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∫01∫0(√1−y2)cos(x2+y2)dxdy evaluate the integral by first converting polar coordinates
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\[\int\limits_{0}^{1}\int\limits_{0}^{\sqrt(1-y^2)}\cos(x^2+y^2)dxdy\]
\(x = r\cos \theta\) \(y = r\sin \theta \) \(dxdy \to rdrd\theta\)
i found the answer (pi/4)*sin(1)
whats your double integral after switching to polar ?
\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{1}\cos(r^2)rdrd\]
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i cant open the link :( can you take a photo of it.
thanks
np :)
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