Cylindrical hoof lateral surface and volume I am trying to figure out the proof for lateral surface and volume of a cylindrical hoof (or, more generally, a cylindrical wedge) given by Wolfram MathWorld but I am having some trouble understanding it. I would be grateful if someone could explain it in details. THanks.
 A: For this problem we will use the variables and diagram as shown on Wolfram Math World .
The first thing we need to do is compute the equation of the plane cutting out the hoof. We have four points on the plane: $(0,0,0)$, $(0,R,0)$, $(0,-R,0)$, and $(R,0,h)$.
We can write any plane in the form $Ax+By+Cz+D=0$. Plugging in $(0,0,0)$ we get $D=0$. Now plugging in $(0,R,0)$ we get $BR=0$, so $B=0$. Finally, using $(R,0,h)$, we get $AR+Ch=0$ so $C=-\dfrac{AR}{h}$. Thus, we get the equation of the plane as 
$$Ax-\frac{AR}{h}z=0$$
and dividing through by $A$ (possible because $A=0$ gives a degenerate answer) and rewriting gives
$$hx=Rz.$$
Now, in order to proceed, we need to perform some integration. First we shall find the volume $V$. If we imagine vertical slices (more precisely the intersection of planes parallel to the $y$ $z$ plane), we get rectangles of height $z(x)=\dfrac{h}{R}x$ and width $y(x)=2\sqrt{R^2-x^2}$. The height here comes from our plane equation and the width comes from the equation for the base of the cylinder: $x^2+y^2=R^2$. Onto our integration!
$$V=\int\limits_{0}^{R}z(x)y(x)\,dx=\frac{2h}{R}\int\limits_{0}^{R}x\sqrt{R^2-x^2}\,dx$$
Letting $x=R\sin t$ we have 
$$
V=\frac{2h}{R}\int\limits_{0}^{\pi/2}R\sin t\sqrt{R^2-(R\sin t)^2}R\cos t\,dt
=2hR^2\int\limits_{0}^{\pi/2}\sin t\cos^2 t\,dt
=\frac{-2hR^2}{3}\cos^3(t)\Biggr|_{0}^{\pi/2}
$$
Hence $V=\dfrac{2}{3}hR^2$ as desired.
To find the lateral surface area $LSA$, we imagine taking flat slices (more precisely the intersection of planes parallel to the $x$ $y$ plane), and we get varying sizes of arcs of a circle of radius $R$. We use the fact that an arc has length $R\theta$ where $\theta$ is the central angle of the arc (for reference you can see this Wolfram Math World article). Since $x=R\cos\frac{\theta}{2}$, we have $z=h\cos\frac{\theta}{2}$, so $\theta=2\cos^{-1}\frac{z}{h}$. Hence
$$LSA=\int\limits_{0}^hR\theta\,dz=2R\int\limits_{0}^h\cos^{-1}\frac{z}{h}\,dz$$
Letting $z=h\cos t$ we get
$$LSA=-2Rh\int\limits_{\pi/2}^0t\sin t\,dt=2Rh\int\limits_0^{\pi/2}t\sin t\,dt=2Rh(-t\cos t+\sin t)\Biggr|_{0}^{\pi/2}$$
Hence $LSA=2Rh$ as desired.
A: The problem of the volume of Archimedes hoof has been carefully laid out and solved in the published article The Method of Archimedes: Propositions 13 and 14 and there are some animations and Mathematica$^{\circledR}$ code here: The Method of Archimedes: Propositions 13 and 14.
I have also done work on the surface area, but this has not been published. Basically, what you do is separate the surface into three part: base, slant plane, and cylindrical collar. Here is an interesting result for you to ponder...
$$A_{plane}+A_{base}=2A_{base}\Phi=\pi\Phi$$
Yes, that's $\Phi$, the golden ratio. (This result is for a hoof in a $2\times2\times2$ cube.) I can also tell you that the surface of the cylindrical collar is $2h=4$.
I'm reluctant to go into further detail on a post that is now two years old. Please, someone let me if there is still interest out there!
