Subject: Calculus

# Disk Integration

Disk Integration or also called disk method is a means of using definite integral to calculate the volume of a solid whose cross sections are similar. But before we go into further details, let us first define two terms: solid of revolution and axis of revolution. If a region in a plane is revolved about a line or an axis, this results to a solid and is called as solid of revolution, and the line or axis is then called axis of revolution. Take for example the figures shown below. When the rectangle is revolved around the line (axis of revolution), a circular disk is formed (see figure below). We can easily calculate its volume by multiplying the area of the disk with its width or thickness or simply equal to V= \pi R^2 w. Now consider a solid of revolution by revolving the following plane region about x-axis. First, we setup a representative rectangle of height R of thickness \delta x on the plane region. Next, we revolve the rectangle about the x-axis. We then have the figure shown below for one circular disk. To solve the whole volume of the solid of revolution, we take into consideration many rectangles revolved about the same axis. Therefore, we will have an n number of disks to approximate the volume of the solid. This approximation appears to become better and better as \left \| \delta x \right \| \to 0 (meaning, as n \to \infty ). So we can define the volume of the solid as

Volume=\lim_{\left \| \delta x \right \| \to 0} \pi \sum_{i=1}^{n}[R(x)]^2\delta x = \pi \int_{a}^{b}[R(x)]^2dx

A similar formula can be derived if the solid is revolved about the y-axis. In summary, to find the volume of solid of revolution with disk method, you can use either of the following,

### Horizontal Axis of Revolution

Volume = \pi \int_{a}^{b}[R(x)]^2dx

### Vertical Axis of Revolution

Volume = \pi \int_{c}^{d}[R(y)]^2dy In the Disk method, we determine the variable of integration by placing a representative rectangle �perpendicular� to the axis of revolution. If the width of the rectangle is \delta x, integrate with respect to x, and if the width of the rectangle is \delta y, integrate with respect to y.

### Example 1

Find the volume of a parabola rotated about the y-axis from y=0 to y=4. The equation of the parabola is given by f(x) = x^2. with respect to x.

Solution: The axis of rotation is along the y-axis, so we will use the following formula;

Volume = \pi \int_{c}^{d}[R(y)]^2dy

Now, R(y) is just the y-dependent version of the function itself. Since f(x)=y=x^2, then x = \sqrt{y} = f(y).

Volume = \pi \int_{c}^{d}[f(y)]^2dy = \pi \int_{c}^{d}ydy

We are interested of the volume from y=0 to y=4, so

Volume = \pi \int_{0}^{4}ydy
=\pi \frac{y^2}{2}|_{0}^{4}
Volume = 8\pi

### Example 2

In the previous example, we rotate the parabola around the y-axis, now find the volume of the same parabola rotated about the x-axis from x=0 to x=3.

Solution: The axis of rotation is along the x-axis, so we will use the following formula;

Volume = \pi \int_{a}^{b}[R(x)]^2dx

Now, R(x) is just the x-dependent version of the function itself. Since f(x)=y=x^2=R(x),

Volume = \pi \int_{a}^{b}[f(x)]^2dx = \pi \int_{a}^{b}x^4dx

We are interested of the volume from x=0 to x=3, so

Volume = \pi \int_{0}^{3}x^4dx
=\pi \frac{y^5}{5}|_{0}^{3}
Volume = 48.6\pi

### Example 3

What solid region can be generated by rotating the line y=ax around x or y axis? The value of a is constant.

ANSWER: Solid cone. ### Example 4

Find the volume of the region generated by rotating the function y=\sin{x} around x-axis from x=0 to x=\pi.

Solution: The axis of rotation is along the x-axis, so we will use the following formula;

Volume = \pi \int_{a}^{b}[R(x)]^2dy

Now, R(x) is just the function itself.

Volume = \pi \int_{a}^{b}[f(x)]^2dx = \pi \int_{a}^{b}\sin^{2}{x}dx

We are interested of the volume from x=0 to x=\pi, so

Volume = \pi \int_{0}^{\pi}\sin^{2}{x}dx

The integral of \sin^{2}{x}dx is just \frac{x}{2}+\frac{\sin{2x}}{4}. Then

Volume=\pi (\frac{x}{2}+\frac{\sin{2x}}{4})|_{0}^{\pi}
Volume = \frac{\pi^2}{2}+1

### Example 5

From example #4, deduce the volume of a region generated by the same function around the same axis from x=0 to x=2\pi without actually computing for the integral. You can use the answer from the previous example.

Solution: Looking at the picture below, we can clearly see that when the point of integration ranges from zero to 2\pi, two chunks of solid regions appear. The first chunk (pink colored) corresponds to the region from zero to \pi. Upon careful observation, the two chunks are the same in dimensions as well as in volume. Thus, the volume of the generated volume only increases twice when the point of integration is extended to 2\pi.

And from example 4, we solved the volume to be

Volume = \frac{\pi^2}{2}+1
.

Since the volume increases twice, the new volume would now be

V = 2(\frac{\pi^2}{2}+1)
V = \pi^2+2
.

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