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# Multiple integrals and change of variables watch

1. A cylindrical hole is drilled through the centre of a ball of radius a > 0 so that the remaining object has a hole of length 2b, 0<b<a. What is the volume of this object?
The equation of the ball is x^2+y^2+z^2 <= a^2, I'm stuck on this, can anyone please help me with this question. Thanks.
2. Well the volume of a cylinder is pi r^2 h, so the volume is 2b pi a^2 , but i can imagine the ends would be curved so im probably wrong
3. (Original post by cj_134)
Well the volume of a cylinder is pi r^2 h, so the volume is 2b pi a^2 , but i can imagine the ends would be curved so im probably wrong
Thanks for the reply, but I'm still not sure how to use the change of variables on this. I have to show that the volune of the piece drilled out is the 2x the double integral of (a^2-x^2-y^2)^0.5 dxdy.
4. I think you need to post the entire question. (The normal way of finding the volume does NOT require multiple integrals, and there's no way I would have guessed that you wanted to show the piece drilled out had a particular volume until your previous post).
5. Ok the final parts for the question are "where R is the planar region x^2+y^2 <= a^2-b^2, and use polar coordinates to evaluate this integral.", (R is on the limits of the double integral) that's everything i have for this question now.

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