Question

Evaluate the iterated integral $\int_{2}^{3} \int_{\frac{3\pi}{2}}^{2\pi} x \cos y \, dy \, dx$. $\int_{2}^{3} \int_{\frac{3\pi}{2}}^{2\pi} x \cos y \, dy \, dx = $ (Simplify your answer.)

          Evaluate the iterated integral $\int_{2}^{3} \int_{\frac{3\pi}{2}}^{2\pi} x \cos y \, dy \, dx$.
$\int_{2}^{3} \int_{\frac{3\pi}{2}}^{2\pi} x \cos y \, dy \, dx = $ (Simplify your answer.)

Evaluate the iterated integral ∫2^3∫(3π)/(2)^2π x cos y dy dx.
∫2^3∫(3π)/(2)^2π x cos y dy dx = (Simplify your answer.)

Added by Claudia G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Zack A

07/06/2023

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∫xcos(y) dy = x∫cos(y) dy = xsin(y) + C Show more…

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Evaluate the iterated integral int_2^3 int_((3pi )/(2))^(2pi ) xcosydydx int_2^3 int_((3pi )/(2))^(2pi ) xcosydydx= (Simplify your answer.) 32 Evaluate the iterated integral xcos y dy dx. 23 2 32 x cos y dy dx=(Simplify your answer.) 23 2

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Transcript

00:01 Okay, let's look at this thing.

00:03 2x minus x squared.

00:07 So y squared is equal to 2x minus x squared.

00:11 X squared plus y squared is equal to 2x.

00:15 So this is actually a circle with center shifted.

00:26 So it's like x minus 1 squared plus y squared is equal to 1.

00:31 So what we're looking at here is this thing right here.

00:37 1, 2, x, y.

00:40 And this is the region.

00:45 So polar coordinates.

00:48 Theta.

00:50 We're going to need theta.

00:52 Theta is 0 to pi over 2.

00:59 Why am i writing pi over 2? and 0 to, so let's look at here.

01:08 This is r squared is equal to 2r cosine theta.

01:14 Okay? so r is 2 cosine theta.

01:18 0 to 2 cosine theta.

01:21 And then 4 squared of x squared plus y squared is r.

01:25 And then another r.

01:28 So we have this thing.

01:29 And then 4 over 3r cubed.

01:32 0 to 2 cosine theta.

01:35 Which is 4 over 3 times 8 cosine cubed theta...

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