Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.4 Volume by Shells - 6.4 Exercises - Page 442: 8

Answer

$\dfrac{104 \pi}{3}$

Work Step by Step

Our aim is to compute the volume of the given curve when it is revolved around $y$-axis by using the shell method. Shell method for computing the volume of the curve: Let us consider two functions $f(x)$ and $g(x)$ (both are continuous functions) with $f(x) \geq g(x)$ on the interval $[m, n]$ and $R$ defines the region which revolves about the y-axis and is known as the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $x=m$ and $x=n$. Then, the volume of the solid can be expressed as: $Volume, V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx$ Here, $2\pi x=\text{Shell Circumference}$ and $[f(x)-g(x)] =\text {Height of the shell}$ We are given that $y=6-x; y=0$ on $[2, 4]$ Thus, $V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx=2 \pi \int_2^4 x(6-x) \ dx$ or, $=2 \pi [ \dfrac{6x^2}{2}-\dfrac{x^3}{3}]_2^4$ or, $=2 \pi [ 3x^2-\dfrac{x^3}{3}]_2^4$ or, $=2 \pi [ 3(16-4)-\dfrac{4^3}{3}+\dfrac{2^3}{3}]_2^4$ or, $=\dfrac{104 \pi}{3}$
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