Volume of Revolution – Shell Method

Shell Method (about the y-axis)

$$V = \int_a^b 2\pi x\, f(x)\, dx$$

When the region is described as a function of $x$ and we revolve around the y-axis, cylindrical shells are usually the cleanest method.

Tip: Radius of each shell = $x$, height of each shell = $f(x)$, thickness = $dx$.

Worked Example

Find the volume of the solid of revolution generated by revolving the region bounded by the graphs of $ y = 2x^{2} + 3 $, $ y = 0 $, and $ x = 2 $ around the y-axis.

1. Identify the region

The region lies under $ y = 2x^{2} + 3 $, above $ y = 0 $, from $ x = 0 $ to $ x = 2 $ (the y-axis closes the region on the left).

2. Choose the method

Revolving around the y-axis → cylindrical shells with respect to $ x $.

3. Set up the integral

$$V = \int_{0}^{2} 2\pi x \bigl(2x^{2} + 3\bigr)\, dx = 2\pi \int_{0}^{2} \bigl(2x^{3} + 3x\bigr)\, dx$$

4. Integrate and evaluate

$$2\pi \left[ \tfrac{1}{2}x^{4} + \tfrac{3}{2}x^{2} \right]_{0}^{2} = 2\pi \left( \tfrac{1}{2}(16) + \tfrac{3}{2}(4) \right) = 2\pi (8 + 6) = 28\pi$$

Final Answer

$ V = 28\pi $

Level 1

Easy

Revolve a region under a quadratic around the y-axis using shells

Find the volume of the solid generated by revolving the region bounded by $ y = x^{2} + 1 $, $ y = 0 $, and $ x = 1 $ around the y-axis.

Show Answer

Setup (shells)

$$V = \int_{0}^{1} 2\pi x (x^{2} + 1)\, dx = 2\pi \int_{0}^{1} (x^{3} + x)\, dx$$

Evaluate

$$2\pi \left[ \tfrac{1}{4}x^{4} + \tfrac{1}{2}x^{2} \right]_{0}^{1} = 2\pi \left( \tfrac{1}{4} + \tfrac{1}{2} \right) = 2\pi \cdot \tfrac{3}{4} = \dfrac{3\pi}{2}$$

$ V = \dfrac{3\pi}{2} $

Find the volume of the solid generated by revolving the region bounded by $ y = 3x^{2} + 2 $, $ y = 0 $, and $ x = 1 $ around the y-axis.

Show Answer

Setup (shells)

$$V = \int_{0}^{1} 2\pi x (3x^{2} + 2)\, dx = 2\pi \int_{0}^{1} (3x^{3} + 2x)\, dx$$

Evaluate

$$2\pi \left[ \tfrac{3}{4}x^{4} + x^{2} \right]_{0}^{1} = 2\pi \left( \tfrac{3}{4} + 1 \right) = 2\pi \cdot \tfrac{7}{4} = \dfrac{7\pi}{2}$$

$ V = \dfrac{7\pi}{2} $

Find the volume of the solid generated by revolving the region bounded by $ y = x^{2} + 4 $, $ y = 0 $, and $ x = 2 $ around the y-axis.

Show Answer

Setup (shells)

$$V = \int_{0}^{2} 2\pi x (x^{2} + 4)\, dx = 2\pi \int_{0}^{2} (x^{3} + 4x)\, dx$$

Evaluate

$$2\pi \left[ \tfrac{1}{4}x^{4} + 2x^{2} \right]_{0}^{2} = 2\pi \left( \tfrac{1}{4}(16) + 2(4) \right) = 2\pi (4 + 8) = 24\pi$$

$ V = 24\pi $

Ready to master volumes of revolution and calculus?

Book a Session with Amy