Spherical Coordinates: Integrating a Hemisphere/Paraboloid

Tater
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Homework Statement
The outermost integral is:
-2 to 2, dx

The middle integral is:
-sqrt(4-x^2) to sqrt(4-x^2), dy

The inner most integral is:
x^+y^2 to 4, dz






The attempt at a solution

Drawing the dydx in a simple 2d (xy) plane, it is circular with a radius of 2. So this means that the period(theta) will go from 0 to 2pi. Drawing in 3d (xyz) yields a hemisphere/paraboloid. Now this is where I'm stuck. I don't know what to do after this or how to really tackle this problem. Do I want to attempt to draw a 'slice' of it in the spherical outline with the variables phi, rho, theta? Do I have to look at it a certain way (2d or 3d)? I just don't see what I can do!

Any help or guidance is greatly appreciated!
 
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You haven't actually said what the question asks you to do..
 
Whoops. Thought I stuck that in there. Anyways, all I have to do is convert it to spherical coordinates (from rectangular to spherical).
 
Tater said:
Whoops. Thought I stuck that in there. Anyways, all I have to do is convert it to spherical coordinates (from rectangular to spherical).

I was afraid of that :wink:

It is indeed a paraboloid, so \rho and \phi will not be independent the way they would if it was a spherical section...

Try finding the relationship between \rho and \phi for the paraboloid's curved and flat surfaces
 
From my weblog
http://buyanik.wordpress.com/2009/05/02/laplacian-in-spherical-coordinates/"
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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