Volumes

The volume of a sphere as an integral

  • We can use definite integrals to calculate volumes of solid.
  • Let's first look at the example of a sphere of radius \(r\).
    • Note that the equation of the sphere of radius \(r\) and centre \((0,0,0)\) is \[ S:x^2+y^2+z^2=r^2. \]
    • The method is to integrate over \(-r\le x\le r\) the function \(A(x)\), which assigns to \(x=x_0\) the area of the cross section of the solid \(S\) and the plane \(x=x_0\).
    • Click here for a demonstration. You can see that this means that we approximate the solid \(S\) with cylinders the base the cross sections.
    • In this particular case, if we fix \(x=x_0\), then from the equation \(x^2+y^2+z^2=r^2\), we get \(y^2+z^2=r^2-x_0^2\).
    • That is, the cross-section at \(x=x_0\) is the circular disk with radius \(\sqrt{r^2-x_0^2}\). That has area \((r^2-x_0^2)\pi\).
    • This implies \(A(x)=(r^2-x^2)\pi\).
    • Using this, we can calculate the volume: \[ V=\int_{-r}^r(r^2-x^2)\pi\,\mathrm dx=\pi(r^2x-\frac13x^3)\Big|_{-r}^r=\pi(r^2(r+r)-\frac13(r^3+r^3))=\frac43r^3\pi. \]

Solids of revolution, part 1

  • More generally, we can consider solids of revolution, that is solids, which we obtain by rotating a region around an axis.
    • If \(f(x)\ge0\) for \(a\le x\le b\), then let \(S\) be the solid obtained by rotating the region under the graph \(y=f(x)\) around the \(x\)-axis.
    • The cross-sections are disks with centre the origin and radius \(f(x)\).
    • Therefore, \(A(x)=f(x)^2\pi\), so that the volume is \(V=\int_a^bf(x)^2\pi\,\mathrm dx\).
    • Warning. In these problems, you have to be careful to make sure you know which axis is the region rotated around. For a given region, you get different solids if you rotate it around the \(x\) and the \(y\)-axes.
  • Let's find the volume of the solid obtained by rotating the region \(R\) under the graph \(y=\sqrt x\) for \(0\le x\le1\).
    • Click here for a demonstration.
    • Since we have \(f(x)=\sqrt x\), the volume will be \[ V=\int_a^bf(x)^2\pi\,\mathrm dx=\int_0^1\pi x\,\mathrm dx=\frac{\pi}{2}. \]

Solids of revolution, part 2

  • More general solids of revolution can be obtained by rotating a region between two curves.
    • If \(0\le f(x)\le g(x)\) for \(a\le x\le b\), then the cross-section is the ring with outer radius \(g(x)\), and inner radius \(f(x)\).
    • Therefore, we have \(A(x)=\pi(g(x)^2-f(x)^2)\), and thus \(V=\int_a^b\pi(g(x)^2-f(x)^2)\,\mathrm dx\).
  • Exercise. Let \(R\) be the region enclosed by the curves \(y=x\) and \(y=x^2\).
    1. Find the volume of the solid of revolution obtained by rotating \(R\) around the \(x\)-axis.
    2. Find the volume of the solid of revolution obtained by rotating \(R\) around the \(y\)-axis.
  • Exercises. 6.1: 2, 6, 8, 10, 12, 16, 48, 49
  • Other solids: 50, 52, 54, 58, 63