Q. 48 - 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 $2\pi {\int }_1^4(-x^3+4x^2+x-4)dx$

and the value of integral is $31.5\pi cubicunits$

Step 1: Given information

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

also 

Step 2: Find the integral and evaluate it

We know that integral can be given as $V=2\pi {\int }_a^{b}r\left(x\right)h\left(x\right)dx$ the axis of rotation is $r\left(x\right)=x+1$ and the height can be given as $h\left(x\right)=x-{(x-2)}^2$

Substituting the values in the integral we get,

$$\begin{gathered} V=2\pi {\int }_1^4(x+1)(x-{(x-2)}^2)dx \\ V=2\pi {\int }_1^4(-x^3+4x^2+x-4)dx \\ V=2\pi {{[\frac{-x^2}{4}+\frac{4}{3}{x}^3+\frac{x^2}{2}-4x]}^4}_1 \\ V=2\pi [15.75] \\ V=31.5\pi \end{gathered}$$