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James Stewart

8 Edition

Chapter 15, Problem 32

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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 $

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Official textbook answer

Video by Alan Ghazarians

Numerade Educator

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02:24

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)\ …

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 $

09:37

Evaluate the iterated integral by converting to polar coordinates.

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

04:14

Evaluate the iterated integral by converting to polar coordinates.

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

02:28

Evaluate the iterated integral by converting to polar coordinates.

$$\int_{-2}^{2} \int_{0}^{\sqrt{4-x^{2}}}\left(x^{2}+y^{2}\right) d y d x$$

03:14

Evaluate the iterated integral by converting to polar coordinates.

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

03:09

03:15

Evaluate the iterated integral by converting to polar coordinates.

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

Transcript

So let’s figure out our downs. So we got why z equal or why Square, in fact is equal to squared of this guy squared, right? So I got y squared is equal to X minus X squared. If we complete the square completing the square here, it’s not too hard on. In fact, we don’t even have to complete this work. Let’s just move the X squared over and let’s convert to polar coordinates. So we get that are far is going to range from to coast on data Zira. And if we look at this guy here, our exes air going to range from zero to two. Okay, so let’s let’s write are integral. So far, zero to two, coz data a square root of R squared D r de theater with an R And if our X is arranging from zero Teo two Well, that just…