# Find surface area of part of cylinder.

I ran into trouble when I'm trying to find a surface area of parts of the cylinder $$x^2+z^2=4$$ bounded by another cylinder $$x^2+y^2=4$$, I simply used a traditional way of double integral, change into polar coordinate calculate $$\iint\limits_{x^2+y^2=4} \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+ \left(\frac{\partial z}{\partial y}\right)^2+1} \,dx\,dy = \int_0^{2\pi}\int_{0}^{2} \frac{2r}{\sqrt{4-(r\cos\theta)^2}} \,dr\,d\theta$$ and eventually this integral diverges. Could anyone tell me where I was wrong ? thanks a lot.

• It doesn't seem like it diverges. Eventually you get something that looks like $$4\sec^2\theta(1-|\sin\theta|) = \frac{4}{1+|\sin\theta|}$$ which has no singularities anywhere. The key is that you get a difference of two diverging things, which if you separated them out before evaluating limits, is not allowed to be considered separately. – Ninad Munshi Jul 15 '20 at 20:35
• Why not parametrize the surface of the cylinder in cylindrical coordinates to start with? (Also, please don't use the differential geometry tag if you don't know what it means.) :) – Ted Shifrin Jul 15 '20 at 22:16
• I'm sorry if I abused the tag of differential geometry. I just thought it can be counted as an elementary-level or introductory-level of Differential geometry problem(curves and surfaces), and thanks for the correction and comment, I cannot just other kinds of parametrization since it has such a requirement. – Richard_Mantle Jul 15 '20 at 22:56

Per my comment, the integral simplifies down to

$$\int_0^{2\pi} \frac{4}{1+|\sin\theta|}\:d\theta$$

after doing the $$r$$ integral. The easiest way to compute this integral then would be to exploit symmetries

$$= \int_0^\pi \frac{8}{1+\sin\theta}\:d\theta = \int_0^\pi \frac{8}{\cos^2\left(\frac{\theta}{2}\right)+\sin^2\left(\frac{\theta}{2}\right)+2\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}\:d\theta$$

$$= \int_0^\pi \frac{8}{\left[\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)\right]^2}\:d\theta = 16\int_0^\pi \frac{\frac{1}{2}\sec^2\left(\frac{\theta}{2}\right)}{\left[1+\tan\left(\frac{\theta}{2}\right)\right]^2}\:d\theta$$

$$= \frac{-16}{1+\tan\left(\frac{\theta}{2}\right)}\Biggr|_0^{\pi^-} = 16$$

And then the final answer for the problem would be $$32$$, doubled to account for both sides of the cylinder.

• thank you very much, sir. it does requires a little analysis near infinity which I might ignore. – Richard_Mantle Jul 15 '20 at 22:04
• Why do have you chosen upper bound $\pi^-$ instead of $\pi^+$? – Sebastiano Jul 15 '20 at 22:20
• I believe this is the right choice since we are actually evaluating this in the interval [0,𝜋/2), I figure out that it can also be done by dividing cos𝜃 at the beginning by changing the definite integral to improper integral to find a trick-less solution. – Richard_Mantle Jul 15 '20 at 22:47
• @Sebastiano because in very general terms, integrals don't care about the action of functions at specific points, they care about the limits to specific points (if you have a finite number of discontinuities, this disclaimer is necessary to make that broad statement). Here, the limit we care about is from below. – Ninad Munshi Jul 16 '20 at 0:06
• Thank you very much. I have before upvoted your answer. Best regards. – Sebastiano Jul 16 '20 at 11:26

Easier to keep this in rectangular coordinates.

Limits

$$x^2 + y^2 = 4\\ x^2 + z^2 = 4\\ y^2 = z^2\\ y = \pm z$$

This suggests that integrating with respect to y and z makes more sense than integrating with respect to x and y.

$$x = \pm \sqrt {4 - z^2}\\ dS = (1, - \frac {\partial x}{\partial y}, - \frac {\partial x}{\partial z})\\ dS =(1,0, -\frac {z}{\sqrt {4-z^2}})\\ \|dS\| = \frac {2}{\sqrt {4-z^2}}$$

$$4\int_{0}^2 \int_{-z}^{z} \frac {2}{\sqrt {4-z^2}} \ dy\ dz$$

we are multiplying by 4 because over the triangle in the yz plane, there is a surface above the plane and a complimentary surface below the plane. There is then an identical surface that appears when $$z < 0$$

• thanks, it's also very helpful – Richard_Mantle Jul 15 '20 at 23:30