Solid of Revolution Problem (semi-circular Groove turned into a Cylinder)

Question

A groove, semi-circular in section and 1cm deep, is turned in a solid cylindrical shaft of diameter 6cm. Find the volume of material removed and the surface area of the groove.

The problem is that i don't understand (or is unclear) the area that i have to consider. Also, which material is getting removed and how/why? I can't visualize this scenario and i'm very confused. I'm aware of the similar "Napking Ring problem", where a cylinder is bored in a sphere etc and i have understood it quite well. But this i can't understand it intuitively and i don't know where to begin really.

The problem is from a book i'm studying and it gives the volume as $25.4cm^3$ and the area as $46.65cm^2$. If anyone could help, offer some advice i would appreciate it very much. Thanks!

Answer 1

The profile of the groove is described by $$r(\theta)=3-\cos\theta,\quad z(\theta)=\sin\theta\qquad\left(-{\pi\over2}\leq\theta\leq{\pi\over2}\right)\ .$$

A horizontal plane at level $z=z(\theta)$ intersects the volume turned away in an annulus of area $$\pi\bigl(3^2-r^2(\theta)\bigr)=\pi(6\cos\theta-\cos^2\theta)\ ,$$ and the vertical distance between the planes $z=z(\theta)$ and $z=z(\theta+d\theta)$ is $$dz=z'(\theta)d\theta=\cos\theta\>d\theta\ .$$ The total volume turned away is therefore given by $$V=2\pi \int_0^{\pi/2}(6\cos\theta-\cos^2\theta)\>\cos\theta\>d\theta={\pi\over3}(9\pi-4)\ .$$ For the surface of the groove we note that $ds=d\theta$ along the profile, since the radius of the groove is $=1$. We therefore obtain $$A=2\int_0^{\pi/2}2\pi r(\theta)\>d\theta=\pi(24-\pi)\ .$$

Answer 2

The groove is carved all around the cylinder. Compute the integrals in cylindrical coordinates. For instance, the volume is given by: $$ 2\int_0^{2\pi}d\phi\int_2^3 rdr \int_0^{\sqrt{1-(3-r)^2}}dz ={\pi\over3}(9\pi-4)\approx 25.42 $$ In the picture below you can see a cross section of the cylinder: the grey half disk represents the removed material and $z$-axis is the axis of symmetry of the cylinder. On the upper part of the half cyrcle $r$ and $z$ are linked by the circle equation: $z=\sqrt{1-(3-r)^2}$.