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The question is as follows:

enter image description here

My work goes like this: ∫∫R sin(x^2 + y^2) dA

= ∫(θ from [0, 2π]) ∫(r from [1, 6]) sin(r^2) (r dr dθ)

= [∫(θ from [0, 2π]) dθ] * [∫(r from [1, 6]) r sin(r^2) dr]

= 2π * [(-1/2) cos(r^2) {for r = 1 to 6}]

= π * (cos 1 - cos 36).

can anyone spot any errors with my work? im submitting it online and its saying its an incorrect answer. Thanks

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It looks fine.:-) – Babak S. Nov 27 '13 at 18:00
Im thinking its gotta be some error with the answer. I will just email the teacher. Thanks. – Jeremy Rowler Nov 27 '13 at 18:04
$cos(36)-cos(1)$ – wfw00d Nov 27 '13 at 18:07
up vote 1 down vote accepted

Your approach is right. Indeed, $$\iint_R\sin(x^2+y^2)dA\Longrightarrow\int_0^{2\pi}d\theta\int\sin(r^2)rdr=\pi\times\cos\theta|_1^6\approx 2.09941$$ The red area is your area in the problem.

enter image description here

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Bull's Eye! Looks like target practice! ;-) +1 – amWhy Nov 28 '13 at 1:23

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