \int_C (x+y^2)dx - 2xydy$ The curve consists of four lines AB, BC, CD and DA. * **Along AB:** * $x = a$, $dx = 0$ * y varies from 0 to b. * $\int_{AB}[(x+y^2)dx-2xydy] = \int_0^b -2ay dy = -2a [\frac{y^2}{2}]_0^b = -ab^2$ * **Along BC:** * $y = b$, $dy = 0$ * x varies from a to -a * $\int_{BC}[(x+y^2)dx-2xydy] = \int_a^{-a} (x+b^2) dx = [\frac{x^2}{2}+b^2x]_a^{-a} = (-a^2 + (-a)b^2) -(\frac{a^2}{2} + ab^2) = -\frac{3}{2}a^2 -2ab^2$ * **Along CD:** * $x = -a$, $dx = 0$ * y varies from b to 0 * $\int_{CD}[(x+y^2)dx-2xydy] = \int_b^0 2aydy = 2a[\frac{y^2}{2}]_b^0 = -ab^2$ * **Along DA:** * $y = 0$, $dy = 0$ * x varies from -a to a * $\int_{DA}[(x+y^2)dx-2xydy] = \int_{-a}^a x^2 dx = [\frac{x^3}{3}]_{-a}^a = \frac{a^3}{3} + \frac{a^3}{3} = \frac{2}{3}a^3$ Adding all four parts together we get: $\int_C \textbf{F}. d\textbf{r} = -ab^2 - \frac{3}{2}a^2 - 2ab^2 + \frac{2}{3}a^3 = -4ab^2 - \frac{3}{2}a^2 + \frac{2}{3} a^3 = -4ab^2 - \frac{7}{6}a^3$ ### Step 2: Calculate $curl \textbf{F}$ $curl \textbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x+y^2 & -2xy & 0 \end{vmatrix} = (-2y -2y) \hat{k} = -4y \hat{k}$ ### Step 3: Calculate $\iint_S curl \textbf{F}.\textbf{n} dS$ For the surface $S$, $\textbf{n} = \hat{k}$. $\\~\\$ $curl \textbf{F}.\textbf{n} = (-4y \hat{k}). (\hat{k}) = -4y$ $\\~\\$ $\iint_S curl \textbf{F}.\textbf{n} dS = \iint_S -4y dydx = -4 \int_0^b \int_{-a}^a y dxdy = -4 \int_0^b y [x]_{-a}^a dy = -8a \int_0^b ydy = -8a [\frac{y^2}{2}]_0^b = -4ab^2 $ $\\~\\$ ### Conclusion Since $\int_C \textbf{F}. d\textbf{r} = \iint_S curl \textbf{F}. \textbf{n} dS$, Stoke's theorem is verified. ## Example 2: Evaluate $\int_C \textbf{F}. d\textbf{r}$ by Stoke's theorem, where $\textbf{F} = y^2 \textbf{i} + x^2 \textbf{j} - (x+z) \textbf{k}$, and C is the boundary of the triangle with vertices at $(0,0,0)$, $(1,0,0)$ and $(1,1,0)$. ### Solution: **Step 1:** Calculate $curl \textbf{F}$ $\\~\\$ $curl \textbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2 & x^2 & -(x+z) \end{vmatrix} = -(2-0) \hat{i} + (2x-2y) \hat{k}= -2 \hat{i} + 2(x-y) \hat{k} $ **Step 2:** Calculate $\iint_S curl \textbf{F}.\textbf{n} dS$ $\\~\\$ Since $z = 0$ we know $\textbf{n} = \hat{k}$. $\\~\\$ $curl \textbf{F}.\textbf{n} = (-2 \hat{i} + 2(x-y) \hat{k}).(\hat{k}) = 2(x-y)$ $\\~\\$ $\iint_S curl \textbf{F}.\textbf{n} dS = \int_0^1 \int_0^x 2(x-y) dydx$ * The limits of integration for y are from 0 to x since we are integrating over a triangle that has a base on the x-axis and a height that increases linearly with x. * Limits of integration for x are from 0 to 1. $\iint_S curl \textbf{F}.\textbf{n} dS = \int_0^1 [2xy-y^2]_0^x dx = \int_0^1 (2x^2 - x^2) dx = \int_0^1 x^2 dx = [\frac{x^3}{3}]_0^1 = \frac{1}{3}$ ### Conclusion $\int_C \textbf{F}. d\textbf{r}= \iint_S curl \textbf{F}.