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Find the surface area of the part of the sphere $x^2 + y^2 + z^2 = a^2$ inside the circular cylinder $x^2 + y^2 = ay$ ($r = a\sin(\theta)$ in polar coordinates), with $a > 0$.

First time posting on this website, sorry for the lack of details on my attempts but I am really not sure where to start on this problem.

A formula that is useful is $A(G) = \int \int \sqrt{f_x^2 + f_y^2 + 1}dA$

$f_x$ is the partial derivative with respect to x, $f_y$ is the partial derivative with respect to y

I know that I need to find an equation which should be $x^2 + y^2 + z^2 = a^2$, and I need to find the limits which is where I am really struggling.

Also according to my professor, I shouldn't have to use any polar coordinate conversions in order to complete this problem, which was my initial approach.

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I'm going to assume that, since the problem asks for the surface area of the sphere found within the cylinder, the total for both hemispheres is what is asked for. So we'll work out the expression for the "upper" hemisphere and double that.

The function we will work with for that hemisphere is $ \ f(x,y) \ = \ z \ = \ \sqrt{a^2 - x^2 - y^2} \ . $ Using direct or implicit differentiation, we find that $ \ f_x \ = \ -\frac{x}{z} \ $ and $ \ f_y \ = \ -\frac{y}{z} \ , $ which gives us the "surface area projection factor",

$$ \ \sqrt{f_x^2 + f_y^2 + 1} \ = \ \sqrt{\frac{x^2}{z^2} + \frac{y^2}{z^2} + 1} \ = \ \frac{\sqrt{x^2 + y^2 + z^2}}{z} \ = \ \frac{a}{\sqrt{a^2 - x^2 - y^2}} \ \ . $$

It is this last result which perhaps led your professor to suggest that polar coordinates were not needed (opinions concerning this may differ...).

To obtain the surface area for the selected portion of the hemisphere, we will want to integrate this function over the circular cross-section of the cylinder in the $ \ xy-$ plane. In Cartesian coordinates, the circle is

$$ x^2 \ + \ y^2 \ = \ ay \ \ \Rightarrow \ \ x^2 \ + \ (y^2 \ - \ ay \ + \ \frac{a^2}{4}) \ = \ \frac{a^2}{4} \ \ \Rightarrow \ \ x^2 \ + \ (y \ - \ \frac{a}{2})^2 \ = \ \frac{a^2}{4} \ , $$

hence it is centered at $ \ ( 0 , \frac{a}{2} ) $ and has radius $ \frac{a}{2} \ . $ (You likely already knew that, since you also gave the polar function for this circle.) The limits of integration are less messy if we integrate in the $ \ x-$ direction first; the equation of the circle yields

$$ x^2 \ = \ \frac{a^2}{4} \ - \ (y^2 \ - \ ay \ + \ \frac{a^2}{4}) \ \ \Rightarrow \ \ x \ = \ \pm \sqrt{ay - y^2} \ . $$

Since the circle is symmetric about the $ \ y-$ axis, we can use just the positive square root and double the result. So the limits of integration in the $ \ x-$ direction are $ \ 0 \ \text{to} \ \sqrt{ay - y^2} \ $ ; in the $ \ y-$ direction, they are $ \ 0 \ \text{to} \ a \ . $

The surface area integral covered both halves of the cross-sectional circle and both hemispheres is then

$$ 2 \ \cdot \ 2 \ \int_0^a \ \int_0^{\sqrt{ay - y^2}} \ \frac{a}{\sqrt{a^2 - x^2 - y^2}} \ \ dx \ dy \ \ . $$

This is not a horrible integral, but also a not too pleasant one, in rectangular coordinates. In polar coordinates, we have

$$ 2 \ \cdot \ 2 \ \int_0^{\pi/2} \ \int_0^{a \sin \theta} \ \frac{a}{\sqrt{a^2 - r^2}} \ \ r \ dr \ d\theta \ \ . $$

Note that this is not a "spherical cap", since the curve of intersection on the spherical surface is not a circle; so there is no convenient shortcut for checking the result.

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This approach does give the correct result for spherical caps. If we pick the one for a circle of radius $ \ \frac{a}{2} \ $ centered on the origin, we can use limits of integration $ \ 0 \ \le \ x \ \le \ \sqrt{\frac{a^2}{4} - y^2} \ $ and $ \ 0 \ \le \ y \ \le \ \ \frac{a}{2} \ . $ We are then integrating over a quarter of the circle, so the area of the full cap is given by

$$ 4 \ \int_0^{a/2} \ \int_0^{\sqrt{\frac{a^2}{4} - y^2}} \ \frac{a}{\sqrt{a^2 - x^2 - y^2}} \ \ dx \ dy \ \ . $$

Using polar coordinates, we have

$$ 4 \ \int_0^{\pi/2} \ \int_0^{a/2} \ \ \frac{a}{\sqrt{a^2 - r^2}} \ \ r \ dr \ d\theta \ \ = \ \ 4a \ \int_0^{\pi/2} d\theta \ \ \int_0^{a/2} \frac{r \ dr}{\sqrt{a^2 - r^2}} \ \ \ $$

$$ = \ \pi a \ \int_{\frac{3}{4} a^2}^{a^2} \ \frac{du}{\sqrt{u}} \ = \ \pi a \ \left( \ 2 \ u^{1/2} \ \vert_{\frac{3}{4} a^2}^{a^2} \ \right) \ = \ 2 \pi a \ ( \ a \ - \ \frac{\sqrt{3}}{2} a ) $$

$$ = \ ( \ 2 \ - \ \sqrt{3} ) \ \pi a^2 \ \ , $$

which agrees with the result from the standard formula for the surface area of a spherical cap.

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