Answer to Question #105607 in Electricity and Magnetism for Rawan

Question

Assume that the vector field A= 5xyax + xy2ay + 4zaz is defined in a region 1≤ x ≤ 3, −2 ≤ y ≤ 4 and −1≤ z ≤ 2. Find the value of the closed-surface integral ∫∫ A.ds over the surface of this region.

Answer 1

The vector field is

$$\textbf{A}= 5xy\textbf{x} + xy^2\textbf{y} + 4z\textbf{z}.$$

The value of the closed-surface integral $\int\int AdS$ over the surface of this region can be found with Gauss's theorem:

$$\Phi=\iint\textbf{A}\text{ d}S=\iiint\text{ div}\textbf{A}\text{ d}V,\\ \space\\ \text{ div}\textbf{A}=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}=\\ \space\\ =\frac{\partial 5xy}{\partial x}+\frac{\partial xy^2}{\partial y}+\frac{\partial 4z}{\partial z}=\\ \space\\ =5y+2xy+4.\\ \space\\ \iiint\text{ div}\textbf{A}\text{ d}V=\iiint(5y+2xy+4)\text{d}x\text{d}y\text{d}z=\\ \space\\ =\int_{-1}^2\Bigg[\int_{-2}^4\Bigg(\int^3_1(5y+2xy+4)\text{d}x\Bigg)\text{d}y\Bigg]\text{d}z=\\ \space\\ =\int_{-1}^2\Bigg[\int_{-2}^4\Bigg((5xy+x^2y+4x)\bigg|^3_1\Bigg)\text{d}y\Bigg]\text{d}z=\\ \space\\ =\int_{-1}^2\Bigg[\int_{-2}^4(18y+8)\text{d}y\Bigg]\text{d}z=\\ \space\\ =\int_{-1}^2\Bigg[(y^2+8y)\bigg|^4_{-2}\Bigg]\text{d}z=\\ \space\\ =\int_{-1}^260\text{ d}z=60z\bigg|^2_{-1}=180.$$