Q. 45 - 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 $2\pi {\int }_1^4(y\sqrt{y}+2y-y^2)dx$

and the value of integral is $12.8\pi$ cubicunits.

Step 1: Given information

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

Also 

Step 2: Find the integral and evaluate it

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

Hence the integral can be given 

$$\begin{gathered} V=2\pi {\int }_a^{b}y(\sqrt{y}+2-y)dy \\ V=2\pi {\int }_1^4(y\sqrt{y}+2y-y^2)dx \\ V=2\pi {{[\frac{2}{5}{y}^{\frac{5}{2}}+y^2-\frac{y^3}{3}]}^4}_1 \\ V=2\pi [6.4] \\ V=12.8\pi \\ V=12.8 \end{gathered}$$