Q. 47 - Calculus

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. 

Step-by-Step Solution

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 1: Given information

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

also we have,

Step 2: 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

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