Triple Integral: Volume of a Solid

iamalexalright
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Homework Statement


Been awhile since I've done them and my memory/reasoning isn't so great apparently...

Use the triple integral to find the volume of the given solid:
The solid enclosed by the cylinder
x^{2} + y^{2} = 9
and the planes y + z = 16 and z = 1. 2. The attempt at a solution
Difficulty is always setting up the bounds of the integral...
-3 \leq y \leq 3
1 \leq z \leq 16 - y
having problems with the xwould it be:
-\sqrt{9 - y^{2}} \leq x \leq \sqrt{9 - y^{2}} ?
 
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Sure. The planes don't intersect inside the cylinder. So you can parametrize the integral over the x,y in the circle defining the cylinder without worrying about the z value. If the planes had intersected inside the circle they would have had to give you a more elaborate description of the region.
 
135*pi, cool! Thanks Dick
 
iamalexalright said:
135*pi, cool! Thanks Dick

That's what I get. :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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