# HeLp: Surface Integral

## Main Question or Discussion Point

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!

mathman
How do you get a negative answer to an integral where the integrand is positive?

saltydog
Homework Helper
Hello Defang. Welcome to PF. This is what I have:

$$g(x,y,z)=x^4+y^4+z^4$$

$$S=\{(x,y,z):x^2+y^2+z^2=1\}$$

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:

$$I=2\int\int_K (x^4+y^4+z^4)ds$$

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

$$K=z(x,y)=\sqrt{1-x^2-y^2}$$

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

$$R=x^2+y^2+1$$

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

$$I=2\int\int_R g(x,y,z(x,y)) \sqrt{(z_x)^2+(z_y)^2+1} dA$$

saltydog
Homework Helper
Defang, I now suspect the function:

$$g(x,y,z)=x^4+y^4+z^4$$

is not spherically symmetrical. Can someone confirm this? If so then my analysis above is not correct. Sorry.

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saltydog
Homework Helper
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.

saltydog
Homework Helper
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.

$$g(x,y,a)=g(x,y,-a)$$

I'm outta' here . . .got some french toast to make . . .

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?

rachmaninoff
"[ 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 ]"

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

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rachmaninoff
$$r^4 = \left( x^2+y^2+z^2 \right) ^2 \neq x^4+y^4+z^4$$

rachmaninoff
Here's my attempt, which is different from saltydog's:

$$\left( x^2 + y^2 + z^2 \right) ^2 = x^4 + y^4 + z^4 + x^2 \left( y^2 + z^2 \right) + y^2 \left( x^2 + z^2 \right) + z^2 \left( x^2 + y^2 \right)$$

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

Which gives us:
\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}

So $$g(R, \theta, \phi ) = R^4 \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right)$$

And the integral we want is
$$\int_0^{2 \pi} \int_0^{\pi} g(1, \theta, \phi) dS = \int_0^{2 \pi} \int_0^{\pi} \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) \sin \theta \; d \theta \; d \phi$$

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rachmaninoff
Just to be sure, you did mean

$$\iint_S x^4 + y^4 + z^4 \; dS$$

and not

$$\iint_S x^4 \hat{i} + y^4 \hat{j} + z^4 \hat{k} \; dS$$

right?

saltydog
Homework Helper
Guys . . . I got:

$$\frac{12\pi}{5}$$

I mean, if I'm wrong I want to be right Ok. What did you get?

rachmaninoff
My calculator tells me its

$$\frac{13 \pi^2}{16}$$

but it's old and the batteries are low, so it could be wrong.

rachmaninoff
Oops! I entered it wrong! The answer is more likely to be

$$\frac{17 \pi}{5}$$.

saltydog
Homework Helper
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.

rachmaninoff
okay... what's your final integral look like? mine was derived above.

-rachmaninoff

saltydog
Homework Helper
rachmaninoff said:
okay... what's your final integral look like? mine was derived above.

-rachmaninoff
Here it is:

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

That's just the top part. Multiply by 2 to get the answer I got (via Mathematica)

saltydog
Homework Helper
Well, I hate to be a pain but I request some closure in this matter. Now:

Rachmaninoff said:

$$\frac{17\pi}{5}$$

I said:

$$\frac{12\pi}{5}$$

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

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!