Disc Method

Given a region under a curve \(y=f(x)\) for the interval \(a \leq x \leq b\), revolving this region around the \(x\)-axis gives a three-dimensional figure called a solid of revolution.

To find the volume of this solid of revolution, consider a thin vertical strip with thickness \(\Delta x\) and height \(y=f(x)\) and consider revolving this thin strip around the \(x\)-axis. This strip generates a thin circular disk with radius \(y=f(x)\) and thickness \( \Delta x\), which has volume

\[ \Delta V = \pi y^2 \Delta x = \pi (f(x))^2 \Delta x.\]

The disk method calculates the volume of the full solid of revolution by summing the volumes of these thin circular disks from the left endpoint \(a\) to the right endpoint \(b\) as the thickness \( \Delta x \) goes to \(0\) in the limit. This gives the volume of the solid of revolution:

\[ V = \int_a^b dV = \int_a^b \pi \big(f(x)\big)^2 dx.\]

Volume of a cylinder:

A cylinder with height \(h\) and base radius \(r\) can be thought of as the solid of revolution obtained by revolving the line \(y=r\) around the \(x\)-axis. Using the disk method, find the volume of the cylinder of height \(h\) and base radius \(r\).

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Using the disk method, we revolve the line \(y=r\) around the \(x\)-axis from \(x=0\) to \(x=h\). Then \[ \Delta V = \pi y^2 = \pi r^2 \Delta x\] and the volume of the right circular cylinder is \[ \begin{align} V &= \int_{0}^{h} dV \\&= \int_{0}^{h} \pi r^2 dx \\ & = \left[ \pi r^2 x \right]^{h}_{0} \\ & = \pi r^2 h. \ _\square \end{align} \]

Volume of a sphere:

A sphere can be thought of as the solid of revolution obtained by revolving a semicircle around the \(x\)-axis. Using the disk method, find the volume of the sphere of radius \(r\).

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We can consider the semicircle to be centered at the origin with radius \(r\), which has equation \(y = \sqrt{r^2 - x^2}\). Then \[ \Delta V = \pi y^2 = \pi \big(r^2 - x^2\big) \Delta x\] and the volume of the sphere is \[ \begin{align} V &= \int_{-r}^{r} dV \\&= \int_{-r}^{r} \pi \big(r^2-x^2\big) dx \\ & = \left[ \pi r^2 x - \frac{\pi x^3}{3} \right]^{r}_{-r}\\ &= \pi r^2 \big(r-(-r)\big) - \left( \frac{\pi r^3}{3} - \frac{\pi (-r)^3}{3} \right)\\ & = 2 \pi r^3 - \left(\frac{2 \pi r^3}{3} \right)\\ & = \frac{4\pi r^3}{3}. \ _\square \end{align}\]

Volume of a right circular cone:

A right circular cone can be thought of as the solid of revolution obtained by revolving a right triangle around the \(x\)-axis. Using the disk method, find the volume of the right circular cone of height \(h\) and base radius \(r\).

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The hypotenuse of the right triangle we revolve around the \(x\)-axis is given by \(y = \frac{r}{h}x \). Then \[ \Delta V = \pi y^2 = \pi \left( \frac{r}{h}x \right)^2 \Delta x\] and the volume of the right circular cone is \[ \begin{align} V &= \int_{0}^{h} dV \\&= \int_{0}^{h} \pi \frac{r^2}{h^2 }x^2 dx \\ & = \left[ \pi \frac{r^2}{h^2} \cdot \frac{1}{3} x^3\right]^{h}_{0}\\ & = \frac{ \pi r^2 h}{3}. \ _\square \end{align}\]

As shown in the examples above, we have the following relationships between the volumes of the cone and sphere as fractional parts of the volumes of their respective circumscribed cylinders:

\[ \begin{align} V_\text{cone} & = \frac{1}{3} V_\text{cylinder}\\\\ V_\text{sphere} &= \frac{2}{3} V_\text{cylinder}. \end{align}\]

When the curve \(\displaystyle{y=\frac{1}{x}}\ \ (1\le x \le \infty)\) is revolved about the \(x\)-axis, a funnel-shaped surface is formed. What is the volume of that revolution?

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Using the disk method gives

\[ \Delta V = \pi y^2 = \pi\cdot \frac{1}{x^2} \Delta x,\]

and the volume of the revolution is

\[ \begin{align} V &= \int_{1}^{\infty} dV \\&= \int_{1}^{\infty} \pi\cdot\frac{1}{x^2}dx \\ & =\left.-\pi\cdot\frac{1}{x}\right|_{1}^{\infty}\\ &= -\pi(0-1)\\ & = \pi. \end{align}\]

It is interesting to note that the volume is finite. \(_\square\)