Q 57 - Calculus

Question

Use definite integrals to find the volume of each solid of revolution described in Exercises $49-61$. (It is your choice whether to use disks/washers or shells in these exercises.)

The region between the graph of $f\left(x\right)=x^2-4x+4$ and the $x-$axis on $\left[0,2\right]$, revolved around the $y-$axis.

Step-by-Step Solution

Answer

The required volume by using shells is $V=\frac{8\mathrm{\pi }}{3}$.

Step 1. Given Information

We have given a function :-

$f\left(x\right)=x^2-4x+4$

We have to find the volume of region of graph of this function and $x-axis$ on $\left[0,2\right]$ revolved around $y-axis$.

Step 2: Find the integral and evaluate it to calculate volume

We know that by using shells the volume is given by :-

$V=2\mathrm{\pi }{\int }_c^d\mathrm{r}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{x}\right)\mathrm{dx}$

Here axis of revolution is $y-axis$. So radius is $r\left(x\right)=x$ and height is given by the function.

So $h\left(x\right)=x^2-4x+4$.

Also the limits will be $0 to 2$.

Then we get the volume as following :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_0^2\mathrm{x}\left({\mathrm{x}}^2-4\mathrm{x}+4\right)\mathrm{dx} \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }{\int }_0^2\left({\mathrm{x}}^3-4{\mathrm{x}}^2+4\mathrm{x}\right)\mathrm{dx} \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }{\left[\frac{{\mathrm{x}}^4}{4}-\frac{4{\mathrm{x}}^3}{3}+\frac{4{\mathrm{x}}^2}{2}\right]}_0^2 \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left(\frac{16}{4}-\frac{32}{3}+\frac{16}{2}\right)-0 \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left(4-\frac{32}{3}+8\right) \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left(12-\frac{32}{3}\right) \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left(\frac{36-32}{3}\right) \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\times \frac{4}{3} \\ \Rightarrow \mathrm{V}=\frac{8\mathrm{\pi }}{3} \end{gathered}$$