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I wish to find the surface area cut from the sphere $x^2+y^2+z^2=2$ and the cylinder $x^2+y^2=1$

Here is what i tried: by uniting the 2 equations we get: $z=1$

The shape looks like a dome, when:

$0\lt z\lt 1$
$0\lt r\lt \sqrt 2$
$0\lt \theta \lt 2\pi$

I'm not sure how to continue from here. any suggestions?

segevp
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  • Do you know about surface integrals, from multivariable calculus? If so, what happened when you tried to set up the surface integral? – Lee Mosher Jan 22 '18 at 14:33

1 Answers1

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The cylinder cuts out two spherical caps from the sphere. The radius of the sphere is $\sqrt 2$, and the two planes defining these caps have equations $z=\pm1$. It follows that the height of these caps is $\sqrt{2}-1$. By a standard theorem (maybe found out by Archimedes) the area of each cap is equal to the area of a cylinder enveloping the sphere around the equator, enclosed between the planes $z=1$ and $z=\sqrt{2}$. It follows that the area per cap is $2\pi\,\sqrt{2}(\sqrt{2}-1)=2\pi(2-\sqrt{2})$.

Christian Blatter
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  • I know understand! if i wanted to solve as a surface integral it should be: $\int_0^{2\pi}\sqrt{2}d\theta\int_1^{\sqrt{2}}dz$. adding up all the "slices" between $1\lt z \lt \sqrt 2$. Am i right? – segevp Jan 22 '18 at 19:48
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    The resulting integral would be $$2\pi\bigl(\sqrt{2}\bigr)^2\int_0^{\pi/4}\sin\theta>d\theta\ ,$$ whereby $\theta=0$ at the north pole. – Christian Blatter Jan 23 '18 at 09:07
  • why is $0\lt \theta \lt \frac{\pi}{4}$? – segevp Jan 23 '18 at 10:07