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I have a sphere $(S)$ of radius $R$ : $x^2 + y^2 + z^2 = R^2$.

Two planes $(P)$ & $(Q)$ :

$(P): x = a$ and $(Q): y = b$

These two planes are cutting the sphere $(S)$ , what is the volume of the upper part of the spherical cap (shaded in yellow) in terms of $R$ , $a$ and $b$

If $(Q)$ would have as equation $y = 0$ then the result is semi-spherical cap, for it I know calculate the volume.

NB: I am familiar to integral enter image description here Thanks for any help, regards!

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    You don't need calculus. Here's a hint: The spherical cap equals a sector of a sphere minus a cone - in the same way that in this image [https://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Chord_in_mathematics.svg/200px-Chord_in_mathematics.svg.png], the area cut off by the chord BX equals the area of the sector of the circle cut off by segments MB and MX minus the area of the triangle MBX. Cones and spherical sectors both have simple formulas. – 2017-02-08
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    I think that is not my case, my case... a spherical cap in vertical position, that means the plane side of it is vertical, this is filled with part of water for example, that volume is what I would like know – 2017-02-08
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    OK, maybe this diagram is better: https://en.wikipedia.org/wiki/File:Kugel-sektor.png. You're looking for the volume of the blue-shaded area above the small circle with radius $a$ in that diagram (which, translating back from that diagram to yours, is where the plane cuts the sphere). That equals the total blue-shaded area (which is a sector) minus the blue-shaded area below the circle (which is a cone). – 2017-02-08
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    https://imgur.com/gallery/7hmSv that is what I would like calculate, this filled with water part but only for this part of spherical cap (shaded in yellow), not with the cylindrical part – 2017-02-08
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    By the way, that is tank, circular cylinder with two spherical cap ends not hemispherical ends.... what you describe is easy to calculate, and without calculus – 2017-02-08
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    +1 for what appears to be a hard question. I couldn't google an answer, though I did find two very related questions on this site, [here](http://math.stackexchange.com/questions/2069227/how-to-find-volume-of-spherical-cap-cut-by-perpendicular-plane) and [here](http://math.stackexchange.com/questions/1673182/how-to-find-the-volume-of-partially-filled-spherical-cap?rq=1), unfortunately neither with an answer. – 2017-02-08
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    @Jens what you found, a question (with the name "khaled" was asked from me long ago but unfortunately no clear answer for me – 2017-02-08
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    Many people gave me answer, but for the whole spherical cap volume, but it's easy... but my question, I think... one could solve with triple integral with cartesian coordinates not spherical – 2017-02-08
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    Ah, you're right - I didn't read carefully enough and thought you were referring to a different region. Sorry. – 2017-02-09

1 Answers 1

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The volume of a sphere can be derived by integration using the disk method. In the same way we can derive the volume of the "sphere corner" you are looking for, but instead of using complete disks (circles), we use circular segments. In your figure, imagine slicing the sphere to the right of $x=a$ into thin slices. Each slice would be a circle and the part of the circle which was in the sphere corner would be a circular segment. We now intend to add these segments together by integrating from the start of the corner ($x=a$) to the end of the corner ($x=\sqrt{R^2-b^2}$).

In the figure below I've shown a circular segment:

enter image description here

From the previous link we find that the area of a circular segment can be expressed as: $$A_{seg}=R_1^{2}*\cos^{-1}(\frac{r_1}{R_1})-r_1* \sqrt{R_1^2-r_1^2}$$

We convert this to our notation by seeing that $r_1 = b$ and that $y=R_1$, giving: $$A_{seg}=y^{2}*\cos^{-1}(\frac{b}{y})-b* \sqrt{y^2-b^2}$$

As $y=\sqrt{R^2-x^2}$, we can now write the integral we need: $$V=\int_{a}^{\sqrt{R^2-b^2}} \, (R^2-x^2)*\cos^{-1}(\frac{b}{\sqrt{R^2-x^2}})-b* \sqrt{(R^2-x^2)-b^2} \, dx$$

Not a pretty integral. And when I enter it into Wolfram-Alpha, the computational time limit is exceeded. However, Wolfram-Alpha can provide the indefinite version of the integral, which is...uh...also not pretty. But, if you have the patience, you can insert the limits of the integral (i.e. $x=a$ and $x= \sqrt{R^2-b^2}$) into Wolfram's result and you have your answer!

EDIT

I've noticed a slight problem. When $x=\sqrt{R^2-b^2}$ we get divisions by zero in the Wolfram result. As a consequence I've decided to actually write the complete Wolfram result, though with simplifications which Wolfram didn't make. Here it is: $$\int \, (R^2-x^2)*\cos^{-1}(\frac{b}{\sqrt{R^2-x^2}})-b* \sqrt{(R^2-x^2)-b^2} \, dx = \\ -\frac{2}{3}bx \sqrt{-b^2+R^2-x^2}+ \frac{1}{3}b(3R^2-b^2)\tan^{-1}(\frac{-x}{\sqrt{-b^2+R^2-x^2}})+ \\ \frac{2}{3}R^{3}\tan^{-1}(\frac{bx}{R \sqrt{-b^2+R^2-x^2}})- \frac{1}{3}x(x^2-3R^2)\cos^{-1}(\frac{b}{\sqrt{R^2-x^2}})$$

That took half an hour. Anyway, if one looks at where the division with zero occurs, it is in the $tan^{-1}$ parts. If one analyses these parts, one finds for the first instance that $tan^{-1}=>-\frac{\pi}{2}$ as $x=>\sqrt{R^2-b^2}$, and for the second instance that $\tan^{-1}=>\frac{\pi}{2}$ as $x=>\sqrt{R^2-b^2}$. The equation above can therefore be used to get the volume V as a function of a and b, provided of course that $a^2+b^2 \lt R^2$.

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    thank you very much, for your answer and your time, my regards – 2017-02-10