Volume using double integral (polar coordiantes)

whynot314
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Homework Statement


use a double integral to find the volume of the solid bounded by.

z=x^2+2y^2 and z=12-2x^2-y^2

I want to change variables using polar coordinates, I know its the top minus the bottom, and the intersection between the two is a circle radius 2.

The Attempt at a Solution



I want to make sure i have the correct set up

\int^{2\pi}_{0} \int^{2}_{0} (12-3r^{2})rdrd\theta
 
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whynot314 said:

Homework Statement


use a double integral to find the volume of the solid bounded by.

z=x^2+2y^2 and z=12-2x^2-y^2

I want to change variables using polar coordinates, I know its the top minus the bottom, and the intersection between the two is a circle radius 2.

The Attempt at a Solution



I want to make sure i have the correct set up

\int^{2\pi}_{0} \int^{2}_{0} (12-3r^{2})rdrd\theta
That looks good !
 
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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