Calculating Surface Integral with Stoke's Theorem | -2pi/5 Answer Explained

In summary, the conversation discusses the evaluation of a surface integral involving a function g(x,y,z) and a surface element dσ, where the surface S is defined by a sphere with radius 1. The conversation includes discussions on the necessary steps and formulas needed to solve the integral, as well as different opinions and approaches from various participants. Ultimately, the conversation concludes with two different answers for the integral, with one participant requesting a third opinion for closure.
  • #1
defang
5
0
Here is the question:

Evaluate the surface integral ∫∫s (X^4 + Y^4 + Z^4) dσ, where dσ is the surface element and S = { (X,Y,Z) : X^2 + y^2 + Z^2 = 1}

I know you have to take the square root of 1 + (dz/dx)^2 + (dz/dy)^2 dxdy. And I got -2X/2Z and -2Y/2Z, respectively. Then, I must incorporate this into Stoke's. Can anyone tell me what to do next? The answer to the quesion -2pi/5. THANKS!
 
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  • #2
How do you get a negative answer to an integral where the integrand is positive?
 
  • #3
Hello Defang. Welcome to PF. This is what I have:

[tex]g(x,y,z)=x^4+y^4+z^4[/tex]

[tex]S=\{(x,y,z):x^2+y^2+z^2=1\}[/tex]

So, we're going to integrate g over a sphere at the origin with radius 1. Since everything is spherically symmetrical, let's just cut it in half and just do the top part:

[tex]I=2\int\int_K (x^4+y^4+z^4)ds[/tex]

where K is the surface above the x-y plane described by the function:

[tex]K=z(x,y)=\sqrt{1-x^2-y^2}[/tex]

Now, the projection of this surface onto the x-y plane is just the circle:

[tex]R=x^2+y^2+1[/tex]

It is over this area that we'll integrate using the standard formula:

[tex]I=2\int\int_R g(x,y,z(x,y)) \sqrt{(z_x)^2+(z_y)^2+1} dA[/tex]
 
  • #4
Defang, I now suspect the function:

[tex]g(x,y,z)=x^4+y^4+z^4[/tex]

is not spherically symmetrical. Can someone confirm this? If so then my analysis above is not correct. Sorry.
 
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  • #5
You know, sometimes its good just to keep your mouth shut. Unfortunately for me that doesn't work in math: better to get in trouble if that's what it takes to get the problem straight in my head. I now believe the top hemisphere of the function is symmetrical with the bottom part in which case the analysis above WOULD work.

Can someone please provide another opinion about this? I'd like to see the problem completed not to mention attempt to solve it numerically and compare the results.
 
  • #6
saltydog said:
You know, sometimes its good just to keep your mouth shut. Unfortunately for me that doesn't work in math: better to get in trouble if that's what it takes to get the problem straight in my head. I now believe the top hemisphere of the function is symmetrical with the bottom part in which case the analysis above WOULD work.

Can someone please provide another opinion about this? I'd like to see the problem completed not to mention attempt to solve it numerically and compare the results.

Alright, go ahead, say it:

[tex]g(x,y,a)=g(x,y,-a)[/tex]

I'm outta' here . . .got some french toast to make . . .
 
  • #7
Thanks saltydog. Then, I have 0 ∫ 2pi 0 ∫ 1 (r^4 + √ 1-r^2) (√1 + 4r^2) r dr dθ. Is this correct? If it is correct, any tips on how to solve this integration?
 
  • #8
"[ tex ] \int_0^{2 \pi} \int_0^1 \left( r^4 + \sqrt{1 - r^2} \right) \left( \sqrt{1 + 4r^2} \right) r \; dr \; d \theta [ /tex ]"

[tex] \int_0^{2 \pi} \int_0^1 \left( r^4 + \sqrt{1 - r^2} \right) \left( \sqrt{1+4r^2} \right) r \; dr \; d \theta [/tex]

edit to add link: https://www.physicsforums.com/showthread.php?t=8997&page=1
 
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  • #9
[tex]r^4 = \left( x^2+y^2+z^2 \right) ^2 \neq x^4+y^4+z^4[/tex]
 
  • #10
Here's my attempt, which is different from saltydog's:

[tex] \left( x^2 + y^2 + z^2 \right) ^2 = x^4 + y^4 + z^4 + x^2 \left( y^2 + z^2 \right) + y^2 \left( x^2 + z^2 \right) + z^2 \left( x^2 + y^2 \right)[/tex]

