Converting this particular double integral to an iterated polar

Question

In class, the professor converted this double integral:

$$\int_{-4}^0 \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} f(x,y) \,dy\,dx$$

into the following iterated polar:

$$\int_{\pi/2}^{\pi} \int_{0}^{4} f(r,\theta) r\,dr\,d\theta$$

My question is this: why is the outer integral for the iterated polar from $\pi$/2 to $\pi$? Shouldn't it extend from $\pi$/2 to $\frac{3{\pi}}{2}$? In other words, shouldn't the iterated polar read:

$$\int_{\pi/2}^{\frac{3{\pi}}{2}} \int_{0}^{4} f(r,\theta) r\,dr\,d\theta$$

Any insight would be deeply appreciated. Thank you.

Edit 1: Reformatted iterated polar to represent proper values.

You're right about the limits. I hope he didn't actually write $f(R,\theta)R,drd\theta$! That should be $f(r,\theta)r,drd\theta$ or $f(R,\theta)R,dRd\theta$. — David C. Ullrich, Nov 24 '17 at 00:08

score 3 · Accepted Answer · answered Nov 24 '17 at 00:09

3

Yes, you are right, most likely It is just a careless mistake.

The region in the quesiton describe the left part of the circle (semicircle) with center at origin with radius $4$. The $x$ value of the region is negative.

Hence, the right $\theta$ value should be from $\frac{\pi}2$ to $\frac{3\pi}{2}$.

answered Nov 24 '17 at 00:09

Siong Thye Goh

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That's a relief. Thank you. – Alex Dymovsky Nov 24 '17 at 00:32

Converting this particular double integral to an iterated polar

1 Answers1