Gauss' Theorem

  • #1

Homework Statement


Let F = <x, z, xz> evaluate ∫∫F⋅dS for the following region:

x2+y2≤z≤1 and x≥0

Homework Equations



Gauss Theorem

∫∫∫(∇⋅F)dV = ∫∫F⋅dS

The Attempt at a Solution



This is the graph of the entire function:

upload_2014-12-17_21-31-15.png

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:

upload_2014-12-17_21-36-46.png


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.
 

Attachments

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,556
767

Homework Statement


Let F = <x, z, xz> evaluate ∫∫F⋅dS for the following region:

x2+y2≤z≤1 and x≥0

Homework Equations



Gauss Theorem

∫∫∫(∇⋅F)dV = ∫∫F⋅dS

The Attempt at a Solution



But only the right half of the circle (from -3π/2 to π/2)...
That is not the correct range for the ##x## positive half of the circle. Do you really mean ##-\frac{3\pi}{2}##?

The integral I set up is the following:

∫∫∫xdzdxdy (x is the dot product of ∇ and F)
Check ##\nabla \cdot \vec F## again.
 
  • Like
Likes lumpyduster
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,514
6,423
This is the graph of the entire function
Well, no, it shows the region (at least, it would if truncated at z = 1 instead of z = 4).
x is the dot product of ∇ and F
Are you sure? I get another term.
-3π/2≤θ≤π/2
Think about that lower bound again. Where is -3π/2 in the plane?

Edit: beaten to the post by LCK (again).
 
  • Like
Likes lumpyduster
  • #4
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.
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).
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,514
6,423
Not sure at all about the lower bound... Should it be 3π/2?
That should work.
 
  • #6
That should work.
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,514
6,423
So I will integrate from π/2 to 3π/2?
No, you said 3π/2 for the lower bound, not the upper bound.
 
  • #8
No, you said 3π/2 for the lower bound, not the upper bound.
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,514
6,423
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 :)
You can take the lower bound to be -π/2 if you prefer. It should give the same result.
 
  • #10
You can take the lower bound to be -π/2 if you prefer. It should give the same result.
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?

Thank you for your help!
 
  • #11
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,514
6,423
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?
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.
 
  • Like
Likes lumpyduster
  • #12
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,556
767
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.
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,514
6,423
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.
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.
 

Related Threads on Gauss' Theorem

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
9
Views
7K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
4K
Replies
7
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
0
Views
2K
Top