Question

If 𝑭 = 𝟔𝒙𝒛𝒊 − 𝟐𝒚 𝟐 𝒋 + 𝒚𝒛𝒌, evaluate ∬ 𝑭 ∙ 𝒏𝒅𝑺 𝑺 where S is the surface of the cube bounded by 𝒙 = 𝟎, 𝒙 = 𝟏, 𝒚 = 𝟎, 𝒚 = 𝟏, 𝒛 = 𝟎, 𝒛 = 𝟏

Answer 1

Use the Divergence Theorem

$$\int\int_S\vec F\cdot\vec n\ ds=\int\int\int_E\text{div}\vec F(x, y, z)dV$$

$$\text{div}\vec F(x, y, z)=\nabla\cdot \vec F$$

$$=\dfrac{\partial}{\partial x}(6xz)+\dfrac{\partial}{\partial y}(-2y^2)+\dfrac{\partial}{\partial z}(yz)$$

$$=6z-4y+y=6z-3y$$

Then by the Divergenve Theorem we have

$$\int\int_S\vec F\cdot\vec n\ ds=\int\int\int_E\text{div}\vec F(x, y, z)dV$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(6z-3y)dzdydx$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\big[3z^2-3yz\big]\begin{matrix} 1 \\ 0 \end{matrix}dydx$$

$$=3\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(1-y)\begin{matrix} 1 \\ 0 \end{matrix}dydx$$

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

$$=\dfrac{3}{2}\displaystyle\int_{0}^{1}dx$$

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

$$=\dfrac{3}{2}$$

$$\int\int_S(6xz\vec i-2y^2\vec j+yz\vec k)\cdot\vec n\ ds=\dfrac{3}{2}$$