# Flux through a surface

1. Jun 18, 2006

### jdstokes

Hi all,

Let $D$ be the top half of a ball of radius $a>0$ and let

$\mathbf{f} = xz \mathbf{i} + y \mathbf{j} + x \mathbf{k}$

Calculate the outward flux across S using the definition of the surface integral.

$\int_{S} \mathbf{f} \cdot \mathbf{n} dS = \int_{cap} \mathbf{f} \cdot \mathbf{n} dS + \int_{disc} \mathbf{f} \cdot \mathbf{n} dS$

$\frac{1}{a}\int_{cap} (xz,y,x) \cdot (x,y,z) d S$
$\frac{1}{a}\int_{cap} x^2z + y^2 + xz d S$
$\frac{1}{a}\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^2\theta\cos\theta + r^2\sin^2\varphi\sin^2\theta + r^2\cos\varphi\sin\theta\cos\theta) (a^2 \sin\theta) d\theta d\varphi$
$a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\sin^2\varphi\sin^3\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi$

$a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi$

I assume that one would then evaluate the integral by parts. But is there an easier method?

Last edited: Jun 18, 2006
2. Jun 18, 2006

### jdstokes

I figured it out using integration by substitution.