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Change the Cartesian integral int[0 to 9]int[0 to y]x dx dy into an equivalent polar integral.
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\[\int\limits_{0}^{9}\int\limits_{0}^{y}x \ dx dy\] I know the first part is \[\int\limits_{0}^{9/\sin }\] but I don't know the second part
Yeah, this is just sort of a silly integral to transform.
any thoughts? I have the first part: \[\int\limits_{?}^{?}\int\limits_{0}^{9/\sin (\theta)}r ^{2}\cos \theta \ dr d \theta \]
@mahmit2012
y=x isn't it?
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I mean for the original integral
yes 0<x<y
I would say you integrate from 45% until 90% so \( \large \frac{\pi}{4}< \theta< \frac{\pi}{2} \)
|dw:1342794156759:dw|
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