Answer to Question #281719 in Math for Khizar

Question

use Stokes theorem evaluate ∮A.dr, where A=y²i+x²j+z²k and C is boundary of the part of the plane x+y+z=1 like in the first octant

Answer 1

$$curl \bold A=\begin{vmatrix} \bold i & \bold j & \bold k \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ y^2 & x^2 & z^2 \end{vmatrix}$$

$$=(2x-2y)\bold k$$

$$\bold n=\langle1, 1, 1\rangle$$

$$curl \bold A\cdot \bold n=2x-2y$$

$$\int_C\bold A\cdot d\bold r=\int\int _Scurl \bold A\cdot d\bold S$$

$$=\int \int_S(2x-2y)dS$$

$$=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-y}(2x-2y)dxdy$$

$$=\displaystyle\int_{0}^1[x^2-2xy]\begin{matrix} 1-y \\ 0 \end{matrix}dy$$

$$=\displaystyle\int_{0}^1((1-y)^2-2y(1-y))dy$$

$$=\displaystyle\int_{0}^1(1-2y+y^2-2y+2y^2)dy$$

$$=[y^3-2y^2+y]\begin{matrix} 1 \\ 0 \end{matrix}=0$$

$$\int_C\bold A\cdot d\bold r=0$$