Question

Find the volume of the solid obtained by revolving the loop of the curve $2a{y}^2=x{(x-a)}^2$about the $x-axis,x>0$

• A. $\frac{1}{2}{\pi }^3{a}^{2}cu.units$
• B. $\frac{\pi a^3}{24}cu.units$
• C. $\frac{1}{2}{\pi }^2{a}^{3}cu.units$
• D. ${\pi }^2{a}^{3}cu.units$

Answer 1

The correct option is B

$\frac{\pi a^3}{24}cu.units$

Step 1: Draw the graph of the given curve.

We have been given an equation of the curve $2a{y}^2=x{(x-a)}^2$.

We need to find the volume of the solid obtained by revolving the loop of the curve $2a{y}^2=x{(x-a)}^2$about the $x-axis,x>0$

The graph of the given curve is as shown in figure.

Step 2: Find the value of the equation of the curve about $x-axis,a>0$ that is for $y=0$.

For $y=0$,

$$\begin{gathered} \Rightarrow 2a{y}^2=x{(x-a)}^2 \\ \Rightarrow 0=x{(x-a)}^2 \\ \Rightarrow x=0or{(x-a)}^2=0 \\ \Rightarrow x=0,x=a \end{gathered}$$

Step 3: Find the required volume.

the volume of the solid obtained by revolving the loop of the curve $2a{y}^2=x{(x-a)}^2$about the $x-axis,x>0$

$$\begin{gathered} \Rightarrow V=\pi {\int }_0^a\left(y^2\right)dx \\ \Rightarrow V=\pi {\int }_0^a\left(\frac{x{(x-a)}^2}{2a}\right)dx \\ \Rightarrow V=\frac{\pi }{2a}{\int }_0^a\left[x(x^2-2ax+a^2)\right]dx \\ \Rightarrow V=\frac{\pi }{2a}{\int }_0^a\left[x^3-2a{x}^2+x{a}^2)\right]dx \\ \Rightarrow V=\frac{\pi }{2a}\left[{\int }_0^a{x}^{3}dx-{\int }_0^{a}2a{x}^{2}dx+{\int }_0^{a}x{a}^{2}dx\right] \\ \Rightarrow V=\frac{\pi }{2a}\left({\left[\frac{x^4}{4}\right]}_0^a-2a{\int }_0^a{x}^{2}dx+a^2{\int }_0^{a}xdx\right) \\ \Rightarrow V=\frac{\pi }{2a}\left({\left[\frac{x^4}{4}\right]}_0^a-2a{\left[\frac{x3}{3}\right]}_0^a+a^2{\left[\frac{x^2}{2}\right]}_0^a\right) \\ \Rightarrow V=\frac{\pi }{2a}\left[\frac{a^4}{4}-\left(2a\times \frac{a^3}{3}\right)+\left(a^2\times \frac{a^2}{2}\right)\right] \\ \Rightarrow V=\frac{\pi }{2a}\left(\frac{a^4}{4}-\frac{2a^4}{3}+\frac{a^4}{2}\right) \\ \Rightarrow V=\frac{\pi }{2a}\times a^4\left(\frac{1}{4}-\frac{2}{3}+\frac{1}{2}\right) \\ \Rightarrow V=\frac{\mathbf{\pi }{\mathbf{a}}^{\mathbf{3}}}{\mathbf{24}}cubicunits \end{gathered}$$

Therefore, option (B) is the correct answer.