Question

Sketch the region \(R\) and evaluate the iterated integral \(\int_{R} \int f(x, y) d A .\) $$ \int_{0}^{\pi} \int_{0}^{\pi / 2} \sin ^{2} x \cos ^{2} y d y d x $$

Short answer

The value of the given double integral over region \(R\) is \(\frac{\pi^{2}}{4}\).

Step-by-step solution

Sketch the region R

The integration limits define the region R. For \(x\), it goes from 0 to \(\pi\), and for \(y\), it ranges from 0 to \(\pi / 2\). This is a rectangle in the first quadrant of the x-y plane with vertices at (0,0), (0, \(\pi/2\)), (\(\pi\), 0) and (\(\pi\), \(\pi/2\)).

Evaluate the Inner Integral

Compute the inner integral first, which is with respect to \(y\): \[\int_{0}^{\pi / 2} \sin^{2}(x) \cos^{2}(y) d y.\] Here, \(\sin^{2}(x)\) is constant and can be factored out. Then apply the power-reducing formula \(\cos ^{2}(y) = \frac{1 + \cos (2 y)}{2}\) to simplify the integral: \[\int_{0}^{\pi / 2} \sin^{2}(x) \left( \frac{1+\cos(2y)}{2} \right) d y = \sin^{2}(x) \left(\frac{y}{2} + \frac{\sin(2y)}{4} \right) \Bigg|_0^{\pi/2} .\] Evaluating at the upper and lower limits of \(y\) gives \(\frac{\pi \sin^{2}(x)}{4}\).

Evaluate the Outer Integral

Now evaluate the integral with respect to \(x\): \[\int_0^{\pi} \frac{\pi \sin^{2}(x)}{4} dx .\] Again let's use the power-reducing identity \(\sin^{2}(x) = \frac{1 - \cos(2x)}{2}\). Substitute this into the integral and integrate: \[\int_0^{\pi} \frac{\pi}{4} \left(\frac{1 - \cos(2x)}{2}\right) dx = \frac{\pi}{8} \left[ x - \frac{\sin(2x)}{2} \right]_0^{\pi} .\] Evaluating the function at the upper and lower \(x\) limits results in \(\frac{\pi^{2}}{4}\).