# Homework Help: Another Surface Integral

1. Oct 15, 2006

### Saketh

Okay - I thought that I figured this stuff out, but I didn't.
The Problem
When $$G(x, y, z) = (1-x^2-y^2)^{3/2}$$, and $$z = \sqrt{1-x^2-y^2}$$, evaluate the surface integral.
My Work
I keep trying this but I end up with the following integral that I cannot evaluate:

$$\int_{-1}^{1} \!\!\! \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (1-x^2-y^2)^{3/2}\sqrt{1+4x^2+4y^2} \,dx \,dy$$.

Conversion to polar coordinates doesn't help much, either. How can I find this surface integral? (Ans: $\frac{\pi}{2}$)

Thanks!

2. Oct 15, 2006

### HallsofIvy

How did you get that differential? $z= \sqrt{1- x^2- y^2}$ is the upper half of the sphere $x^2+ y^2+ z^2= 1$.

Here's one way to treat it: think of the sphere as a 'level surface' of the function $f(x,y,z)= x^2+ y^2+ z^2$. Then $\nabla f= 2xi+ 2yj+ 2zk$. "Normalize that to the xy-plane by dividing through by the coefficient of k: $\frac{x}{z}i+ \frac{y}{z}j+ 1$ and take the length to get
$$dS= \sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}dxdy$$
$$= \frac{\sqrt{x^2+ y^2+ z^2}}{z}dxdy= \frac{1}{z}dxdy$$
Since $z= \sqrt{1- x^2- y^2}$, your integral becomes
$$\int_{x= -1}^1\int_{y= -\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{(1-x^2-y^2)^{3/2}}{(1-x^2-y^2)^\frac{1}{2}}dydx= \int_{x= -1}^1\int_{y= -\sqrt{1-x^2}}^{\sqrt{1-x^2}}(1- x^2- y^2) dydx$$

Another way: using spherical coordinates, with $\rho= 1$ gives parametric equations for the hemispher:
$x= cos(\theta)sin(\phi)$, $y= sin(\theta)sin(\phi)$, and $z= cos(\phi)$, with $0\le \theta \le 2\pi$, $0 \le \phi \le \frac{\pi}{2}$.
You can write the "position vector" of a point on the hemisphere as
$$\vec{r}= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}[/itex] Then the partial derivatives are [tex]\vec{r}_\theta = -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}$$
$$\vec{r}_\phi = cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{j}- sin(\phi)\vec{k}$$
The "fundamental vector product" is the cross product of those:
$$-cos(\theta)sin^2(\phi)\vec{i}- sin(\theta)sin^2(\phi)\vec{j}- sin(\phi)cos(\phi)\vec{k}$$
and its length gives the differential of surface area in those parameters:
$$dS= sin^2(\phi)d\theta d\phi$$

Of course, $1- x^2- y^2= 1- cos^2(\theta)sin^2(\phi)- sin^2(\theta)sin^2(\phi)= 1- sin^2(\phi)= cos^2(\phi)$ so $(1-x^2-y^2)^\frac{3}{2}= cos^3(\phi)$

In terms of those parameters your integral is
$$\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\frac{\pi}{2}}cos^3(\phi)sin^2(\phi) d\phi d\theta$$

3. Oct 15, 2006

### Saketh

I do not understand the "level surface," normalization, and "dividing through by the coefficient of k." Although mathematically I understand what you are doing, the concepts behind it confuse me.
My textbook has not introduced gradients. It introduces divergence first, so technically I am not supposed to understand your gradient and normalization technique yet.

I use the following integral to solve these types of problems:
$$\int \!\!\! \int_R G(x, y, z) \sqrt{1 + \left( \frac{\partial f}{\partial x} \right ) ^2 + \frac{\partial f}{\partial y} \right ) ^2} \,dA$$
where $R$ is the projected surface (usually on the xy-plane), $f$ is one of the coordinates as a function of the other two (usually $$z = f(x, y)$$), and $dA$ is the differential area element on the project (usually $$\,dx \,dy$$). This integral is based on the idea that
$$\hat{\textbf{n}} = \frac{-i\frac{\partial f}{\partial x} - j\frac{\partial f}{\partial y} + k}{\sqrt{1 + \left( \frac{\partial f}{\partial x} \right ) ^2 + \left( \frac{\partial f}{\partial y} \right )^2}}$$

EDIT: Never mind. Though I still don't understand your procedure, I really messed up in taking the partial derivatives of z. Therein was the problem.

Last edited: Oct 15, 2006