Q. 56 - Calculus

Question

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

Step-by-Step Solution

Answer

The volume is $57.6\pi cubicunits$

Step 1: Given information

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

Step 2: Find the integral and evaluate it

We know that integral can be given as $V=2\pi {\int }_c^{d}r\left(y\right)h\left(y\right)dy$. The axis of revolution is x-axis

hence radius is $r\left(y\right)=y$ and $h\left(y\right)=\sqrt{y}-2$

Substituting the values in the equation we get,

$$\begin{gathered} V=2\pi {\int }_0^{4}y(\sqrt{y}-2)dy \\ V=2\pi {\int }_0^4(y\sqrt{y}-2y)dy \\ V=2\pi {{[\frac{2}{5}{y}^{\frac{5}{2}}-y^2]}^4}_{0}V=2\pi [28.8] \\ V=57.6\pi \end{gathered}$$