Triple Integral Evaluation: Where Did I Go Wrong?

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Homework Statement



http://img19.imageshack.us/img19/2559/triple.th.jpg

Homework Equations





The Attempt at a Solution



I get 16pi/3 (sqrt(2) -1) as the final result, but when I input the answer to the computer, it doesn't accept it. Am I doing a wrong integration/calculation anywhere? I double checked and I can't to find anything wrong.
 
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It seems that when you multiplied by 2pi, you said 16/3* 2pi=16pi/3...that's not really true is it?:wink:
 
oh so it should be 32pi/3 then.. thanks for pointing out the silly mistake
 
You went from
-\frac{1}{3}\left(4^{3}{2}-8^{3}{2})-\frac{8}{3}
to
\frac{16}{3}(\sqrt{2}-1)
instead of
\frac{8}{3}(2\sqrt{2}-1)
 
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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