In finding the surface area of an oblique cylinder, C, as shown here. I have parametrised the surface as suggested in this question (link). $$r(\phi,s)=(r\cos(\phi)+s\cos(v), r\sin(\phi), s\sin(v)) (0\le\phi\le2\pi, 0\le s\le a)$$

We find the surface area by finding the gram matrix and end up with the integral: $$Area=\int_C\sqrt{r^2-r^2\sin^2(\phi)\cos^2(v)}d\phi dt$$ $$=r\int_0^a\int_0^{2\pi}\sqrt{1-\sin^2(\phi)\cos^2(v)}d\phi dt$$ $$=ar\int_0^{2\pi}\sqrt{1-\sin^2(\phi)\cos^2(v)}d\phi$$

And I don’t see how to get further. Is there a substitution or something else that works here?


You are asking an elementary question (which are the hardest questions to answer). The slant height of a such a cylinder is $a$ and thus the lateral surface area is $a2\pi r$. Given the angle $v$, the slant height, $a$, is:

$$ \sin(v) = \frac{h}{a} \rightarrow a = \frac{h}{\sin(v)} $$

Thus the lateral surface area is:

$$ a2\pi r = \frac{2\pi rh}{\sin(v)} $$

To prove this using calculus is fairly elementary, all you need to do is take small "slivers" of rectangles which go up the side of the cylinder. This rectangle with have an area of:

$$ dA = bh = rd\theta a = a rd\theta $$

Where $a = \frac{h}{\sin(v)}$ (via geometry). Then you integrate around the entire cylinder:

$$ A = \int dA = ar\int\limits_0^{2\pi}d\theta = 2\pi ar = \frac{2\pi rh}{\sin(v)} $$

And you cannot "break this down" any further because if you try (i.e. you say--how do I know you can run the full slant height) all you will end up with is a double integral of the form:

$$ A = r\int\limits_0^{2\pi}d\theta\int\limits_0^a da = r\int\limits_0^{2\pi}d\theta\int\limits_0^{\frac{h}{\sin(v)}} da $$

  • $\begingroup$ I don't quite understand your proof - what is $b$ when you say dA=bh? $\endgroup$ – Yimin3256 Mar 25 '16 at 4:17
  • $\begingroup$ @Yimin3256 $b$ is the "base" and $h$ is the "height"--which is which isn't important. One of them is the slant height, $a$, and one is the differential length along the circle $rd\theta$--together that is $ard\theta$. $\endgroup$ – Jared Mar 25 '16 at 4:23

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