Tricky Surface Integral Homework | Solved Step by Step"

[jegues]

Homework Statement

See figure attached for problem statement.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

I tried to solve it without using divergence theorem, just a straight forward surface integral.

I got up to this point and got stuck. I did try this before using divergence theorem and it didn't seem any easier, which way am I suppose to do this integral?

Thanks again!

 

[lanedance]

can't read it, though maybe speherical polar coords could be useful?

 

[jegues]

can't read it, though maybe speherical polar coords could be useful?

Here's a better size pic of the problem statement, sorry it's so small in the original post.

I'm still stuck on this one.

Is spherical the right route to take? Should I be applying Divergence theorem?

Thanks again!

 

[lineintegral1]

Try parameterizing the surface rather than computing the integral directly. The divergence theorem will work ONLY if the bottom of the hemisphere is closed. It doesn't look like that is what the problem is stating.

 

[jegues]

Try parameterizing the surface rather than computing the integral directly. The divergence theorem will work ONLY if the bottom of the hemisphere is closed. It doesn't look like that is what the problem is stating.

Yes but I could always add the surface z=0 to close and remove it after applying divergence theorem.

Anyways, what do you mean by parameterizing the surface?

 

[HallsofIvy]

Yes but I could always add the surface z=0 to close and remove it after applying divergence theorem.

Anyways, what do you mean by parameterizing the surface?

Any smooth surface, a two dimensional object, can be written with x, y, and z as functions of two parameters.
Here, your figure is the upper hemisphere of the sphere x^2+ y^2+ z^2= a^2
and the simplest way to parameterize it is to use spherical coordinates with the \rho[/b] coordinate fixed at a:<br /> x= a cos(\theta)sin(\phi)<br /> y= a sin(\theta)sin(\phi)<br /> z= a cos(\phi)<br /> <br /> To use that to integrate on the surface, write the &quot;position vector&quot; of any point on the sphere as <br /> \vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= a cos(\theta)sin(\phi)\vec{i}+ a sin(\theta)sin(\phi)\vec{j}+ a cos(\phi)\vec{k}<br /> <br /> Differentiating with respect to the two parameters gives two vectors in the tangent plane at each point:<br /> \vec{r}_\theta= -a sin(\theta)sin(\phi)\vec{i}+ a cos(\theta)sin(\phi)\vec{j}<br /> \vec{r}_\phi= a cos(\theta)cos(\phi)\vec{i}+ a sin(\theta)cos(\phi)\vec{j}- a sin(\phi)\vec{k}<br /> <br /> The cross product of those two vectors, called the &quot;fundamental vector product&quot; for the surface,<br /> a^2cos(\theta)sin^2(\phi)\vec{i}+ a^2sin(\theta)sin^2(\phi)\vec{j}+ a^2sin(\phi)cos(\phi)\vec{k}<br /> if multiplied by d\theta d\phi give d\vec{S}, the &quot;vector differential of surface area&quot; and its length a^2 sin(\phi), times d\theta d\phi is the &quot;differential of surface area&quot; in those parameters.