Lecture #6: Surfaces of Revolution

Surfaces of Revolution

Given a curve \(f(x) \ge 0\) over the interval \([a, b]\), a surface of revolution is the 2D figure obtained by revolving the graph of \(f(x)\) around an axis.

\[ f(x) = \sqrt{1 - x^2} \,\,\, \textrm{over}\, [-1, 1] \]

Common Surfaces of Revolution

\[ f(x) = \sqrt{1 - x^2} \,\,\, \textrm{over}\, [-1, 1] \]

A sphere has a volume of \(V = \dfrac{4}{3} \pi r^2\) and a surface area of \(SA = 4 \pi r^2\).

Common Surfaces of Revolution

What is the lateral surface area of a cylinder?

Center of Gravity of a Curve

Center of Gravity of a Curve

\[ \begin{aligned} \overline{x}_P = \frac{\displaystyle\sum_{i=1}^n m_i\overline{x}_i}{\displaystyle\sum_{i=1}^n m_i} &\approx \frac{\displaystyle\sum_{i=1}^n x_i^* \sqrt{1 + f'(x_i^*)^2} \Delta x}{\displaystyle\sum_{i=1}^n \sqrt{1 + f'(x_i^*)^2} \Delta x} \\[1em] \overline{y}_P = \frac{\displaystyle\sum_{i=1}^n m_i\overline{y}_i}{\displaystyle\sum_{i=1}^n m_i} &\approx \frac{\displaystyle\sum_{i=1}^n f(x_i^*) \sqrt{1 + f'(x_i^*)^2} \Delta x}{\displaystyle\sum_{i=1}^n \sqrt{1 + f'(x_i^*)^2} \Delta x} \end{aligned} \]

Center of Gravity of a Curve

Suppose that \(f(x)\) is continuous and differentiable on the interval \([a, b]\). Then the center of gravity of the graph of \(f(x)\) is given by:

\[ \begin{aligned} \overline{x} &= \dfrac{\displaystyle\int_a^b x \sqrt{1 + f'(x)^2} dx}{\displaystyle\int_a^b \sqrt{1 + f'(x)^2} dx} & &\textrm{and} & \overline{y} &= \dfrac{\displaystyle\int_a^b f(x) \sqrt{1 + f'(x)^2} dx}{\displaystyle\int_a^b \sqrt{1 + f'(x)^2} dx} \end{aligned} \]

Surfaces of Revolution

If a segment of a plane curve \(\mathcal{C}\) is revolved around an axis that does not intersect the curve (expect possibly at its endpoints), the are \(SA\) of the resulting surface of revolution is given by the product of the length of \(\mathcal{C}\) times the distance \(d\) traveled by the centroid of \(\mathcal{C}\).

\[ SA = (\textrm{distance traveled by the center of gravity}) \times (\textrm{arc length}) \]

Surfaces of Revolution

The surface area of a function \(f(x) > 0\) over \(0 < a < x < b\) when revolved around the \(x\)-axis is:

\[ \begin{aligned} SA &= 2 \pi \overline{y} \times (\textrm{arc length}) \\ &= 2\pi \left( \dfrac{\displaystyle\int_a^b f(x) \sqrt{1 + f'(x)^2} dx}{\cancel{\displaystyle\int_a^b \sqrt{1 + f'(x)^2} dx}} \right) \left( \cancel{\displaystyle\int_a^b \sqrt{1 + f'(x)^2} dx} \right) \\ &= 2\pi \displaystyle\int_a^b f(x) \sqrt{1 + f'(x)^2} dx \end{aligned} \]

Surfaces of Revolution

The surface area of a function \(f(x) > 0\) over \(0 < a < x < b\) when revolved around the \(y\)-axis is:

\[ \begin{aligned} SA &= 2 \pi \overline{x} \times (\textrm{arc length}) \\ &= 2\pi \left( \dfrac{\displaystyle\int_a^b x \sqrt{1 + f'(x)^2} dx}{\cancel{\displaystyle\int_a^b \sqrt{1 + f'(x)^2} dx}} \right) \left( \cancel{\displaystyle\int_a^b \sqrt{1 + f'(x)^2} dx} \right) \\ &= 2\pi \displaystyle\int_a^b x \sqrt{1 + f'(x)^2} dx \end{aligned} \]

Example: Surfaces of Revolution

Example: Prove that the (lateral) surface area of a cylinder of height \(h\) and base radius \(r\) is \(SA = \pi r h\).

Example: Surfaces of Revolution

Example: Prove that the (lateral) surface area of a cone of height \(h\) and base radius \(r\) is \(SA = \pi r l = \pi r \sqrt{r^2 + h^2}\).