How Do I Solve This Challenging Surface Integral Problem?

[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:

<br /> \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<br />.

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

Thanks!

 

[HallsofIvy]

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:

<br /> \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<br />.

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

Thanks!

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]<br /> Then the partial derivatives are <br /> \vec{r}_\theta = -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}<br /> \vec{r}_\phi = cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{j}- sin(\phi)\vec{k}<br /> The &quot;fundamental vector product&quot; is the cross product of those: <br /> -cos(\theta)sin^2(\phi)\vec{i}- sin(\theta)sin^2(\phi)\vec{j}- sin(\phi)cos(\phi)\vec{k}<br /> and its length gives the differential of surface area in those parameters:<br /> dS= sin^2(\phi)d\theta d\phi<br /> <br /> 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)<br /> <br /> In terms of those parameters your integral is<br /> \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\frac{\pi}{2}}cos^3(\phi)sin^2(\phi) d\phi d\theta

 

[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:
<br /> \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<br />
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
<br /> \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}}<br />

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.