Question

Evaluate the double integral \(\int_{R} \int f(r, \theta) d A,\) and sketch the region \(R\). $$ \int_{0}^{\pi / 2} \int_{0}^{3} r e^{-r^{2}} d r d \theta $$

Short answer

The value of the double integral is \(-\frac{1}{2}\pi e^{-9} + \frac{\pi}{4}\), and the region \(R\) is a quarter disk with radius 3 in the first quadrant.

Step-by-step solution

Integrate with respect to r

Start by integrating the inner integral (with respect to \(r\)) using the power rule for integration, which states \(\int x^n dx = \frac{1}{n+1}x^{n+1}\). However, this is an atypical case, and applying the power rule directly is not possible. Here, a substituion is necessary. Let \(u = r^2\), so \(du = 2r dr\). Therefore, \(r dr = \frac{1}{2} du\). The new limits of the integral are from \(u = 0\) to \(u = 9\), and the integral becomes \(\frac{1}{2} \int_{0}^{9} e^{-u} du\). The integral of \(e^{-u}\) is \(-e^{-u}\), yielding \(-\frac{1}{2} [e^{-u}]_0^9\), which simplifies to \(-\frac{1}{2}e^{-9} + \frac{1}{2}\).

Integrate with respect to \(\theta\)

Then, take this result and carry out the integration with respect to \(\theta\). This integral becomes \(\int_{0}^{\pi / 2} [-\frac{1}{2}e^{-9} + \frac{1}{2}] d\theta\). The integrals of the constants with respect to \(\theta\) are \(-\frac{1}{2}e^{-9} \theta + \frac{1}{2} \theta\), evaluated between 0 and \(\pi / 2\). This simplifies to \(-\frac{1}{2}\pi e^{-9} + \frac{\pi}{4}\).

Sketch the region R

The region \(R\) is all points (r, \(\theta\)) such that 0 ≤ r ≤ 3 and 0 ≤ \(\theta\) ≤ \(\pi / 2\). This is a quarter of a disk with radius 3 in the first quadrant of the polar coordinate plane.