# Evaluate the integral by converting to polar coordinates. \int_{0}^{1} \int_{y}^{\sqrt{2 - y^2}}...

## Question:

Evaluate the integral by converting to polar coordinates.

{eq}\displaystyle \int_{0}^{1} \int_{y}^{\sqrt{2 - y^2}} 4(x + y) dx dy {/eq}

## Integral in polar coordinates:

We will solve the problem where we will change to polar form {eq}x=r\cos\theta\\ y=r\sin\theta\\ dxdy=rdrd\theta {/eq} here we will find the radius and then solve the integral by using the power rule.

To solve the problem we will change the integral to polar form:

{eq}\int_{0}^{1}\int_{y}^{\sqrt{2-y^{2}}}4(x+y)dxdy {/eq}

Now changing to polar form:

{eq}x=r\cos\theta\\ y=r\sin\theta\\ dxdy=rdrd\theta {/eq}

The integral will be:

{eq}=4\int_{0}^{\frac{\pi}{4}}\int_{0}^{\sqrt{2}}r(\cos \theta +r\sin\theta)rdrd\theta\\ =\frac{8\sqrt{2}}{3}\left [ \sin\theta -\cos\theta \right ]\\ =\frac{8\sqrt{2}}{3} {/eq}

Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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