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

For revolving around the x-axis, the integral is \(S = 2\pi\int_a^b f(\theta) | \sin\theta | \sqrt{1+ \left(\frac{f'(\theta) \cos\theta - f(\theta) \sin\theta}{f(\theta) \cos\theta}\right)^2} f(\theta) \cos\theta\ d\theta\). For revolving around y-axis, the integral is \(S = 2\pi\int_c^d f(\theta) | \cos\theta | \sqrt{1+ \left(\frac{f'(\theta) \sin\theta + f(\theta) \cos\theta}{f(\theta) \sin\theta}\right)^2} f(\theta) \sin\theta\ d\theta\).

Step-by-step solution

Identify Polar Coordinates

In polar coordinates, the function is \(r = f(\theta)\). A point is defined by the distance \(r\) from the origin and the angle \(\theta\) it makes with the x-axis.

Convert To Cartesian Coordinates For Revolving Around the X-Axis

The conversion from polar to Cartesian coordinates results in: \(x = r \cos\theta = f(\theta) \cos\theta\) and \(y = r \sin\theta = f(\theta) \sin\theta\). The surface area \(S\) we are looking for is described by the formula: \(S = 2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx\), but we need to find the derivative in terms of \(\theta\). Therefore we differentiate \(y\) with respect to \(x\), apply the resulting derivative to the surface area formula, and then achieve the integral.

Formulate the Integral for Revolving Around the X-Axis

After the differentiation, we find the integral that gives us the surface area for the revolution around the x-axis: \(S = 2\pi\int_a^b f(\theta) | \sin\theta | \sqrt{1+ \left(\frac{f'(\theta) \cos\theta - f(\theta) \sin\theta}{f(\theta) \cos\theta}\right)^2} f(\theta) \cos\theta\ d\theta\).

Convert To Cartesian Coordinates For Revolving Around the Y-Axis

Again, the conversion from polar to Cartesian coordinates results in: \(x = r \cos\theta = f(\theta) \cos\theta\) and \(y = r \sin\theta = f(\theta) \sin\theta\). Now we refer to the surface area formula with respect to the y-axis: \(S = 2\pi\int_c^d x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy\), where we need to find the derivative in terms of \(\theta\). So we differentiate \(x\) with respect to \(y\), apply the resulting derivative to the surface area formula, and then achieve the integral.

Formulate the Integral for Revolving Around the Y-Axis

With the derivative, we get the integral that gives us the surface area for revolution around the y-axis: \(S = 2\pi\int_c^d f(\theta) | \cos\theta | \sqrt{1+ \left(\frac{f'(\theta) \sin\theta + f(\theta) \cos\theta}{f(\theta) \sin\theta}\right)^2} f(\theta) \sin\theta\ d\theta\).