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I am working on a problem from MIT Single Variable Calculus in problem set 2, part 2, question 1 part (a) (https://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/assignments/ps2b.pdf).

The problem concerns the surface area of a spherical cap (http://mathworld.wolfram.com/SphericalCap.html). Now the first part of the problem has me a bit flummoxed as here is what is required:

Find the formula for the height $h$ of the spherical cap in terms of $R$ (the radius of the sphere) and $r$ (the radius of the disk formed by passing a horizontal plane through the sphere).

Based on the link above/Wikipedia I know that the following formula involves all three variables of interest ($h$, $r$, and $R$):

$$R = \frac{r^2+h^2}{2h}.$$

But the problem is, when I try to isolate $h$ to put the equation in terms of $R$ and $r$, I get some really nasty results due to $h$ being quadratic in the numerator.

The best form I can get it in is:

$$h=\frac{r^2}{2R-h}.$$

I feel like I am missing something obvious here, or perhaps the above equation ($h=\frac{r^2}{2R-h}$) is precisely what is required? I don't want to move forward on this problem until I know that I have a correct formula for $h$ as it is leveraged for the remainder of the problem. Any advice would be great.

EDIT:

I believe I have figured this out.

First, we get an expression which relates all three variables of interest.

(1) For a spherical cap we know that $R^2$ (the radius of the sphere squared) is such that $R^2=(R-h)^2 + r$ by the Pythagorean theorem.

(2) Solving for $h$ gives:

$$R^2-r^2 = (R-h)^2\\ \sqrt{R^2-r^2} = \pm (R-h).$$ But the problem states that the plane slices the sphere at or above the center of the sphere. Hence we know that the height $h$ is at most $R$ and thus $h \leq R \Rightarrow (R-h) \geq 0$ and thus we can take the non-negative result.

(3) This gives a final result of

$$ \sqrt{R^2-r^2} = R-h \\ R - \sqrt{R^2-r^2} = h.$$

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If your formula is correct, just expand it :

$$h(2R-h) = r^2 \Leftrightarrow h^2 - 2Rh + r^2=0$$

You can now solve this as a quadratic equation with respect to $h$. It will then be :

$$D= 4R^2 - 4r^2$$

And if $D \geq 0$ then you'll have two solutions :

$$h= \frac{-2R \pm \sqrt{4R^2 - 4r^2}}{2}$$

Take into account that since we're talking about dimensions over the real space, you only want the positive solution, such that $h>0$.

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  • $\begingroup$ This definitely got me on the right track so marking as accepted answer, thank you. However, I omitted some details from the problem statement (i.e. the plane slices the sphere at or above the center). Using this answer and a bit more thinking/geometric analysis I edited my question above with the solution I believe I was looking for so that others may reference it. $\endgroup$ – ClownInTheMoon Dec 19 '17 at 1:25

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