Question

Give the integral formulas for the area of the surface of revolution formed when the graph of \(r=f(\theta)\) is revolved about (a) the \(x\) -axis and (b) the \(y\) -axis.

Short answer

The integral formulas for the area of a surface of revolution formed from \(r=f(\theta)\) are: When revolved about the x -axis, \(A_x = 2 \pi \int_{\theta_1}^{\theta_2} f(\theta) \sin(\theta) \sqrt{1 + (f'(\theta))^2} d\theta\), and when revolved about the y -axis, \(A_y = 2 \pi \int_{\theta_1}^{\theta_2} f(\theta) \cos(\theta) \sqrt{1 + (f'(\theta))^2} d\theta\)

Step-by-step solution

The general formula for surface area of revolution

The surface area \(A\) of a curve described by \(y=f(x)\), \(a \leq x \leq b\), revolved around the x-axis is given by the integral \(A = 2 \pi \int_{a}^{b} y \sqrt{ 1 + (f'(x))^2 } dx \). This formula extends to polar coordinates with some modification.

Revolve around the x-axis

When the curve is revolved around the x-axis, consider \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Through some manipulation we get, \(y = f(\theta) \sin(\theta)\). Therefore when the curve is revolved about the x-axis, the surface area \(A_x\) is given by \(A_x = 2 \pi \int_{\theta_1}^{\theta_2} f(\theta) \sin(\theta) \sqrt{1 + (f'(\theta))^2} d\theta\).

Revolve around the y-axis

Similarly, derived from the manipulation for the y-coordinate, \(x = f(\theta) \cos(\theta)\), when the curve is revolved about the y-axis, the surface area \(A_y\) is given by \(A_y = 2 \pi \int_{\theta_1}^{\theta_2} f(\theta) \cos(\theta) \sqrt{1 + (f'(\theta))^2} d\theta\).