Volumes of Solids of Revolution — Disks & Washers

In AP Calculus BC, one of the major applications of definite integrals is finding volumes of solids of revolution—3D shapes formed when a region (2D area) is rotated around an axis.

Solids of Revolution

When you rotate a 2D area around an axis (usually the x-axis or y-axis), the shape sweeps out a solid.

Example: Rotating the area under a curve around the x-axis forms something like a bowl or cone.

Disk Method

Use this when:

• The region touches the axis of rotation.

• No inner radius → solid is filled.

Formula (rotating around the x-axis)

If the region bounded by $$y=f(x) , f(x)\ge0$$ is rotated around the x-axis, the volume is:

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

Why this works?

• At each x-value, rotation forms a circular disk with:

  • radius = f(x)

  • area = $$\pi(f(x))^{2}$$

• Add up (integrate) all disks → total volume.

Example

Rotate $$y=\sqrt{x}$$​ from x=0 to x=4 around the x-axis.

V=$$\pi\int_{0}^{4}(\sqrt{x})^{2}dx=\pi\int_{0}^{4} x dx=\pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=\pi\cdot8=8\pi$$

Washer Method

Use this when:

• Region does NOT touch axis of rotation.

• You must subtract inner volume from outer volume.

Formula (x-axis rotation)

Suppose:

• Outer radius: R(x)

• Inner radius: r(x)

V=$$\pi\int_{a}^{b}[R(x)^{2}-r(x)^{2}]dx$$

Interpretation

• $$\pi{R(x)^{2}}$$ → area of outer circle

• $$\pi{r(x)^{2}}$$ → area of inner hole

• Subtract → area of washer.

Example

Region between:

• $$y=4$$

• $$y=x^2$$

Rotate around x-axis.

Outer radius: R(x)=4
Inner radius: r(x)=$$x^2$$

V=$$\pi\int_{-2}^{2}(4^2-(x^2)^2)dx=\pi\int_{-2}^{2}(16-x^4)dx$$

Rotation Around the y-axis

Same ideas but solve radii as functions of y.

Disk (around y-axis)

If x=g(y), the formula becomes:

V=$$\pi\int_{a}^{b}[g(y)]^2dy$$

Washer (around y-axis)

V=$$\pi\int_{a}^{b}[R(y)^2 – r(y)^2]dy$$

Important:
You must express the functions in terms of y, or switch to the shell method (next topic if you want it).