Question

In Exercises \(11-22,\) evaluate the iterated integral. $$ \int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r d r d \theta $$

Short answer

The value of the iterated integral is \( \frac{\pi}{2} \).

Step-by-step solution

Inner Integration

To calculate the inner integral \( ∫_0^{2 \cos \theta} r dr \), we will use power rule for integration, \( ∫x^n dx = \frac{x^{n+1}}{n+1} \). Here, \( n = 1 \) and so the integral becomes \(\frac{r^2}{2}\). Now put the limits from 0 to \(2 \cos \theta\), we get \(\frac{(2 \cos \theta)^2}{2} - \frac{0^2}{2} = 2 \cos^2 \theta\). This is the result of the inner integral.

Outer Integration

To calculate the outer integral, we will now integrate the result from step one \( ∫_0^{\pi/2} 2 \cos^2 \theta d\theta \). Using the double-angle identity, \( 2 \cos^2 \theta = 1 + \cos 2 \theta \), we replace this to make the integral simpler and perform the integration: \( ∫_0^{\pi/2} 1 + \cos 2 \theta d\theta = \theta + \frac{1}{2} \sin 2\theta |_0^{\pi/2} = \frac{\pi}{2}\). This is the result of the outer integral.