Question

Evaluate the double integral \(\int_{R} \int f(r, \theta) d A,\) and sketch the region \(R\). $$ \int_{0}^{\pi / 4} \int_{0}^{4} r^{2} \sin \theta \cos \theta d r d \theta $$

Short answer

The value of the double integral is \(\frac{32}{6}\) and the region \(R\) is a sector of a circle in the first quadrant, bounded by the angles \(\theta = 0\) and \(\theta = \frac{\pi}{4}\), within a radius of 4.

Step-by-step solution

Integrate with respect to \(r\)

The first step is to evaluate the inner integral with respect to \(r\). So, the integral becomes: \[ \int_{0}^{4} r^{2} dr\] After evaluating the above integral, we get \(\frac{1}{3}r^3\). Now, we can substitute the limits of \(r\) (from 0 to 4) to get \(\frac{64}{3}\). Thus, our given double integral simplifies to: \[ \int_{0}^{\pi / 4} \frac{64}{3} \sin \theta \cos \theta d\theta \]

Integrate with respect to \(\theta\)

Now, we will integrate with respect to \(\theta\). Remember that \(\sin 2\theta = 2\sin\theta\cos\theta\), so we can simplify the integral as: \[\int_{0}^{\pi / 4} \frac{64}{6} \sin 2\theta d\theta\] The integral of \(\sin 2\theta\) from 0 to \(\frac{\pi}{4}\) is \(\frac{1}{2}\). Therefore, the whole integral evaluates to: \(\frac{32}{6}\).

Sketch the region \(R\)

The region \(R\) corresponding to the integration limits is a sector of a circle. The boundary of the sector is defined by the radial lines \(\theta = 0\) and \(\theta = \frac{\pi}{4}\), and the circle of radius 4. It is located in the first quadrant of the polar coordinate system, enclosed between the angle 0 and \(\frac{\pi}{4}\), going from the origin to a radius of 4.