## Volume of a Solid |

Let \(S\) be the solid between the planes \(x=a\) and \(x=b\) such that \(A(x)\) is the area of its vertical cross-section at \(x\). If \(A\) is a continuous function on \([a,b]\), then the volume of \(S\) is \[ \int_a^b A(x)\;dx.\]

Break \([a,b]\) into \(n\) subintervals \([x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]\) where \(x_i=a+i \Delta x\)
and \(\Delta x=(b-a)/n\). Choose \(x_i^*\) from \([x_{i-1},x_i]\), \(i=1,2,\ldots,n\). Partition \(S\) into
\(n\) solids \(S_1,S_2,\ldots,S_n\) by intersecting \(S\) by the vertical planes at \(x_1,x_2,\ldots,x_{n-1}\).
Then the volume of \(S_i\) is approximately \(A(x_i^*) \Delta x\). Therefore the volume of \(S\) is
\[V(S)\approx \sum_{i=1}^n A(x_i^*) \Delta x.\]
This approximation of \(V(S)\) gets better as \(n\to \infty\). Thus
\[V(S)=\lim_{n\to \infty} \sum_{i=1}^n A(x_i^*) \Delta x =\int_a^b A(x)\;dx.\]

**Example.** Prove that the volume of a right circular cylinder with radius \(r\) and height \(h\) is
\(\pi r^2h\).

*Solution.* Consider the right circular cylinder with radius \(r\) that is centered at the origin \((0,0,0)\)
and lies on \(\left[-\frac{h}{2}, \frac{h}{2} \right]\) on the \(x\)-axis. Its vertical cross-section at \(x\)
is a disk which has radius \(r\) and consequently the area \(A(x)=\pi r^2\). Since the cylinder lies on
\(\left[-\frac{h}{2}, \frac{h}{2} \right]\) on the \(x\)-axis, its volume is
\[\int_{-\frac{h}{2}}^{\frac{h}{2}} \pi r^2 \;dx= \left. \pi r^2x \right\vert_{-\frac{h}{2}}^{\frac{h}{2}}
=\pi r^2\frac{h}{2}-\pi r^2\frac{-h}{2}=\pi r^2h. \]

**Example.** Prove that the volume of a sphere of radius \(r\) is \(\frac{4}{3}\pi r^3\).

*Solution.* Consider the sphere of radius \(r\) that is centered at the origin \((0,0,0)\). Its vertical
cross-section at \(x\) is a disk which has radius \(\sqrt{r^2-x^2}\) and consequently the area
\(A(x)=\pi (\sqrt{r^2-x^2})^2=\pi (r^2-x^2)\). Since the sphere lies on \([-r,r]\) on the \(x\)-axis, its volume
is
\[\int_{-r}^r \pi (r^2-x^2) \;dx= \left. \pi\left( r^2x-\frac{x^3}{3} \right) \right\vert_{-r}^r
=\pi\left( r^3- \frac{r^3}{3}\right)-\pi\left( -r^3+ \frac{r^3}{3}\right) =\frac{4}{3}\pi r^3. \]

**Example.** Prove that the volume of a pyramid of height \(h\) whose base is a square of side \(a\) is
\(\frac{1}{3}a^2h\).

*Solution.* Consider the square pyramid with height \(h\) whose base is on the \(xy\)-plane centered at
the origin \((0,0,0)\). Its horizontal cross-section at \(z\) is square region with side, say \(s\). By equating
the ratio of the base and height of two similar triangles \(\triangle ADE\) and \(\triangle ABC\) (see figure),
we have
\[\frac{s/2}{h-z}=\frac{a/2}{h} \implies s=\frac{a(h-z)}{h}=a\left( 1-\frac{z}{h} \right).\]
The horizontal cross-section at \(z\) has the area \[A(z)=s^2=a^2\left( 1-\frac{z}{h} \right)^2=a^2\left( 1-2\frac{z}{h}+\frac{z^2}{h^2} \right).\]

Since the pyramid lies on \([0,h]\) on the \(z\)-axis, its volume is \[\int_0^h a^2\left( 1-2\frac{z}{h}+\frac{z^2}{h^2} \right) \;dz= \left. a^2\left( z-\frac{z^2}{h}+\frac{z^3}{3h^2} \right) \right\vert_0^h = a^2\left( h-\frac{h^2}{h}+\frac{h^3}{3h^2} \right)=\frac{1}{3}a^2h. \]

Last edited