Question

Consider the region between $f\left(x\right)=4-x^2$and the line y = 5 on [0, 2]. For each line of rotation given in Exercises 39–42, use the shell method to construct definite integrals to find the volume of the resulting solid.

Short answer

The integral can be given as $V=2\pi {\int }_0^2(2-x)(x^2+1)dx$

and the value of integral is$\frac{20}{3}\pi cubicunits$

Step-by-step solution

Given information

We are given a function $f\left(x\right)=4-x^2$and

Find the integral and evaluate it

We know that integral can be given as

$V=2\pi {\int }_a^{b}r\left(x\right)h\left(x\right)dx$where r and h are the radius and height of the function

Now the revolution is around the line x=2

Hence the radius can be given as

$r\left(x\right)=2-x$and the height can be given as $h\left(x\right)=x^2+1$

and the interval is [0,2]

Substituting the values in integral can be given as

role="math" localid="1650609078445" $$\begin{gathered} V=2\pi {\int }_0^2(2-x)(x^2+1)dx \\ V=2\pi {\int }_0^2(2x^2+2-x^3-x)dx \\ V=2\pi {{[\frac{2x^3}{3}+2x-\frac{x^4}{4}-\frac{x^2}{2}]}^2}_0 \\ V=2\pi \left[\frac{10}{3}\right] \\ V=\frac{20\pi }{3}cubicunits \end{gathered}$$