NEW! Section 6.3 – Volume by Cylindrical Shells
Instructions
- The first set of videos below explain the concepts in this section.
- This page also includes exercises that you should attempt to solve yourself. You can check your answers and watch the videos explaining how to solve the exercises.
- You can find additional practice problems for this section on the Practice Problems page for this Module.
Concepts
- Using the method of cylindrical shells to find the volume of a rotational solid
- Determining whether to use cylindrical shells, disks, or washers to find the volume of a rotational solid
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Find the volume of the solid obtained by rotating the region bounded by \(y=x-x^2\), \(y=0\), about the \(y\) axis.
\(V=\dfrac{\pi}{6}\)
2. Find the volume of the solid obtained by rotating the region bounded by \(y=\displaystyle{{1}\over{x}}\), \(y=-2\), \(x=1\), \(x=4\) about the \(y\) axis.
\(V=36\pi\)
3. Using two different methods, find an integral that gives the volume of the solid obtained by rotating region bounded by \(y=x^2\), \(x=0\) and \(y=2\) about the line \(y=3.\)
Washers: \(\displaystyle V=\int_0^{\sqrt{2}} \pi \left[ \left(3-x^2\right)^2-1\right]\, dx\)
Shells: \( \displaystyle V=\int_0^2 2\pi (3-y)\sqrt{y}\, dy\)
Shells: \( \displaystyle V=\int_0^2 2\pi (3-y)\sqrt{y}\, dy\)
4. Using two different methods, find an integral that gives the volume of the solid obtained by rotating the region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x+y=2\) about the \(x\)-axis.
Washers: \(\displaystyle V=\int_0^1 \pi \left( \sqrt{x} \right)^2\, dx+ \int_1^2 \pi \left( 2-x \right)^2\, dx\)
Shells: \( \displaystyle V=\int_0^1 2\pi y\left(2-y-y^2\right) \, dy\)
Shells: \( \displaystyle V=\int_0^1 2\pi y\left(2-y-y^2\right) \, dy\)