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Find double integral of 6xy+6ydA with respect to the bounds x^2+y^2=16,y=0,y=3x.
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My work: \[\int\limits_{0}^{4}\int\limits_{0}^{3x}6xy+6y dy dx\] = 2034 Unfortunately, it's incorrect
=**2304
did you graph the region over which we are integrating
lets do that first
|dw:1426981390888:dw|
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one way you can do it
|dw:1426981797145:dw|
Polar? I thought the calculation wouldn't be fun. Because the function is 6xy + 6y
i think you can do it rectangularly
you have to find that intersection point `c`, where does x^2+y^2 = 16 intersect y = 3x, that x value
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(2/5)*sqrt(10)
Oh! That makes a lot of sense! I see my problem. Thanks so much!! Would rectangularly be the best way to solve it by the way?
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