Question

Evaluate 𝐹 ∙ 𝑛𝑑𝑆, where 𝐹 = 4𝑥𝑧𝑖 − 𝑦 ² 𝑗 + 𝑦𝑧𝑘 and S is the surface of the bounded by 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1.

Answer 1

The Divergence Theorem

$$\iint\vec F\cdot d\vec S=\iiint \text{div}\vec F dV$$ S\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V \ \ \ \ \ \ \ \ \ \ \ \ \

The divergence of $\vec F$ is

$$\text{div}\vec F=\nabla\cdot \vec F$$

$$=(i\dfrac{\partial}{\partial x}+j\dfrac{\partial}{\partial y}+k\dfrac{\partial}{\partial z})(4xzi− y^2j + yzk)$$

$$=\dfrac{\partial}{\partial x}(4xz)+\dfrac{\partial}{\partial y}(-y^2)+\dfrac{\partial}{\partial z}(yz)=4z-2y+y$$

$$=4z-y$$

Then

$$\iint\vec F\cdot d\vec S=\iiint \text{div}\vec F dV$$ S\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V \ \ \ \ \ \ \ \ \ \ \ \ \

$$=\displaystyle\int_{x=0}^{1}\displaystyle\int_{y=0}^{1}\displaystyle\int_{z=0}^{1}(4z-y)dzdydx$$

$$=\displaystyle\int_{x=0}^{1}\displaystyle\int_{y=0}^{1}[2z^2-yz]\begin{matrix} 1\\ 0 \end{matrix}dydx$$

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

$$=\displaystyle\int_{x=0}^{1}[2y-\dfrac{y^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}dx=\displaystyle\int_{x=0}^{1}\dfrac{3}{2}dx$$

$$=\dfrac{3}{2}[x]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{3}{2}$$

$$\iint\vec F\cdot d\vec S=\dfrac{3}{2}$$ S\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \