Question

Consider the region between$f\left(x\right)={(x-2)}^2$ and g(x) = x on [1, 4]. For each line of rotation given in Exercises 45–48, use the shell method to construct definite integrals to find the volume of the resulting solid.

Short answer

The integral can be given as

$2\pi {\int }_c^d(-y^2+y\sqrt{y}+\sqrt{y}+y+2)dy$

And the value of the integral is$19.13\pi$.

Step-by-step solution

Given information

We are given functions$f\left(x\right)={(x-2)}^2$ and g(x) = x

also we have,

Find the integral and evaluate it

We know that the integral can be given as $V=2\pi {\int }_c^{d}r\left(y\right)h\left(y\right)dy$as the axis of rotation is y=-1 the radius can be given as $r\left(y\right)=(y+1)$and the height can be given as $h\left(y\right)=(\sqrt{y}+2-y)$

Hence the integral can be given as

$$\begin{gathered} V=2\pi {\int }_c^d(y+1)(\sqrt{y}+2-y)dy \\ V=2\pi {\int }_c^d(-y^2+y\sqrt{y}+\sqrt{y}+y+2)dy \\ V=2\pi {{[\frac{-y^3}{3}+\frac{2}{5}{y}^{\frac{5}{2}}+\frac{2}{3}{y}^{\frac{3}{2}}+\frac{y^2}{2}+2y]}^4}_{1}V=2\pi [12.8-3.23] \\ V=2\pi [9.56] \\ V=19.13\pi \end{gathered}$$