Example Find the surface integral using divergence theorem．

$$\iint_{S} (xz^2{\bf i} + (x^2 y - z^3){\bf j} + (2xy + y^2 z){\bf k})\cdot{\bf n}dS$$

Here, $S$ is composed of two surfaces $S_{1} : z = \sqrt{a^2 - x^2 - y^2}, x^2 + y^2 \leq a^2$ and $S_{2} : z = 0, x^2 + y^2 \leq a^2$

Answer ${\bf F} = xz^2{\bf i} + (x^2 y - z^3){\bf j} + (2xy + y^2 z){\bf k}$. So,

$${\rm div}{\bf F} = \nabla \cdot {\bf F} = z^2 + x^2 +y^2$$

Let the inside of the upper sphere $S$ be $V$. Then by the divergence thereom,

$$\iint_{S}{\bf F}\cdot {\bf n}dS = \iiint_{V}{\rm div}{\bf F}dV$$

$$= \iiint_{V}(x^2 + y^2 + z^2)dV$$

To evaluate this integral, we use the polar coordinates. Note that $\theta$ is the angle from the $x$-axis，$\phi$ is the angle from the $z$-axis.

$$x = \rho \sin{\phi}\cos{\theta}$$

$$y = \rho \sin{\phi}\sin{\theta}$$

$$z = \rho \cos{\phi}$$

Now

$$x^2 + y^2 + z^2 = \rho^2 \sin^{2}{\phi}\cos^{2}{\theta} + \rho^2 \sin^{2}{\phi}\sin^{2}{\theta} + \rho^2 \cos^{2}{\phi} = \rho^2$$

Jacobian is given by the followings:

$$J = \begin{array}{\vert lll\vert} x_{\rho} & x_{\phi} & x_{\theta... ...os{\theta}\\ \cos{\phi} & -\rho \sin{\phi} & 0 \end{array} = \rho^2 \sin{\phi}$$

Since $S$ is upper sphere, $\theta$ ranges $0 \leq \theta \leq 2\pi$ and the $\phi$ ranges $0 \leq \phi \leq \pi$. Finally, the radius goes from $0 \leq \rho \leq a$. Thus, we have

$$\iiint_{V}(x^2 + y^2 + z^2)dV = \int_{\theta = 0}^{2\pi}\int_{0}^{\pi}\int_{0}^{a}\rho^2 \rho^2 \sin{\phi}d\rho d\phi d\theta$$

$$= \int_{0}^{2\pi}d\theta \int_{0}^{\pi}\sin{\phi}d\phi \int_{0}^{a}\rho^4d \rho$$

$$= (2\pi)(2)(\frac{a^5}{5}) = \frac{4 \pi a^5}{5}$$