A solid \(S\) of revolution is obtained by rotating a region about a line. First suppose \(S\) is obtained by rotating a region about the \(x\)-axis or a line on the \(xy\)-plane parallel to the \(x\)-axis. If \(S\) is between \(x=a\) and \(x=b\) such that \(A(x)\) is the area of its vertical cross-section at \(x\) and \(A\) is a continuous function on \([a,b]\), then the volume of \(S\) is \[V(S)= \int_a^b A(x)\;dx.\] Disk Method: If the vertical cross-section is a disk with radius \(r(x)\), then \(A(x)=\pi r^2\) and \[V(S)= \int_a^b \pi r^2 \;dx.\] Washer Method: If the vertical cross-section is a washer (annular ring) with the inner radius \(r_{in}(x)\) and the outer radius \(r_{out}(x)\), then \(A(x)=\pi \left( r_{out}^2-r_{in}^2 \right)\) and \[V(S)= \int_a^b \pi \left( r_{out}^2-r_{in}^2 \right) \;dx.\]

Example. Find the volume of the solid obtained by rotating the region bounded by \(y=e^{-x}\), \(x=0\), \(x=1\), and \(y=0\) about the \(x\)-axis.

Solution. The cross-section of the solid at \(x\) perpendicular to the \(x\)-axis is a disk with the radius \(e^{-x}\) and the area \(\pi \left( e^{-x}\right)^2=\pi e^{-2x}\). Since the solid is in between \(x=0\) and \(x=1\), its volume is \[ \int_0^1 \pi e^{-2x} \;dx= \left.\pi \frac{e^{-2x}}{-2}\right\vert_0^1=-\frac{\pi}{2} \left( e^{-2}-e^0 \right)=\frac{\pi\left(e^2-1 \right)}{2e^2}. \]

Example. Find the volume of the solid obtained by rotating the region bounded by \(y=\frac{x^2+16}{8}\), \(y=1\), \(x=1\), and \(x=4\) about the line \(y=-1\).

Solution. The cross-section of the solid at \(x\) perpendicular to the \(x\)-axis is a washer with the inner radius \(r_{in}=1-(-1)=2\) and the outer radius \(r_{out}=\frac{x^2+16}{8}-(-1)=\frac{x^2}{8}+3\). Since the solid is in between \(x=1\) and \(x=4\), its volume is \[\begin{align*} \int_1^4 \pi \left( r_{out}^2-r_{in}^2 \right) \;dx &= \int_1^4 \pi \left( \left(\frac{x^2}{8}+3\right)^2-2^2 \right) \;dx\\ &= \int_1^4 \pi \left( \frac{x^4}{64}+\frac{3}{4}x^2+5 \right) \;dx\\ &= \int_1^4 \frac{\pi}{64} \left( x^4+48x^2+320 \right) \;dx\\ &=\left.\frac{\pi}{64} \left( \frac{x^5}{5}+16x^3+320x \right) \right\vert_1^4\\ &=\frac{\pi}{64} \left[\left( \frac{4^5}{5}+16\cdot 4^3+320\cdot 4 \right)- \left( \frac{1^5}{5}+16\cdot 1^3+320\cdot 1 \right)\right]\\ &=\frac{10863 \pi}{320}. \end{align*}\]

The above techniques will be similarly applied to a solid \(S\) obtained by rotating a region about the \(y\)-axis or a line on the \(xy\)-plane parallel to the \(y\)-axis:
If \(S\) is between \(y=c\) and \(y=d\) such that \(A(y)\) is the area of its vertical cross-section at \(y\) and \(A\) is a continuous function on \([c,d]\), then the volume of \(S\) is \[V(S)= \int_c^d A(y)\;dy.\] We apply the disk method and the washer method similarly to find \(A(y)\).

Example. Find the volume of the solid obtained by rotating the region bounded by \(xy=1\), \(x=1\), \(x=2\), and \(y=0\) about the \(y\)-axis.

Solution. The cross-section of the solid at \(y\) perpendicular to the \(y\)-axis is a washer with the inner radius \(r_{in}=1\) and the outer radius \(r_{out}=2\) when \(0\leq y\leq 0.5\) and \(r_{out}=\frac{1}{y}\) when \(0.5\leq y\leq 1\). Since the solid is in between \(y=0\) and \(y=1\), its volume is \[\begin{align*} \int_0^1 \pi \left( r_{out}^2-r_{in}^2 \right) \;dy &= \int_0^{0.5} \pi \left( 2^2-1^2 \right) \;dy +\int_{0.5}^1 \pi \left( \left( \frac{1}{y}\right)^2-1^2 \right) \;dy\\ &= \int_0^{0.5} 3\pi \;dy +\int_{0.5}^1 \pi \left( y^{-2}-1 \right) \;dy\\ &=\left.3\pi y \right\vert_0^{0.5} +\left. \pi\left( -\frac{1}{y}-y \right) \right\vert_{0.5}^1\\ &=3\pi \cdot 0.5+ \pi\left[\left( -1-1 \right)- \left( -\frac{1}{0.5}-0.5\right)\right]\\ &=1.5\pi +0.5\pi\\ &=2\pi. \end{align*}\]

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