Thread: Tripple integration question spherical or cylindrical coordinates

Use cylindrical or spherical coordinates, whichever seem more appropriate,
to evaluate the triple integral where V lies above the paraboloid and below the plane

I am having trouble with the bounds and am not sure whether to use cylindrical or spherical coordinates.

I guess I can see it in cylindrical:

z is going...

from r^2 (the paraboloid) up to 6y i.e. 6r sin theta (the plane roof)

... for every bit that r is going...

from zero (the z axis) along to as far as where r^2 = 6r sin theta => r = 6 sin theta

... for every bit that theta is going from zero (the +ive x axis) round to pi (-ive). So,



int from 0 to pi int from 0 to 6sin theta int from r^2 to 6rsin theta rz dz dr dtheta - Wolfram|Alpha

The integral graph there (going 0 to pi on the horizontal) looks about right.

Those zero sins aren't going to work, but see also,

integrate 1944 sin^6 theta - Wolfram|Alpha

which gives 121.5 pi. If this isn't wrong I'll put a pic for the computation similar to

http://www.mathhelpforum.com/math-he...tml#post621761

when I'm home, could be a while.



Anyway, hope that helps.


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Yeah that looks good to me. Thanks! That last integral is nasty.