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: 9

Answer

$\pi$

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 $y=3x; y=3$ on $[0, 1]$ Thus, $V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx=2 \pi \int_0^1 x[3-3x] \ dx$ or, $= 6\pi \int_0^1 (x-x^2) \ dx$ or, $= 6 \pi [ \dfrac{x^2}{2}-\dfrac{x^3}{3}]_0^1$ or, $= 6 \pi [ \dfrac{(1)^2}{2}-\dfrac{(1)^3}{3}]_0^1$ or, $= \pi$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.