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James Stewart

8 Edition

Chapter 15, Problem 30

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Evaluate the iterated integral by converting to polar coordinates.

$ \displaystyle \int_0^a \int_{-\sqrt{a^2 – y^2}}^{\sqrt{a^2 – y^2}} (2x + y)\ dx dy $

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Official textbook answer

Video by Alan Ghazarians

Numerade Educator

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03:13

Evaluate the iterated integral by converting to polar coordinates.

$ \displaystyle \int_0^2 \int_0^{\sqrt{2x – x^2}} \sqrt{x^2 + y^2}\ dy dx $

03:15

Evaluate the iterated integral by converting to polar coordinates.

$\int_{0}^{a} \int_{-\sqrt{a^{2}-y^{2}}}^{\sqrt{a^{2-y^{2}}}}(2 x+y) d x d y$

02:56

Evaluate the iterated integral by converting to polar coordinates.

$ \displaystyle \int_0^2 \int_0^{\sqrt{4 – x^2}} e^{-x^2 – y^2}\ dy dx $

11:38

Evaluate the iterated integral by converting to polar coordinates.

$\int_{0}^{a} \int_{-\sqrt{a^{2}-y^{2}}}^{0} x^{2} y d x d y$

09:37

Evaluate the iterated integral by converting to polar coordinates.

$\int_{0}^{2} \int_{0}^{\sqrt{2 x-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x$

02:21

Evaluate the iterated integral by converting to polar coordinates.

$$\int_{0}^{a} \int_{0}^{\sqrt{a^{2}-y^{2}}} y d x d y$$

04:14

Evaluate the iterated integral by converting to polar coordinates.

$$\int_{0}^{2} \int_{y}^{\sqrt{2 y-y^{2}}} x d x d y$$

03:08

Transcript

So we need to see how this is ranging. So we see that our range for X given by this hmm and arrange for why is given by this. So if we really take a look at this, we can see that this portion here is this guy. This portion here is this guy. And why is going from zero to a Okay, so we’ve got a semicircle, So let’s change coordinates for a semicircle. Uh, we’re going from zero to pi zero to a got to co sign data minus R sine data. R D R D data. Okay, switch these as well. Oh, all right. So let’s go ahead and integrate this r squared coastline data. Um,…