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I am trying to find the area of a surface, but I can't describe the domain of integration correctly. The surface is part of the cylinder $x^2+z^2=a^2$ inside $x^2+y^2=a^2$. Here is what I have done so far:

I tried to parametrize the cylinder $S:\{(x,y,z)=(a\cos{u},v,a\sin{u}):0 \le a^2\cos^2{u}+v^2 \le a^2\}$. But I can't integrate using this condition, it gets too hairy. Is there another way to visualize this problem?

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  • $\begingroup$ You have $z^2=a^2-x^2$ and then parametrize $x$ and $y$ with regular polar coordinates. That might work. I.e. parameterize the surface with respect to the region you are integrating over. $\endgroup$
    – jdods
    May 23, 2016 at 1:30
  • $\begingroup$ You mean $x = r\cos{u}$; $y = r\sin{u}$; $z = \sqrt{a^2 - r^2 \cos^2{u}}$? That leads to a very difficult integral as well... $\endgroup$ May 23, 2016 at 1:52
  • $\begingroup$ I was wrong, no need to parameterize with polar coordinates, see answer below. $\endgroup$
    – jdods
    May 23, 2016 at 2:54

1 Answer 1

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The cylinder you want to find the surface area is $z=f(x,y)=\sqrt{a^2-x^2}$ so the differential of surface area is $dS=\sqrt{1+\left(\frac{-2x}{\sqrt{a^2-x^2}}\right)^2}dxdy$. Note that we are only taking the upper half of the surface, so will double our answer in the end.

$$ \begin{aligned} \iint_SdS&=\iint_D \sqrt{1+\left(\frac{-x}{\sqrt{a^2-x^2}}\right)^2} dx dy \\ &=\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \frac{a}{\sqrt{a^2-x^2}} dy dx \\ &=\int_{-a}^a2adx = 4a^2. \end{aligned}$$ Then we double the answer to get $8a^2$ for the surface area desired.

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