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Evaluate the double integral and sketch the region R.
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\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{1\sin \Theta}\Theta r dr d \theta\]
^ This involves change of variables with polar coordinates
The limits for theta are 0 to pi/2. The limits for r depend on theta. They always start at 0 and go to sin(theta)
I integrated the problem correctly with respect to dr, but dtheta is where I'm stuck at
sin(0) is 0. sin(pi/2) is 1, so as theta increases, the upper limit for r goes from 0 to 1.
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This is how far I got: \[\int\limits_{0}^{\pi/2} \theta r^2/2 d \theta\]
|dw:1334280701565:dw|
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