Mumbai University > Electronics and Telecommunication > Sem 3 > Applied Maths 3

Marks: 6M

Year: Dec 2013

By stoke's theorem, $\int\bar{f}d\bar{r} = \int\int_{s}\bar{N}(\bar{V}\cdot \bar{F})ds$

Here s is the region bounded by 2 circles

Now in xy plane $\bar{N}k$

ds = dx dy

$\bar{V} \times f = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 + y^2 & x^2 - y^2 & 0 \end{vmatrix} \\ = \hat{i} \bigg(0 - \frac{\partial}{\partial z}x^2 - y^2\bigg) - \hat{j}\bigg(\frac{\partial}{\partial x}(0) - \frac{\partial}{\partial z}x^2 + y^2\bigg) + \hat{k}\bigg(\frac{\partial}{\partial x}x^2 - y^2 - \frac{\partial}{\partial y}x^2 + y^2\bigg) \\ \therefore \bar{V} \times f = \hat{k}(2x - 2y) \\ \therefore \hat{N}(\bar{V} \times f) = \hat{k} * \hat{k}(2x - 2y) = 2x - 2y$

If we use polar co-ordinates the equations of the circles are r = 2, r = 4, x = rcos$\theta$, y = rsin$\theta$

Therefore, $2x - 2y = 2[rcos\theta - rsin\theta] = 2r[cos\theta - sin\theta] \\ dxdy = rdrd\theta$

$\therefore \int\int\limits_{s}N(\bar{V} \times F)ds = \int\limits_{\theta = 0}^{2\pi}\int\limits_{r = 2}^{4} 2r(cos\theta - sin\theta)rdrd\theta \\ \therefore \int\int\limits_{s}N(\bar{V} \times F)ds = 2 \int\limits_{\theta = 0}^{2\pi}\int\limits_{r = 2}^{4} r^2 (cos\theta - sin\theta)drd\theta \\ \therefore \int\int\limits_{s}N(\bar{V} \times F)ds = 2 \int\limits_{\theta = 0}^{2\pi}(cos\theta - sin\theta) \int\limits_{r = 2}^{4}r^2 dr \\ \therefore \int\int\limits_{s}N(\bar{V} \times F)ds = 2[sin\theta - cos\theta]_{0}^{2\pi} \bigg[\frac{r^3}{3}\bigg]_{2}^{4} \\ \therefore \int\int\limits_{s}N(\bar{V} \times F)ds = 2(0) \frac{64 - 8}{3} \\ \therefore \int\int\limits_{s}N(\bar{V} \times F)ds = 0$

$\therefore \oint[(x^2 + y^2)\hat{i} + (x^2 - y^2)\hat{j}]d\bar{r} = 0$