How do you find the volume obtained by rotating $0≤y≤8\sqrt{x - x^2}$ about the $y$-axis within $0≤x≤π$?

Question

I can't seem to obtain a simple way expression of $x$ in terms of $y$, and the integration is very complex otherwise.

Answer 1

The curve $y=8\sqrt{x-x^2}$ is half of an ellipse. It extends from $x=0$ to $x=1$. The limit of $\pi$ must be an error. Rotation about the $x$ axis looks more sensible than about the $y$ axis. The integral is quite difficult as you noticed. So altogether the problem as stated might not be what was intended.

To obtain the volume when the half-ellipse is rotated about an external axis we can use Pappus' 2nd Centroid Theorem : the volume of revolution is the area under the ellipse times the distance travelled by the centroid. By symmetry the centroid is located at $x=\frac12$; its $y$ co-ordinate is more difficult to obtain but is not required.

The area under the ellipse is $$A=\int_0^1 y dx =\int_0^1 8\sqrt{x-x^2} dx$$ You are right - this integral is not easy to evaluate. Start by getting it into the form $\int \sqrt{1-u^2}du$ then use the substitution $u=\sin(t)$. $$4(x-x^2)=1-(1-4x+4x^2)=1-(2x-1)^2$$

Let $2x-1=\sin t$ then $2dx=\cos t dt$. When $x=0, t=-\pi/2$; when $x=1, t=+\pi/2$. Then

$$A=\int_0^1 2\sqrt{4(x-x^2)} (2dx)=\int_{-\pi/2}^{+\pi/2} 2\sqrt{1-\sin^2t} \cos t dt=\int_{-\pi/2}^{+\pi/2} 2\cos^2 tdt=\int_{-\pi/2}^{+\pi/2} (1+\cos 2t) dt= [t+\frac12 \sin 2t]_{-\pi/2}^{+\pi/2} = \pi$$

(Check : The area of an ellipse with semi-major and semi-minor axes $a, b$ is $\pi ab$. Here $a=4$ and $b=\frac12$ giving area $2\pi$ for the whole ellipse.)

The distance travelled by the centroid during revolution is $2\pi(\frac12)=\pi$. Therefore the volume of revolution is $$V=\pi A=\pi^2$$

Rotation about $x$ axis

In this case we can use as elements of volume $dV$ vertical thin disks of radius $y$ and thickness $dx$ : $$dV=\pi y^2 dx$$ $$V=\int_0^1 \pi.64(x-x^2)dx=32\pi [x^2-\frac23x^3]_0^1=32\pi (1-\frac23)=\frac{32}{3}\pi$$

(Check : The volume of an ellipsoid is $\frac43 \pi abc$. Here $a=c=4, b=\frac12$ giving $V=\frac43 \pi (4.\frac12.4)=\frac{32}{3}\pi$.)