Volumes
Let \(S\) be a three-dimensional solid, placed so that it lies between the vertical planes \(x = a\) and \(x = b\). If the cross sectional area of \(S\) in the plane \(S_x\), through \(x\) and perpendicular to the \(x\)-axis, is \(A(x)\), where:
\(A\) is a continuous function, then the volume of \(S\) is \(V=\int_a^bA(x)dx.\)
If \(S\) is a solid of revolution, then all cross-sections are disks, and \(A(x)=\pi r_{x}^{2}\) where \(r\) is the radius of the cross-sectional disk in the plane \(S_x\) through \(x\).
Rotation about the \(x\)-axis or horizontal line
\[V=\int_{x=a}^{x=b}\pi r^2 dx\] or \[V=\int_{x=a}^{x=b}\pi(r_{outer}^2-r_{inner}^2)dx\]
Example
Find the volume of the solid obtained by rotating the region bounded by the curves \(y=\sqrt{x-1} , y=0\), and \(x=5\) about the \(x\)-axis.
Solution
The cross-sectional area perpendicular to the \(x\)-axis is \(A(x)=\pi (y(x))^2\).
The volume of a slice with thickness \(dx\) is \(dV=A(x)dx=\pi(y(x))^{2}dx .\)
The volume of a given solid of revolution is \[\begin{aligned}
\text{V}& =\int_{1}^{5}A(x)dx=\int_{1}^{5}\pi y^{2} dx=\pi\int_{1}^{5}(\sqrt{x-1})^{2} dx \\
&=\pi\int_{1}^{5}(x-1)dx=\pi\Bigg(\frac{x^{2}}{2}-x\Bigg)\Bigg|_{1}^{5} \\
&=\pi\Bigg[\Bigg(\frac{5^{2}}{2}-5\Bigg)-\Bigg(\frac{1^{2}}{2}-1\Bigg)\Bigg]=8\pi
\end{aligned}\]
Rotation about the \(y\)-axis or vertical line
\[V=\int_{y=a}^{y=b}\pi r^2 dy\]
or
\[V=\int_{y=a}^{y=b}\pi(r_{outer}^2-r_{inner}^2)dy\]
Example
Find the volume of the solid obtained by rotating the region bounded by the curves \(x=y^{2}+1\) and \(x=3\) about the line \(x=3\).
Solution
The cross-sectional area perpendicular to the \(y\)-axis is \[\begin{aligned}
A(y)& =\pi(3-x(y))^{2} \\
&=\pi(3-(1+y^2))^2 \\
&=\pi(2-y^2)^2
\end{aligned}\]
The volume of a slice with thickness \(dy\) is \(dV=A(y)dy=\pi(x(y))^2dy .\)
The volume of the solid object is \[\begin{aligned}
\text{V}& =\int_{-\sqrt{2}}^{\sqrt{2}}A(y)dy=\int_{-\sqrt{2}}^{\sqrt{2}}\pi(2-y^{2})^{2}dy \\
&=\pi\int_{-\sqrt{2}}^{\sqrt{2}}(4-2y^{2}+y^{4})dy \\
&=\pi\Bigg(4y-\frac{2y^{3}}{3}+\frac{y^{5}}{5}\Bigg)\Bigg|_{1}^{5}=\frac{64\sqrt{2}\pi}{15}
\end{aligned}\]