Attempt 1 this item has been answered 9 attempts remaining...

Attempt 1: This item has been answered. 9 attempts remaining. Match each double integral in polar with the graph of the region of integration. 1. $\int_0^1 \int_{-\pi/2}^{3\pi/4} f(r, \theta) r d\theta dr$ 2. $\int_0^{2\pi} \int_1^2 f(r, \theta) r dr d\theta$ 3. $\int_{3\pi/2}^{2\pi} \int_0^2 f(r, \theta) r dr d\theta$ 4. $\int_0^{2\pi} \int_0^1 f(r, \theta) r dr d\theta$ 5. $\int_1^2 \int_{3\pi/4}^{7\pi/4} f(r, \theta) r d\theta dr$ 6. $\int_{-\pi/4}^{3\pi/4} \int_0^2 f(r, \theta) r dr d\theta$ 7. $\int_0^1 \int_0^{3\pi/4} f(r, \theta) r d\theta dr$ 8. $\int_{3\pi/4}^{3\pi/2} \int_0^1 f(r, \theta) r dr d\theta$

Transcript

00:01 Let's go through these one by one.

00:01 So we've got the integral of f of r, sorry, double integral of f of r theta, r dr d theta, from three to four on r, and from zero to two pi.

00:18 Anything ever says zero to two pi, that means on a polar plane, it's going to be a full rotation, a full rotation.

00:27 So we need to look for ones with like a disc circular shape.

00:32 So using that sort of knowledge here, it's going to be e or f.

00:37 Because our r here is we're taking the integral from 3 to 4.

00:44 So this is 3 and this is 4.

00:47 We're taking the integral that goes around the 3 and then all the way around 4 as well.

00:54 So the only option there would leave us with f.

00:56 So this is what you'd call, we'd call this a circular annulus with inner radius 3 and outer radius 4 centred at the origin, covering the range of angles from 0 to 2.

01:10 So our answer here is going to be f.

01:15 Okay, doing the same for 2.

01:17 This time we have the integral, again, full rotation, but instead it's not an annulus, it's just a disc.

01:27 So if it's a disk, we're just covering 0 to 3, going all the way around.

01:36 And the only one that would correspond to would have to be e.

01:44 Okay, 3.

01:46 We've got the integral from 3 pi over 4, 3 pi over 2, and from 0 to 3.

02:00 F of r theta r dr theta.

02:07 This integral represents a sector of a disk.

02:10 So you can see that 0 to 3 pi is the same as the previous one.

02:15 It's sort of chopped off.

02:17 It's going to be centred at the origin, radius 3, but it's only going to cover these angles here.