## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 14 - Section 14.4 - Double Integrals in Polar Form - Exercises - Page 778: 18

#### Answer

$$\pi$$

#### Work Step by Step

Our aim is to integrate the triple integral as follows: $\int^1_{-1} \int^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}}\dfrac{2}{(1+x^2+y^2)} \space dy \space dx =\int^{\pi}_0 \int^1_0 \dfrac{2r}{1+(r \cos \theta )^2+(r \sin \theta )^2} \space dr \space d\theta$ or, $= \int^{\pi}_0 \int^1_0 \dfrac{2r}{(1+r^2)^2}dr d\theta$ or, =4 \times $\int^{\pi/2}_0 [\dfrac{-1}{1+r^2}]^1_0d\theta$ or, $=2 \times \int^{\pi/2}_0 d\theta$ or, $=\pi$

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