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WebWork 16, Problem 2

nfitzen

I’m having trouble with WebWork 16, Problem 2.

My problem version is as follows:

Evaluate the double integral \iint_{R}(3x-y)\mathrm{d}A, where R is the region in the first quadrant enclosed by the circle x^2+y^2=4 and the lines x=0 and y=x, by changing to polar coordinates.

Because we’re constrained to the first quadrant (r\geq0, \theta\in[0,\frac{\pi}{2}]), we obtain the following in service of our integration bounds:

x=0\implies\theta=\frac{\pi}2, \\ y=x\implies\theta=\frac{\pi}4.

Given our disk of radius 2, this means that our integration variables are bounded as follows:

r\in [0,2], \\ \theta\in \left [\frac\pi 4,\frac\pi 2 \right ].

Ergo, our double integral can be expressed as the following iterated integral, by substituting in x=\cos\theta,y=\sin\theta:

\int_{\frac\pi 4}^{\frac\pi 2} \int_0^2 (3\cos\theta-\sin\theta)r\mathrm{d}r\mathrm{d}\theta \\ = \int_{\frac\pi 4}^{\frac\pi 2}(3\cos\theta-\sin\theta) \left[\frac{r^2}{2}\right ]_0^2 \mathrm{d}\theta \\ = 2\int_{\frac\pi 4}^{\frac\pi 2} 3\cos\theta-\sin\theta\; \mathrm{d}\theta \\ = 2[3\sin\theta + \cos\theta]_{\pi/4}^{\pi/2} \\ = 2(3-\frac{3\sqrt 2}{2} - \frac{\sqrt 2}{2}) \\ = \boxed{6-4\sqrt 2}\approx 0.343146.

This answer was marked as incorrect in the WebWork. Is there something I’m doing wrong?

Plugging my iterated integral into Desmos also gave me that answer (as a decimal approx.). Am I getting the integral bounds wrong, or is the WebWork not accepting the correct answer?

nfitzen

Well, I figured out my issue. x=r\cos\theta, not \cos\theta, and similarly with y.

Colby_Howell

This helps! The bounds for \theta are \frac{\pi}{4} to \frac{\pi}{2}, I had that wrong. Thank you!!