Surface Integrals help

nathanfsu · #1 $Old$ November 29th, 2008, 08:45 PM

Hi, I'm having a little trouble to solve these problems. May anyone help me please?

What is the area of the spherical surface X^2+Y^2+Z^2=9 inside the cylinder
X^2+Y^2=3x ?

What is the area of the spherical surface X^2+Y^2+Z^2=9 inside the cylindrical X^2+Y^2=4 ?

What is the area of the spherical surface X^2+Y^2+Z^2=16 inside the cylindrical Y^2+Z^2=4Z ?

Thanks, and sorry for my English because it is not my native language

nathanfsu · #2 $Old$ November 30th, 2008, 05:39 AM

anyone?

shawsend · #3 $Old$ November 30th, 2008, 05:46 AM

When you draw it, the approach becomes immediately clear right? Just integrate under that section of intersection:

$S=2\mathop\int\int\limits_{\hspace{-15pt}D} \sqrt{(f_x)^2+(f_y)^2+1}\;dA$

where:

$f(x,y)=\sqrt{9-y^2-x^2}$

$D=\left\{(x,y): (x-3/2)^2+y^2=(3/2)^2\right\}$

Just complete the square in $x^2+y^2=3x$ to get D.

I haven't actually tried to do the integration though. May need to use spherical or other means to evaluate it. Also, it's easy to draw. Here's the Mathematica code:

Code:

pic1 = ContourPlot3D[{x^2 + y^2 + z^2 == 9, 
    x^2 + y^2 == 3*x}, {x, -3, 3}, 
   {y, -3, 3}, {z, -3, 3}, ContourStyle -> 
    {Opacity[0.4], LightPurple}]

nathanfsu · #4 $Old$ November 30th, 2008, 06:15 AM

hm my problem is that I can't solve the integration, I tried polar and spherical coordinates, but I think I'm not using the right integration limit

Any Help

shawsend · #5 $Old$ November 30th, 2008, 06:36 AM

So what's the integral you get? We can just integrate over the first quadrant right since it's symmetric and just multiply by 2 and that's just the top so multiply by 2 again but for now, just the top over the first quadrant I get:

$S_{uq1}=3\int_0^3\int_{0}^{\sqrt{3x-x^2}}\frac{1}{\sqrt{9-(x^2+y^2)}}dydx$

where we'll set $S_{uq1}$ be the surface area over the top part just over the first quadrant.

The x^2+y^2 part looks amendable to polar coordinates. Mathematica reports for that integral $S_{uq1}=9/2(\pi-2)$ . Don't see immediately how to do the integration though.

nathanfsu · #6 $Old$ November 30th, 2008, 06:48 AM

Yeah, I got the same integration. The problem is that after I tried to put in polar coordinates, the integration that came is very difficult to solve without cas
X=1,5 + R.Cos(A)
Y=R.Sen(A)
0<=R<=1.5
0<=A<=pi

it would be:

2*intg*intg R/(Sqrt(-Rcos(A)-R^2+6.75))drda
I stop here; don't know how to solve, or if this is right

shawsend · #7 $Old$ November 30th, 2008, 09:00 AM

Hey, this ain't no hill:

$S_{uq1}=3\int_0^3\int_{0}^{\sqrt{3x-x^2}}\frac{1}{\sqrt{9-(x^2+y^2)}}dydx$

You can do the first part right:

$\int_0^{\sqrt{3x-x^2}} \frac{dy}{\sqrt{(9-x^2)-y^2}}$

That's just an inverse sine but I ran into problems with that so I converted it to an inverse tan. When I simplify it and apply the limits, I get for the outer integral::

$\int_0^3 \arctan\left(\sqrt{\frac{3}{x}}\right) dx$

Now, I think we can integrate that by parts. You'll need to check all this cus' I went through it quick but I think the principle is sound.

nathanfsu · #8 $Old$ November 30th, 2008, 10:47 AM

Thanks man

chiph588@ · #9 $Old$ November 30th, 2008, 08:32 PM

Shouldn't the outer integral be $\int_0^3 \arctan\left(\sqrt{\frac{x}{3}}\right) dx$ ?

shawsend · #10 $Old$ December 1st, 2008, 04:07 AM

Quote:

Originally Posted by chiph588@ $View Post$

Shouldn't the outer integral be $\int_0^3 \arctan\left(\sqrt{\frac{x}{3}}\right) dx$ ?

I'm not sure; I didn't go through it rigorously . . . Perhaps Nathan could confirm this.

nathanfsu · #11 $Old$ December 1st, 2008, 07:38 AM

I did on my HP that integral, and the result was
-ATAN(SQRT(3X)/3)

How do you do to post math symbols ?