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FIind the area of the part of the cylinder $x^2+y^2=2ay$ that lies inside the sphere $x^2+y^2+z^2=4a^2$

The intersection of the cylinder with the sphere gives us the shape $2ay+z^2=4a^2$. So if I were to find the surface area of this shape I am done. However it is almost impossible to express the surface area element since I get a $0$ in the denominator.

Can somebody help me by using the suggested approach (calculate the surface area of $2ay+z^2=4a^2$ to get the answer or explain why that method is not possible?

mathnoob123
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1 Answers1

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Note that on the x-y plane this is simply a circle radius $a$ centered at (0,a). The bounds for the height (z) is the sphere $z=f(x,y)= \sqrt{4a^2-(x^2+y^2)}$. Thus the surface area $\int \int dS$ in cylindrical coordinates is $2\int_{-\frac{\pi}{2}}^\frac{\pi}{2} \int_0^{2sin{\theta}} \sqrt{1+f_x^2+f_y^2}dx dy$ over the circle.

pacosta
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