1
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ the cylinder is standing on its circular end in the explanation. $\endgroup$ – stanley dodds Mar 25 '15 at 16:13
  • $\begingroup$ 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.) $\endgroup$ – Akiva Weinberger Mar 25 '15 at 17:07
  • $\begingroup$ Fairly certain that the semiminor axis is always going to be $r$ here, but I'm not sure. $\endgroup$ – Akiva Weinberger Mar 25 '15 at 17:11
  • $\begingroup$ yes, that is correct. thank you for the ellipse formula also. semiminor axis = r and semimajor axis = $\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$ $\endgroup$ – stanley dodds Mar 25 '15 at 17:19
0
$\begingroup$

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 $=\pi r\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.