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Section 5.2 Volumes by Slicing

Figure 5.2.1.
How can we find the volume when the above region is revolved around the x-axis?

Subsection 5.2.1 Volumes of Solids of Revolution -- Disk Method

If we cut the previous graph up into small, evenly spaced verticle strips, and rotate them around the x-axis, we would get a series of disks. The equation for the volume of a disk is \(V_{disk}=\pi r^2 h\text{.}\) We can apply this to our graph by using the following:

  • \(r=f(x_k^*)\) where \(f(x_k^*)\) is the height of each individual "strip"

  • \(h=\Delta x\) where \(\Delta x \) is the width of each individual "strip"

  • This gives us the equation: \(V_{n,disk}=\pi(f(x_k^*))^2 \Delta x\)

So how do we find the total volume of all disks?

\begin{align*} V_{total}\amp=V_{1,disk}+V_{2,disk}+...+V_{n,disk}\\ \amp = \pi(f(x_1^*))^2 \Delta x +\pi(f(x_2^*))^2 \Delta x +...+\pi(f(x_n^*))^2 \Delta x \\ \amp=\displaystyle\sum_{k=1}^{n} \pi(f(x_k^*))^2 \Delta x \end{align*}

To find the exact volume of a solid of revolution, we must take the limit of the above equation, turning this into a definite integral.

\begin{align*} V \amp = \displaystyle\lim\limits_{n\to\infty} \displaystyle\sum_{k=1}^{n} \pi(f(x_k^*))^2 \Delta x\\ \amp = \int_a^b \pi (f(x))^2dx \end{align*}
Example 5.2.1.

Let's find the volume of the function \(y=\sqrt{x}\) from 1 to 9 revolved about the x-axis:

\begin{align*} V\amp=\int_a^b\pi(f(x))^2dx\\ \amp=\int_1^9\pi(\sqrt{x})^2dx\\ \amp=\pi\int_1^9xdx\\ \amp=\pi\left[\dfrac{x^2}{2}\right]_1^9\\ \amp=\pi\left(\dfrac{(9)^2}{2}-\dfrac{(1)^2}{2}\right)\\ \amp=\pi\left(\dfrac{80}{2}\right)\\ \amp=40\pi \end{align*}

Subsection 5.2.2 Volumes of Solids of Revolution -- Washer Method

Figure 5.2.2.
How do we find the volume of the solid of revolution in this case?

Similarly to the disk method, we need to cut up the functions into even subintervals. In this case though, when we revolve these subintervals around the x-axis, we end up with a washer instead of a disk.

What is the volume of the washer? We can think of it as the volume of a larger disk minus the volume of a smaller disk.

  • \(r_1\) = radius of big disk

  • \(r_2\) = radius of small disk

  • \(\displaystyle V_{washer} = \pi r_1^2 h - \pi r_2^2 h \)

  • \(\displaystyle V_{washer} = \pi \left( r_1^2 - r_2^2 \right) h \)

Now, we need to convert the variables to our equation to find the total volume:

  • \(\displaystyle r_1 = f(x_k^*)\)

  • \(\displaystyle r_2 = g(x_k^*)\)

  • \(\displaystyle h=\Delta x\)

So, \(V_n = \pi \left( (f(x))^2 -(g(x))^2 \right) \Delta x\)

Finally, the exact volume:

\begin{equation*} V = \int_a^b \pi \left( (f(x))^2 - (g(x))^2 \right) dx \end{equation*}
Example 5.2.2.

Let's set up the integral to find the volume of the image above revolved around the x-axis. The area is bound above by \(f(x)=sin(x)+3\text{,}\) below by \(g(x)=cos(x)+2\text{,}\) to the left by \(x = 1\text{,}\) and the right by \(x = 4\text{.}\)

\begin{align*} V \amp = \int_a^b \pi \left( (f(x))^2 - (g(x))^2 \right) dx\\ \amp = \int_1^4 \pi \left( (sin(x)+3)^2 - (cos(x)+2)^2 \right) dx\\ \amp = \int_1^4 \pi \left( (sin^2(x)+6sin(x) + 9) - (cos^2(x)+4cos(x)+4) \right) dx\\ \amp = \int_1^4 \pi \left(sin^2(x)+6sin(x) + 9 - cos^2(x)-4cos(x)-4 \right) dx\\ \amp = \pi \int_1^4 sin^2(x)+6sin(x) - cos^2(x)-4cos(x)+5 dx \end{align*}

Subsection 5.2.3 Volumes of Solids of Revolution -- Shell Method

Figure 5.2.3.
What if we wanted to revolve a function around the y-axis? We can approach this in a similar manner as the previous two methods. Again, we will split the function into equally sized subintervals along the x-axis. If we revolve one of these subintervals around the y-axis, we get a cylindrical shell.

If we "peel" open this shell, we roughly get a box shaped like a stick of gum. We can find the volume of the box by using the equation: \(V_{shell} \approx V_{box} = l \cdot w \cdot h\text{.}\) Now we need to changes this equations into terms of the fucntion:

  • \(w = \Delta x\) (the thickness of the shell)

  • \(l = 2 \pi r = 2 \pi x\) (r is equal to x -- this becomes the radius)

  • \(h = f(x)-g(x) \) (the distance between the two function)

So, \(V_{shell} \approx 2 \pi x \left(f(x) - g(x)\right)\Delta x \)

Now the only thing left to do is to plug this into a definite integral:

\begin{equation*} V = \int_a^b 2\pi x \left(f(x) - g(x)\right) dx \end{equation*}
Example 5.2.3.

Let's set up the integral to find the volume of the image above revolved around the y-axis. The area is bound above by \(f(x)=x^2\text{,}\) on the left by \(x=0\text{,}\) and on the right by \(x=2\text{.}\)

\begin{align*} V \amp = \int_a^b 2\pi x \left(f(x) - g(x)\right) dx\\ \amp = \int_0^2 2\pi x \left(x^2\right) dx\\ \amp = 2\pi \int_0^2 x^3 dx \\ \amp = 2 \pi \left[ \dfrac{x^4}{4}\right]_0^2\\ \amp = 2 \pi \left( \dfrac{(2)^4}{4}- \dfrac{(0)^4}{4}\right)\\ \amp = 2 \pi \left( \dfrac{16}{4} \right)\\ \amp = 8 \pi \end{align*}