Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.2 Volumes - 5.2 Exercises - Page 374: 16

Answer

$$ \frac{29\pi }{30}$$

Work Step by Step

$y= x^2,\ \ x=y^2 ; \quad$ about the $x=-1$ First, we find the intersection points \begin{aligned} x^4&=x\\ x^4-x&=0\\ x(x^3-1)&=0 \end{aligned} Then \begin{aligned} x=0,\ \ x=1 , \to \ \ (0,0), (1,1)\end{aligned} Here the oute radius is $\sqrt{y} +1 $ and the inner is $ y^2+1$ \begin{aligned} A(y)&= \pi [f^2(y)-g^2(y)]\\ &= \pi[ (\sqrt{y}+1)^2- (y^2+1)^2]\\ &= \pi [-y^4-2y^2+y+2\sqrt{y}] \end{aligned} Then the volume of the solid given by \begin{aligned} V&= \int_a^bA(y)dy\\ &= \pi\int_0^1 (-y^4-2y^2+y+2\sqrt{y})dy\\ &=\pi (-\frac{y^5}{5}-\frac{2y^3}{3}+\frac{y^2}{2}+\frac{4}{3}y^{\frac{3}{2}})\bigg|_0^1\\ &= \frac{29\pi }{30} \end{aligned}
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