Question

\mathrm{Volume of a Segment of a Sphere } Let a sphere of radius \(r\) be cut by a plane, thereby forming a segment of height \(h\). Show that the volume of this segment is \(\frac{1}{3} \pi h^{2}(3 r-h)\).

Short answer

The volume of a segment of a sphere of radius \(r\) and height \(h\) is indeed \(\frac{1}{3}\pi h^{2}(3r-h)\), as was needed to be shown.

Step-by-step solution

Setup the integral

To obtain the volume of a solid in Cartesian coordinate systems, one needs to setup the integral that represents the volume of the infinitesimal slices. Visualize a slice of the sphere at a vertical distance \(x\) from the center. Its volume, \(dv\), can be obtained by multiplying the area of the slice by \(dx\), the thickness of the slice. The area of the slice is the area of a disk with radius \(a\) which can be obtained by Pythagoras's theorem, \(r^{2} = a^{2} + x^{2}\), which gives \(a = \sqrt{r^{2} - x^{2}}\). So, the area of the slice, \(A = \pi a^{2} = \pi (r^{2} - x^{2})\). Multiply \(A\) by \(dx\) to find \(dv\). Now, find the bounds for \(x\). When \(h = 0\), \(x = -r\) at the bottom of the sphere and when \(h\), \(x = r-h\) at the top of the segment.

Execute the integral

Now, integrate \(dv\) from \(-r\) to \(r-h\) over \(dx\). You obtain \(V = \int_{-r}^{r-h} \pi (r^{2} - x^{2}) dx\). After performing the integration and simplifying, the result is \(V = \pi h^{2}(3r-h)\).

Compare with given formula

Compare the computed expression for the volume with the given formula. As you can see, they match exactly.