Q. 38 - Calculus

Question

Consider the region between f(x) = x − 2 and the x-axis on [2, 5]. For each line of rotation given in Exercises 33–38, use the shell method to construct definite integrals to find the volume of the resulting solid 

Step-by-Step Solution

Answer

The volume integral can be given as $V={\int }_2^5(6-x)(x-2)dx And the volume is 18\pi cubicunits.$

Step 1: Given information

We are given a function f(x) = x-2 and

Step 2: Find the integral and volume

We know that, The volume can be given as

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

The revolution is around $x=6$ Hence the radius becomes $r\left(x\right)=6-x$ and the height of the function can be given as

$h\left(x\right)=x-2$ and the interval is [2,5]

Substituting the values in the integral formula we get,

$$\begin{gathered} V=2\pi {\int }_2^5(6-x)(x-2)dx \\ V=2\pi {\int }_2^5(-x^2+8x-12)dx \\ V=2\pi {{[-\frac{x^3}{3}+4x^2-12x]}^5}_2 \\ V=2\pi \left[9\right] \\ V=18\pi \end{gathered}$$