Question

Use the integration capabilities of a graphing utility to approximate to two decimal places the area of the surface formed by revolving the curve about the polar axis. $$ r=\theta, \quad 0 \leq \theta \leq \pi $$

Short answer

The answer will be the numerical result obtained from the graphing utility, rounded to two decimal places. The specific answer cannot be provided in this format as a graphing utility is needed to calculate it.

Step-by-step solution

Identify the formula

For a curve given in polar coordinates, the surface area \( A \) of the surface formed by revolving the curve from \( \theta=a \) to \( \theta=b \) about the polar axis is given by the formula: \[A = 2\pi \int_a^b r \sin(\theta) d\theta\]Here, \( r = \theta \) and the limits of integration are from 0 to \( \pi \).

Substitute the function and the limits into the formula

The formula becomes:\[A = 2\pi \int_0^\pi \theta \sin(\theta) d\theta\]This is the integral that needs to be evaluated.

Evaluate the integral using a graphing utility

To evaluate the integral using a graphing utility, input the function \( \theta \sin(\theta) \) and find the definite integral from 0 to \( \pi \). Round the numerical result to two decimal places.