\textbf{n} dS$ which verifies Stoke's theorem. ## Example 3: Verify Stoke's theorem for the vector field $\textbf{F} = (2x-y)\textbf{i} - yz \textbf{j} - y^2z \textbf{k}$, over the upper half of the surface $x^2 + y^2 + z^2 = 1$ bounded by its projection on the x-y plane. ### Solution: **Step 1: Define the Surface** Let S be the upper half surface of the sphere $x^2+y^2+z^2 = 1$. The boundary of S is a circle in the x-y plane of radius 1 and centre O. The equation of C is $x^2 + y^2 = 1$, $z = 0$, whose parametric form is $x = cos(t)$, $y = sin(t)$, $z = 0$, $0 \leq t \leq 2\pi$. **Step 2: Calculate $\int_C \textbf{F}. d\textbf{r}$** $\int_C \textbf{F}. d\textbf{r} = \int_C [(2x-y)\textbf{i} -yz \textbf{j} - y^2z \textbf{k}].(\textbf{i}dx + \textbf{j}dy + \textbf{k}dz) = \int_C [(2x-y)dx - yzdy - y^2z dz]$ $\\~\\$ Since $z = 0$, this simplifies to: $\\~\\$ $\int_C [(2x-y)dx - yzdy - y^2z dz] = \int_C (2x-y)dx$ $\\~\\$ Substituting $x = cos(t)$ and $y = sin(t)$ and $dx = -sin(t) dt$ we get: $\\~\\$ $\int_C (2x-y)dx = \int_0^{2\pi} (2cos(t)-sin(t))(-sint)dt = \int_0^{2\pi} (-2sintcost + sin^2(t))dt = \int_0^{2\pi} (-sin(2t) + \frac{1}{2}(1-cos(2t))dt = [\frac{1}{2}cos(2t) + \frac{1}{2}t - \frac{1}{4}sin(2t)]_0^{2\pi} = (\frac{1}{2} - 0 + \pi - 0 - (\frac{1}{2} - 0 + 0 - 0)) = \pi$ **Step 3: Calculate $curl \textbf{F}$** $\\~\\$ $curl \textbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2x-y & -yz & -y^2z \end{vmatrix} = (-2yz + 2yz)\hat{i} + (0-0)\hat{j} + (0+1)\hat{k} = \hat{k}$ **Step 4: Calculate $\iint_S curl \textbf{F}.\textbf{n} dS$** Since we are dealing with the x-y plane here, we can say $\textbf{n} = \hat{k}$. $\\~\\$ $curl \textbf{F}.\textbf{n} = \hat{k}. \hat{k} = 1$ $\\~\\$ $\iint_S curl \textbf{F}.\textbf{n} dS = \iint_S (1) dS = \iint_S dS$. To calculate the integral, we will change it from a surface integral into a double integral over the projection of S onto the x-y plane. $\\~\\$ $\iint_S curl \textbf{F}.\textbf{n} dS = \iint_S dS = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dy dx$ * We are integrating over a circle with radius 1 so limits of integration for y are from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$ * Limits of integration for x are from -1 to 1 $\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dy dx = \int_{-1}^1 [y]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dx = \int_{-1}^1 2\sqrt{1-x^2} dx$ This integral will need to be solved using a trigonometric substitution. * Let $x = sin(\theta)$ * $dx = cos(\theta)d\theta$ Substituting into the integral, we get: $\\~\\$ $\int_{-1}^1 2\sqrt{1-x^2} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\sqrt{1-sin^2(\theta)} cos(\theta) d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2cos^2(\theta) d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + cos(2\theta)) d\theta = [\theta + \frac{1}{2}sin(2\theta)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = (\frac{\pi}{2} + \frac{1}{2}sin(\pi)) - (-\frac{\pi}{2} + \frac{1}{2}sin(-\pi)) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$ **Conclusion:** $\int_C \textbf{F}. d\textbf{r} = \iint_S curl \textbf{F}.\textbf{n} dS$, which verifies Stoke's theorem.