Question

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

Short answer

The value of the double iterated integral is \( \frac{1}{16} (\pi)^2 - \frac{2}{3} \)

Step-by-step solution

Identify the first integral to evaluate

The given expression is a double integral. The inner integral with respect to \( r \) needs to be evaluated first. We start from: \( \int_{0}^{\pi / 2} \left(\int_{0}^{\sin \theta} \theta r d r\right) d \theta \)

Evaluate the inner integral

To calculate the inner integral, we regard \( \theta \) as a constant because it's with respect to \( r \). The integral of \( r \) with respect to itself is \( r^2/2 \). Therefore:\( \int_{0}^{\sin \theta} \theta r d r = \frac{1}{2} \theta (sin\theta)^2 \)

Substitute the inner integral

Replace the inner integral in the overall integral with the result obtained from the previous step. We get:\( \int_{0}^{\pi / 2} \frac{1}{2} \theta (sin\theta)^2 d \theta \)

Identify the second integral to evaluate

Now we need to evaluate the remaining integral, which is with respect to \( \theta \). The integral is from 0 to \( \pi/2 \) of \( \frac{1}{2} \theta \left[\sin(\theta)\right]^2 \)

Evaluate the remaining integral

To integrate this function, we use integration by parts, with \( u = \theta \) and \( dv = \frac{1}{2}[\sin(\theta)]^2 d\theta \). Therefore:\( \int_{0}^{\pi / 2} \frac{1}{2} \theta (sin\theta)^2 d \theta = \frac{1}{16} (\pi)^2 - \frac{2}{3}\)

Conclusion

The result of the double integral \( \int_{0}^{\pi / 2} \int_{0}^{\sin \theta} \theta r d r d \theta \) is \( \frac{1}{16} (\pi)^2 - \frac{2}{3} \)