Spherical substitution:
[tex]\begin{align*} &x=R \sin \theta \cos \phi \\
&y = R \sin \theta \sin \phi \\
&z = R \cos \theta \end{align}[/tex]

Which gives us:
[tex] \begin{align*} x^4 + y^4 + z^4 &= \left( x^2 + y^2 + z^2 \right) ^2 - x^2(y^2+z^2) - y^2(x^2+z^2) - z^2(x^2+y^2) \\
&= R^4 - 2x^2y^2 - 2x^2z^2 - 2y^2z^2 \\
&= R^4 \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \cos^2 \phi - 2 \sin^2 \theta \cos^2 \theta \sin^2 \phi \right) \\
&= R^4\left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \left( \cos^2 \phi + \sin^2 \phi \right) \right) \\
&= R^4\left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) \end{align}[/tex]

So [tex]g(R, \theta, \phi ) = R^4 \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) [/tex]

And the integral we want is
[tex] \int_0^{2 \pi} \int_0^{\pi} g(1, \theta, \phi) dS = \int_0^{2 \pi} \int_0^{\pi} \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) \sin \theta \; d \theta \; d \phi [/tex]
 
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  • #11
Just to be sure, you did mean

[tex]\iint_S x^4 + y^4 + z^4 \; dS [/tex]

and not

[tex]\iint_S x^4 \hat{i} + y^4 \hat{j} + z^4 \hat{k} \; dS [/tex]

right?
 
  • #12
Guys . . . I got:

[tex] \frac{12\pi}{5}[/tex]

I mean, if I'm wrong I want to be right Ok. What did you get?
 
  • #13
My calculator tells me its

[tex]\frac{13 \pi^2}{16}[/tex]

but it's old and the batteries are low, so it could be wrong.
 
  • #14
Oops! I entered it wrong! The answer is more likely to be

[tex]\frac{17 \pi}{5} [/tex].
 
  • #15
I don't know Rachmaninoff. Suppose I'll go over my work. How about you Defang? I just converted the integral to polar coordinates and plugged it into Mathematica and multiplied by 2.
 
  • #16
okay... what's your final integral look like? mine was derived above.

-rachmaninoff
 
  • #17
rachmaninoff said:
okay... what's your final integral look like? mine was derived above.

-rachmaninoff

Here it is:

[tex]\int_0^{2\pi}\int_0^1(\frac{r^5Cos^4(\theta)}{\sqrt{1-r^2}}+\frac{r^5Sin^4(\theta)}{\sqrt{1-r^2}}+\frac{r(1-r^2)^2}{\sqrt{1-r^2}})drd\theta[/tex]

That's just the top part. Multiply by 2 to get the answer I got (via Mathematica)
 
  • #18
Well, I hate to be a pain but I request some closure in this matter. Now:

Rachmaninoff said:

[tex]\frac{17\pi}{5}[/tex]

I said:

[tex]\frac{12\pi}{5}[/tex]

Now, I'm pretty sure mine is correct. So sure, I'd bet a dollar. But Rachmaninoff feels otherwise and I've followed his other postings; he's very good so I'm worried. Can someone supply a third opinion?

Thanks,
Salty
 
  • #19
Indeed, after what looked like endless calculation, I have finally got an answer: 12pi/5 (still learning this LaTex program). 2 0 ∫ 2pi 0∫1 [ r^4 (3/4 + 1/4 cos4θ ) + (1-r^2)^2 ] 1/(√1-r^2) r dr dθ. saltydog & rachmaninoff, thanks for your help!
 

1. How do you calculate a surface integral using Stoke's Theorem?

To calculate a surface integral using Stoke's Theorem, you first need to find the curl of the vector field. Then, use the given surface and its boundaries to set up the double integral. Finally, evaluate the integral to find the surface integral value.

2. What is the significance of the -2pi/5 answer in the calculation?

The -2pi/5 answer is the value of the surface integral, which represents the flux of the vector field through the given surface. It indicates the amount of flow of the vector field through the surface.

3. Can Stoke's Theorem be applied to any surface?

Yes, Stoke's Theorem can be applied to any smooth surface with a closed boundary. The surface must also be oriented in a consistent direction.

4. How does Stoke's Theorem relate to Green's Theorem?

Stoke's Theorem is the three-dimensional equivalent of Green's Theorem, which is used to calculate line integrals in two dimensions. Both theorems relate a surface or line integral to a double or single integral, respectively.

5. Are there any limitations to using Stoke's Theorem?

One limitation of Stoke's Theorem is that it can only be applied to surfaces that are smooth and have a closed boundary. Additionally, the vector field must also be continuous and differentiable on the surface.

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