Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Discovery Project - Volumes of Hyperspheres - Page 1095: 1

Answer

$A=\pi\,r^2$

Work Step by Step

We can express the area of a circle by the double integral: $$\iint_{R}dA$$ R represents the region inside a circle on the $xy$-plane centered at the origin with radius r. We can rewrite this integral as an iterated integral by adding bounds and get: $$\int_{-r}^{r}\bigg(\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}dy\bigg)dx$$ Cleaning up the inner integral, we get: $$\int_{-r}^{r}\bigg(\bigg[y\bigg]_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\bigg)dx=\int_{-r}^{r}2\sqrt{r^2-x^2}\,dx$$ Because of symmetry, we can rewrite this integral as: $$4\int_{0}^{r}\sqrt{r^2-x^2}\,dx$$ Using the substitution $x=r\sin\theta$ and changing the bounds accordingly, we can rewrite the integral as: $$4\int_{0}^{\pi/2}\sqrt{r^2-r^2\sin^2\theta}\,r\cos\theta\,d\theta$$ Simplifying using the Pythagorean trigonometric identity, we get: $$4r^2\int_{0}^{\pi/2}\cos^2\theta\,d\theta$$ Solving, we get: $$4r^2\bigg[\frac{1}{2}+\frac{1}{4}\sin2x\bigg]_{0}^{\pi/2}\\=4r^2\bigg(\frac{\pi}{4}\bigg)=\pi r^2$$
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