Triple integral (spherical coordinate)

  • Thread starter naspek
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  • #1
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find the volume of the solid D that lies above the cone z = (x^2 + y^2)^1/2
and below the sphere z = (x^2 + y^2 + z^2)

i've done the integration until i need to substitute cos phi = u..
however.. i dont know to change the range..

http://imageshack.us/photo/my-images/839/spherical.jpg/"
 
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  • #2
LCKurtz
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[Edit] Never mind the below paragraph. Your work is correct; I misread the equation at first. All you have to do is put in the correct u limit in your last equation. Since u = cos(φ) and φ goes from 0 to pi/4, your upper u limit is 1/sqrt(2). Be careful with your lower limit for u, it isn't 0. Just finish your integration.


I don't think your integral is set up correctly. I see a sphere centered at (0,0,1/2) with radius 1/2 intersecting the cone at z = 1/2. So you apparently want the volume under the top half of the sphere but above the cone. Are you working this in two parts, the hemisphere and the cone with a flat base? If so, for the cone, ρ goes from 0 to z=1/2 which is ρcos(φ)=1/2 or ρ = (1/2)sec(φ). And you have to add the hemisphere.
 
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