Triple Integral of z over [0, 2*pi] for r [1, 2]

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SUMMARY

The discussion focuses on calculating the triple integral of z over the specified limits where z ranges from (r-1) to sqrt(1-(r-2)^2), r from 1 to 2, and theta from 0 to 2*pi. The user consistently arrives at an incorrect result of (17/4)*pi, while the correct answer is pi/2. The issue lies in the order of integration, specifically the antiderivation sequence of dz, dr, and d(theta), which affects the final outcome of the integral.

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[SOLVED] Triple integral

Homework Statement



Calculate the triple integral of z when z [(r-1), sqrt(1-(r-2)^2)], r [1, 2], tetha [0, 2*pi]

2. The attempt at a solution

I've tried again and again, and I always get (17/4)*pi, while the answer is pi/2. Is there anything wrong with antiderivating in this order: dz, dr, d(tetha)?
 
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