Q. 40 - Calculus

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. 

Step-by-Step Solution

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 1: Given information

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

Step 2: Find the integral and evaluate it

We know that integral can be given as

$V=2\pi {\int }_a^{b}r\left(x\right)h\left(x\right)dx$ 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

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