Question

Each of the definite integrals in Exercises 11–16 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis, computed with the shell method. Find this region.

$2\pi {\int }_0^1\left(3y-3y^2\right)dy$

Short answer

The region is $y=1-\frac{x}{3}$.

Step-by-step solution

Step 1. Given Information. 

We are given,

$2\pi {\int }_0^1\left(3y-3y^2\right)dy$

Step 2. Finding the region.

Rewriting the integral $2\pi {\int }_0^1\left(3y-3y^2\right)dy$as,

$2\pi {\int }_0^1\left(3y-3y^2\right)dy=2\pi {\int }_0^{1}y(3-3y)dy$

The shell represented by the integral is,

$2\pi y_k{f}^{-1}\left(y_k\right)\Delta y$

Compare $2\pi y_k{f}^{-1}\left(y_k\right)\Delta y$with the function inside the integral $2\pi {\int }_0^1\left(3y-3y^2\right)dy$as,

$$\begin{gathered} x=3-3y \\ y=1-\frac{x}{3} \end{gathered}$$

Therefore, the region is $y=1-\frac{x}{3}$.