1. ## Surface Integrals help

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

2. anyone?

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

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

where:

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

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

Just complete the square in $\displaystyle x^2+y^2=3x$ 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}]

4. 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

5. 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:

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

where we'll set $\displaystyle S_{uq1}$ 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 $\displaystyle S_{uq1}=9/2(\pi-2)$. Don't see immediately how to do the integration though.

6. 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

7. Hey, this ain't no hill:

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

You can do the first part right:

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

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::

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

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.

8. Thanks man

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

10. Originally Posted by chiph588@
Shouldn't the outer integral be $\displaystyle \int_0^3 \arctan\left(\sqrt{\frac{x}{3}}\right) dx$ ?
I'm not sure; I didn't go through it rigorously . . . Perhaps Nathan could confirm this.

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

How do you do to post math symbols ?