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DryRun
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Homework Statement
Find the surface area of the portion of the cylinder [itex]x^2+y^2=4y[/itex] lying inside the sphere [itex]x^2+y^2+z^2=16[/itex]
Homework Equations
Double integral for finding surface area.
[itex]\iint_S d\sigma[/itex]
The Attempt at a Solution
At first, i thought it was just the area of the surface of sphere found inside the cylinder which goes up and down infinitely and is parallel to the z axis. But then it occurred to me that it was about finding the lateral surface area of the cylinder, which is a tad more complicated.
See attached graph for cylinder projection on the x-y plane. The sphere has a radius of 4, so it completely encloses the cylinder in the x-y plane, until the latter shoots through the sphere.
Here is my plan to find the lateral area of the cylinder: split the cylinder in half across y-z plane and then project one half on the y-z plane and then finally multiply the answer by 2 to get the whole surface area.
But if i substitute the equation of the cylinder into the equation of the sphere, i get [itex]4y+z^2=16[/itex] which gives me the equation of points of intersection at the surface between the sphere and cylinder. See second attached graph. I can use that equation to find the limits later on.
So, considering the equation of the cylinder directly. Since i will be projecting onto the y-z plane: [itex]x=\sqrt{4y-y^2}[/itex]. Then, i find ##x_y## and ##x_z## and the usual steps to finding the surface area gives:
[tex]\iint_Ω \frac{2}{\sqrt{4y-y^2}}\,.dydz[/tex]
[tex]\int^{z=4}_{z=-4} \int^{y=\frac{16-z^2}{4}}_{y=0} \frac{2}{\sqrt{4y-y^2}}\,.dydz[/tex]
I'm stuck with the integral.
I tried with polar coordinates but it just gets more complicated. Any advice?
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