Calculating Volumes with Cylindrical Shells Examples 1

Calculating Volumes with Cylindrical Shells Examples 1

We will now look at some more examples of calculating volumes via the cylindrical shell method. We will use the formula $\int_{a}^{b} 2\pi x f(x) \: dx$ extensively.

Example 1

Calculate the volume of the solid by rotating the region between the functions $f(x) = x$ and $g(x) = x^2$ about the $y$-axis.

The graph below depicts the region we're rotating about the y-axis:

Screen%20Shot%202014-03-16%20at%205.18.23%20PM.png

Notice that we are essentially taking the volume obtained when rotating f(x) around the y-axis and subtracting the volume obtained when rotating g(x) around the y-axis.

First, let's find the points of intersection which are rather easy as f(x) = g(x) when x = 0 or x = 1. So the upper and lower bounds of our integral will be 0 and 1. Now applying our cylindrical shells formula we obtain:

(1)
\begin{align} \mathbf{Volume} = 2\pi \int_0^1 x f(x) \: dx - 2\pi \int_0^1 x g(x) \: dx \\ \mathbf{Volume} = 2\pi \int_0^1 x[f(x) - g(x)] \: dx \\ \mathbf{Volume} = 2\pi \int_0^1 x[x - x^2] \: dx \\ \mathbf{Volume} = 2\pi \int_0^1 x^2 - x^3 \: dx \\ \mathbf{Volume} = 2\pi [\frac{x^3}{3} - \frac{x^4}{4}] \bigg |_0^1 \\ \mathbf{Volume} = 2\pi [\frac{1}{3} - \frac{1}{4} ] \\ \mathbf{Volume} = \frac{\pi}{6} \end{align}

Example 2

Calculate the volume of the solid by rotating the region of area trapped between the function $f(x) = x - x^2$ and $g(x) = 0$ around the line $x = 2$.

This time we are rotating the region bounded by $f$ and the $x$-axis along the line $x = 2$ as illustrated in this diagram:

Screen%20Shot%202014-03-16%20at%205.30.31%20PM.png

This time the radius of our shells are $2 - x$ as show by the graph above. Additionally, the function $f$ has $x$-intercepts at $x = 0$ and $x = 1$, so the area trapped by the curve $f$ and the $x$-axis is bounded on the interval $[0, 1]$. Hence applying the formula we get:

(2)
\begin{align} \mathbf{Volume} = 2\pi \int_0^1 (2 - x)(x - x^2) \: dx \\ \mathbf{Volume} = 2\pi \int_0^1 (x^3 - 3x^2 + 2x) \: dx \\ \mathbf{Volume} = 2\pi [\frac{x^4}{4} - x^3 + x^2 ] \bigg |_0^1 \\ \mathbf{Volume} = 2\pi [\frac{1}{4} - 1+ 1 ] \\ \mathbf{Volume} = \frac{\pi}{2} \end{align}

Example 3

Calculate the volume obtained by rotating the region bounded by $xy = 1$, $x = 0$, $y = 1$, and $y = 3$ around the $x$-axis using cylindrical shells.

We first note that we will be rotating the following region about the x-axis:

Screen%20Shot%202014-04-20%20at%204.53.24%20AM.png

We note that this area is defined by taking the function $x = 1/y = y^{-1}$ and integrating it on the interval 1 ≤ y ≤ 3. The radius of our shells will be [[$ y

(3)
\begin{align} V = \int_1^3 2 \pi y \cdot y^{-1} \: dy \\ V = 2 \pi \int_1^3 1 \: dy \\ V = 2 \pi y \bigg |_1^3 \\ V = 2 \pi (3 - 1) \\ V = 2 \pi (2) \\ V = 4 \pi \end{align}
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