Use spherical coordinates to find the volume of the solid enclosed by the sphere x^2 + y^2 +z^2 =4a^2 and the plane z=0 and z=a.

I'm absolutely clueless about this question:confused:. Need Help again:p

thank you very much.

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- Sep 12th 2007, 08:19 AMkittycatspherical coordinates -triple integral
Use spherical coordinates to find the volume of the solid enclosed by the sphere x^2 + y^2 +z^2 =4a^2 and the plane z=0 and z=a.

I'm absolutely clueless about this question:confused:. Need Help again:p

thank you very much. - Sep 12th 2007, 08:54 AMThePerfectHacker
Assuming a>0 the equation $\displaystyle x^2+y^2+z^2 = 4a^2 = (2a)^2$ is a pshwre with radius $\displaystyle 2a$. We pass a plane $\displaystyle z=a$ through it meaning half the length of the radius, see below.

So know that $\displaystyle 0\leq r\leq 1$ and $\displaystyle 0\leq \theta\leq 2\pi$ now we need the bounds on $\displaystyle \phi$. That means we need to find the red angle.

This can be done with some geometry. I drew a red line to show the triangle. Notice we get a small right triangle inside. The hypotenuse is twice as long as one of its sides, this is a $\displaystyle 30-60-90$ triangle. So the red angle is $\displaystyle 60^o = \frac{\pi}{3}$. - Sep 12th 2007, 08:43 PMkittycat
Hi,

I still can't conclude this question.

May I ask why http://www.mathhelpforum.com/math-he...0f1f2041-1.gif ?

What is the rho limits in here?

My friend said this question need to decompose into two pieces Gsub1 and Gsub2 in order to find the volume of the solid. Is it true?

Please teach me. Thank you very much. - Sep 12th 2007, 09:00 PMJhevon
at the moment, i think your friend is correct, one way is to find two volumes and subtract one from the other. i'm trying to think of an easier way though, where we can find one volume.

let me describe the current method to you. We will define the volume $\displaystyle G_1$ as the upper hemisphere of the sphere we are considering. that is, the half of the sphere that is above the plane z = 0 (which is the xy-plane). obviously here, $\displaystyle 0 \le \rho \le 2a$, $\displaystyle 0 \le \phi \le \frac {\pi}{2}$ and $\displaystyle 0 \le \theta \le 2 \pi$

Now we define $\displaystyle G_2$ as the volume of top of the sphere that is bounded by the plane z = a. similar to one of your recent problems (see here). Now i leave finding the limits for this figure to you.

when you are done, the desired volume is given by $\displaystyle G_1 - G_2$

i will try to upload a figure so that you can conceptualize what i am talking about. hopefully i can draw one in MSpaint

EDIT: Here is the figure, it took forever to draw in paint. I have MatLab, but I don't remember how to plot 3d figures with it, so i have to do all this hard work. - Sep 13th 2007, 06:35 AMkittycat
Hi Jhevon,

thank you very much for your help and effort.

I have the limits of variables for Gsub2.

rho is from asectheta to 2a

phi is from o to pi/3

theta is from 0 to 2pi

Am I right ? I am asking that because I can't get the final answer, ( 11pi*a^3)/3, as given by my teacher. - Sep 13th 2007, 09:29 AMJhevon
i'm sorry. i didn't realize you changed your response. when you edit your post, it does not show up that a change has been made to your post, so i had no idea you gave me your information. if you see me taking a long while to respond after doing something like editing your post, you should send me a quick PM just to see if i am there.

your data for phi is incorrect

EDIT: O wait, my bad, phi is ok. so scratch what i said just above this line. I don't see why your not getting the answer. Did you remember that for volume you integrate over the function 1? that is, $\displaystyle f(\rho \cos \theta \sin \phi, \rho \sin \theta \sin \phi, \rho \cos \phi) = 1$

EDIT 2: I just did the problem (with your limits, of course, since they are correct) and got the required answer