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Find the surface area of two cylinders $$y^2 + z^2 = 1$$ and $$x^2 + y^2 = 1$$

I have so far set the two equations to equal $$x= \pm z$$ and $$y= \sqrt{(1-z^2)}$$ I am a little confused on how to set up the integration problem. So far I have $$1/\sqrt{(1-z^2)}dy$$ from 0 to 1 and am not certain if that is the correct approach.

asikgudquestin
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  • @Semsem I have so far set the two equations to equal x= +-z and y= sqrt(1-z^2). I am a little confused on how to set up the integration problem. So far I have 1/sqrt(1-z^2)dy from 0 to 1 and am not certain if that is the correct approach. – asikgudquestin Apr 15 '14 at 20:04
  • Please add the above comment to your question – Semsem Apr 15 '14 at 20:29

2 Answers2

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The surface area is $$S=2\int\int_D \sqrt{1+f_x^2+f_y^2}dxdy$$ where $z=f=\sqrt{1-y^2}$ and so $f_y=\frac{-y}{\sqrt{1-y^2}}$ so $$S=2\int\int_D \sqrt{1+\frac{y^2}{1-y^2}}dxdy=2\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2} \frac{1}{\sqrt{1-y^2}}dxdy \\=2\int_{-1}^12dy=8 $$

Semsem
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  • how did you did it , meaning some of the previous steps can you please, show by which you arrived to the above expression how did you got the equation of the common enclosed domical surface. Can you clarify the concept to solve the problem. Please –  Jan 22 '21 at 16:40
  • This answer is only partly correct - really surface area is $16$, twice more. Take, please, look at https://math.stackexchange.com/questions/4638205/which-of-the-following-answers-actually-calculates-the-surface-area-of-the-stein/ – zkutch Feb 17 '23 at 07:55
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Another way to find this is integrating on the lengths of the intersections of the surface of interest and the planes of the form $y=k$ for $-1\leq k \leq 1$. I haven't tried it in a while but here it goes. The sections look like parenthesis joined by flat top and bottom. By parameterizing one of the cylinders with $\theta$; the length of the part that looks like parenthesis is $4\theta$ (here $\theta$ is in radians of course :)). The top and bottom segments add up to $4\cos(\theta)$. Integrating from $\theta=0$ to $\theta=\frac{\pi}{2}$: $$\int_0^\frac{\pi}{2} 8(\theta + \cos(\theta))d\theta=\pi^2+8$$

I am not 100% sure about this answer but it does compare nicely to the surface area of the unrestricted drum which is $6\pi$.

user143427
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