# Homework Help: Gauss' Theorem

1. Dec 17, 2014

### lumpyduster

1. The problem statement, all variables and given/known data
Let F = <x, z, xz> evaluate ∫∫F⋅dS for the following region:

x2+y2≤z≤1 and x≥0

2. Relevant equations

Gauss Theorem

∫∫∫(∇⋅F)dV = ∫∫F⋅dS

3. The attempt at a solution

This is the graph of the entire function:

Thank you Wolfram Alpha.

But my surface is just the half of this paraboloid where x is positive. So I thought if I looked down the x-axis I would get something like this:

But only the right half of the circle (from -3π/2 to π/2)...

The integral I set up is the following:

∫∫∫xdzdxdy (x is the dot product of ∇ and F)

I converted to polar coordinates

∫∫∫r2cosθdzdrdθ

Bounds:
r2≤z≤1

0≤r≤1

-3π/2≤θ≤π/2

I ended up getting

-(4/15)∫cosθdθ
-3π/2≤θ≤π/2

(4/15)[sin(π/2)-sin(-3π/2)] = 0

Answer should be 4/15 according to the back of the book.

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2. Dec 17, 2014

### LCKurtz

That is not the correct range for the $x$ positive half of the circle. Do you really mean $-\frac{3\pi}{2}$?

Check $\nabla \cdot \vec F$ again.

3. Dec 17, 2014

### haruspex

Well, no, it shows the region (at least, it would if truncated at z = 1 instead of z = 4).
Are you sure? I get another term.
Think about that lower bound again. Where is -3π/2 in the plane?

Edit: beaten to the post by LCK (again).

4. Dec 17, 2014

### lumpyduster

Not sure at all about the lower bound... Should it be 3π/2? I was going to also say 0 to π, but if my upper bound is correct then that would be wrong.

5. Dec 17, 2014

### haruspex

That should work.

6. Dec 17, 2014

### lumpyduster

So I will integrate from π/2 to 3π/2? I guess I'm confused how to take into account that x≥0 when switching to polar coordinates. I just know I want to go from x=0 to x=1 on the Cartesian coordinate system. Would 0 to π also work?

7. Dec 17, 2014

### haruspex

No, you said 3π/2 for the lower bound, not the upper bound.

8. Dec 17, 2014

### lumpyduster

Yeah I probably sound like a total noob here, but I've never done an integral where my lower bound was greater than my upper bound... Will try and report back if necessary :)

9. Dec 17, 2014

### haruspex

You can take the lower bound to be -π/2 if you prefer. It should give the same result.

10. Dec 18, 2014

### lumpyduster

I just don't know how to get these bounds though.... Can I say that since 0≤x≤1 and since x=rcosθ and r=1, that arccos(0)≤θ≤arccos1?

11. Dec 18, 2014

### haruspex

Yes, but that doesn't pin down the bounds on theta. For example, π/2 to 0 would seem to be right if you base it only that. I don't think you can do better than consider what the physical region looks like and deduce the range for theta from that. Just bear in mind that running theta continuously from the lower bound to the upper bound (together with the ranges on r and z) must reach each point of the region exactly once, and no points outside the region.

12. Dec 18, 2014

### LCKurtz

Sorry to be so late getting back to this thread, but I think something is getting missing in translation. Usually in describing limits on an integral when you say from $a$ to $b$ you would assume $a$ is the lower limit and $b$ is the upper limit, and $a < b$. You integrate in the positive direction. So you would say from $0$ to $\pi / 2$ to describe the first quadrant, not $\pi /2$ to $0$. If you want to describe the right side of the circle you could go from $-\pi /2$ to $\pi /2$ or from $3\pi /2$ to $5\pi /2$ or something similar.

13. Dec 18, 2014

### haruspex

I think you misunderstand the point I was making. I was saying that in order to find the limits on theta for integration it is not sufficient to consider the range of values that cos(theta) can take. You might miss part of the range (since over the range of theta some cos(theta) values are repeated), or include regions which are not in the range (like going from 0 to 5pi/2), or, as you note, get the upper and lower limits reversed.