Volume of ice cream cone triple integral

Panphobia
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Homework Statement


Find the triple integral for the volume between a hemisphere centred at ##z=1## and cone with angle ##\alpha##.

The Attempt at a Solution



What I tried to do first was to get the radius of the hemisphere in terms of the angle ##\alpha##. In this case the radius is ##\tan \alpha##. I already figured out this integral in cylindrical polar coordinates and Cartesian coordinates. I am having a lot of trouble with spherical coordinates. I am trying to get ##\rho## for the hemisphere by drawing the projection of the shape on the xz plane and trying to get a formula for a radial ray that hits the hemisphere. The formula for a circle that is shifted up by one is ##x^2+(z-1)^2=\tan^2 \alpha##, this is the part where I am stuck trying to find the equation for ##\rho## in terms of the angle ##\phi## of the radial ray. The limits for ##\theta## and ##\phi## are really easy, but again ##\rho## I just can't seem to get.
 
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Do you need the integral in spherical coordinates? It is messy, and the other coordinate systems are much easier.

You are missing y2 in your hemisphere formula.
You can find z as function of the two angles, plug it into the equation for the hemisphere (in the right place) and solve for ρ.
 
mfb said:
Do you need the integral in spherical coordinates? It is messy, and the other coordinate systems are much easier.

You are missing y2 in your hemisphere formula.
You can find z as function of the two angles, plug it into the equation for the hemisphere (in the right place) and solve for ρ.
So I will have to use the quadratic formula to solve for ##\rho##?
 
I disagree with mfb on this problem. In fact spherical coordinates is the natural and easiest way to work this problem. You didn't state it in your problem but I assume you are talking about the upper half of the sphere of radius ##1## centered at ##(0,0,1)##. Its equation is ##x^2+y^2+(z-1)^2 = 1## or, expanding it, ##x^2 + y^2 + z^2 = 2z##. If you change that to spherical coordinates you should get ##\rho = 2\cos\phi##. So the ##\rho,~\theta,~\phi## limits are all very easy.
 
How could you fix the radius of the half-sphere to 1?
 
mfb said:
How could you fix the radius of the half-sphere to 1?
I couldn't. I scanned it too quickly and assumed it was the "standard" problem you normally see in calculus books.
 
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