surface area of 'cylinder' with the top cut at an angle

I don't know what the name for this shape is, so in essence it is a cylinder, radius at base $r$, which has had a wedge of the top cut off at an angle so that rather than a circle the upper face is an ellipse. its height at the top of the slanted ellipse is $h_{max}$, and the height at the bottom is $h_{min}$. the volume was easy to calculate, $\pi r^2 \frac{h_{min}+h_{max}}2$. the surface area is harder: the circle is just $\pi r^2$. I am sure that I could calculate the area of the ellipse, I just haven't got round to it, but the area of the once-rectangle is a challenge, as the upper edge is a wave. I assume it involves trigonometry, but I don't know what the formula is. help please? (apologies if my explanation is not clear)

• the cylinder is standing on its circular end in the explanation. – stanley dodds Mar 25 '15 at 16:13
• Remember that the area of an ellipse is $\pi ab$, where $a$ and $b$ are the semimajor and semiminor axes. (It's basically what happens when you take a circle of radius $1$ — and hence of area $\pi$ — and stretch it by a factor of $a$ horizontally and a factor of $b$ vertically.) – Akiva Weinberger Mar 25 '15 at 17:07
• Fairly certain that the semiminor axis is always going to be $r$ here, but I'm not sure. – Akiva Weinberger Mar 25 '15 at 17:11
• yes, that is correct. thank you for the ellipse formula also. semiminor axis = r and semimajor axis = $\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$ – stanley dodds Mar 25 '15 at 17:19

sorry, hadn't really thought it through. although it is a wave, it goes above the average height just as much as it goes below it, so it is similar to the volume calculation: $2\pi r \frac{h_{min}+h_{max}}2$. I haven't calculated the surface area of the ellipse yet. EDIT: thanks for ellipse formula. semiminor axis $=r$ and semimajor axis $=\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$ area of ellipse $=\pi r\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$
• right, the height of the lateral surface is proportional to $\cos \alpha$ going from $h_{min}$ to $h_{max}$. – G Cab Jun 28 at 23:11