# 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 $$0\le x\le1$$.
• Since we have $$f(x)=\sqrt x$$, the volume will be $V=\int_a^bf(x)^2\pi\,\mathrm dx=\int_0^1\pi x\,\mathrm dx=\frac{\pi}{2}.$
• 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.