Volume of solid revolving around y axis by two curves

Question

Find the volume of the solid formed by revolving the region bounded by $y=(x-2)^2$ and $y=x$ about the y axis. I first find the area bound and then multiply it by $2\pi$ . First I get $\int_0^1((x-2)^2 -x )dx=7/2$ and then multiply it with $2\pi$ so my answer was $7\pi$ but options are

1. $(45\pi)/2 $
2. $23\pi$
3. $(47\pi)/2$
4. $24\pi$

please tell the solution

Answer 1

That region can be seen in the picture below. If $(x,y)$ belongs to it then:

• if $0\leqslant y\leqslant1$, then $x$ can be any number from $2-\sqrt y$ to $2+\sqrt y$;
• if $1\leqslant y\leqslant4$, then $x$ can by any number from $y$ to $2+\sqrt y$.

So, using the disk method, your volume is equal to$$\pi\left(\int_0^1\left(2+\sqrt y\right)^2-\left(2-\sqrt y\right)^2\,\mathrm dy+\int_1^4\left(2+\sqrt y\right)^2-y^2\right)=\frac{45}2\pi.$$So, the correct option is the first one.

An even simpler approach consists in using the shell method. The volume is equal to$$2\pi\int_1^4x\bigl(x-(x-2)^2\bigr)\,\mathrm dx=\frac{45}2\pi.$$