L13: Volume of Solids of Revolution

Lecture on finding volumes of solids generated by revolving a region about an axis, using the disc method.

Definition

If the area under a curve in a plane is rotated through one complete revolution ($360^{\circ}$ or $2\pi$ radians) about a fixed line in its plane, the solid formed is known as a solid of revolution.

Usually, the $x$-axis, $y$-axis, or any line parallel to these axes is taken to be the axis of revolution.

To get a solid of revolution we start out with a function $y = f(x)$ on an interval $[a, b]$.

The Disc Method

The volume is found by assuming the solid is composed of lots of thin circular discs all aligned perpendicular to the axis of revolution.

A typical disc has:

  • radius $r$ equal to the function value at that point
  • thickness $\delta x$ (or $\delta y$)

Volume of one disc: $$V_{\text{disc}} = \pi r^{2} \cdot (\text{thickness})$$

The total volume is obtained by summing these discs and taking the limit, which gives a definite integral.

General Formulas

Revolving about the $x$-axis

If $y = f(x)$ is continuous and $f(x) \geq 0$ on $[a, b]$, the volume obtained by revolving $360^{\circ}$ about the $x$-axis is:

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

Revolving about the $y$-axis

If $x = g(y)$ is continuous and $g(y) \geq 0$ on $[c, d]$, the volume obtained by revolving $360^{\circ}$ about the $y$-axis is:

$$V = \pi \int_{c}^{d} \bigl[g(y)\bigr]^{2},dy$$

Introductory Example (About the $x$-axis)

Problem: Given $y = 2x$ for $0 \leq x \leq 3$, rotated about the $x$-axis.

Setup:

  • The region is a triangle under the line $y = 2x$ from $x = 0$ to $x = 3$.
  • When rotated about the $x$-axis, it forms a cone.
  • A typical disc perpendicular to the $x$-axis has radius $r = y = 2x$ and thickness $\delta x$.

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

Evaluation: $$V = \pi \left[\frac{4x^{3}}{3}\right]_{0}^{3} = \pi \left(\frac{4(27)}{3} - 0\right) = 36\pi \text{ unit}^{3}$$

Worked Example (About the $y$-axis)

Problem: Find the volume when the region bounded by $y = x^{2}$ and the $y$-axis, for $0 \leq x \leq 2$, is rotated about the $y$-axis.

Setup:

  • Express $x$ as a function of $y$: $x = \sqrt{y}$.
  • When $x = 0$, $y = 0$; when $x = 2$, $y = 4$. So $y$ ranges from $0$ to $4$.
  • A typical disc perpendicular to the $y$-axis has radius $r = x = \sqrt{y}$ and thickness $\delta y$.

Integral: $$V = \pi \int_{0}^{4} \bigl(\sqrt{y}\bigr)^{2},dy = \pi \int_{0}^{4} y,dy$$

Evaluation: $$V = \pi \left[\frac{y^{2}}{2}\right]_{0}^{4} = \pi \left(\frac{16}{2} - 0\right) = 8\pi \text{ unit}^{3}$$

Example 1 — $y = x^{3}$ about the $x$-axis

Problem: Find the volume of the solid obtained by rotating the region bounded by $y = x^{3}$ on the interval $0 < x < 2$ about the $x$-axis.

Setup: $$V = \pi \int_{0}^{2} \bigl(x^{3}\bigr)^{2},dx = \pi \int_{0}^{2} x^{6},dx$$

(Solution left as exercise / continued in next lecture.)

Example 2 — $y = \sqrt{x}$ about the $x$-axis

Problem: Find the volume of the solid generated by revolving the region defined by $y = \sqrt{x}$, $x = 3$, and the $x$-axis, about the $x$-axis.

Setup:

  • Bounds: $[0, 3]$
  • Radius: $r = y = \sqrt{x}$

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

(Solution left as exercise / continued in next lecture.)

Example 3 — $x = \sin(y)$ about the $y$-axis

Problem: Find the volume of the solid generated by rotating the region under $x = \sin(y)$, $0 \leq y \leq \pi$, and the $y$-axis, about the $y$-axis.

Setup: $$V = \pi \int_{0}^{\pi} \bigl(\sin y\bigr)^{2},dy$$

Evaluation: Using the half-angle identity $\sin^{2}y = \frac{1}{2} - \frac{1}{2}\cos(2y)$:

$$V = \pi \int_{0}^{\pi} \left(\frac{1}{2} - \frac{1}{2}\cos(2y)\right) dy$$

$$= \pi \left[\frac{y}{2} - \frac{1}{4}\sin(2y)\right]_{0}^{\pi}$$

$$= \pi \left(\frac{\pi}{2} - 0\right) = \frac{\pi^{2}}{2}$$

Example 4 — $y = 3e^{-x}$ about the $x$-axis

Problem: Find the volume of the solid obtained when the region bounded by $y = 3e^{-x}$, for $0 \leq x \leq 1$, is revolved about the $x$-axis.

Setup:

  • Radius: $r = y = 3e^{-x}$

Integral: $$V = \int_{0}^{1} \pi \bigl(3e^{-x}\bigr)^{2},dx = \int_{0}^{1} \pi \cdot 9e^{-2x},dx$$

(Solution left as exercise / continued in next lecture.)

Key Takeaways

  1. Sketch the region and identify the axis of revolution.
  2. Determine whether to integrate with respect to $x$ or $y$:
    • About $x$-axis → integrate with respect to $x$ (vertical discs)
    • About $y$-axis → integrate with respect to $y$ (horizontal discs)
  3. Express the radius as the appropriate function value.
  4. Set up the definite integral $V = \pi \int r^{2},d(\text{variable})$.
  5. Evaluate the integral.

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