Question

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}} d y d x $$

Short answer

The value of the double integral over the region is \( \frac{\pi a^{3}}{12} \)

Step-by-step solution

Sketch the Region

The domain of integration is given by \(0 \leq x \leq a\) and \(0 \leq y \leq \sqrt{a^{2}-x^{2}}\). This set of inequalities describes a semi-circle with radius \(a\) in the positive xy-plane.

Transform into Polar Coordinates

The region defined can be better expressed in polar coordinates. The transformation into polar coordinates is accomplished by the equations \(x = r \cos(\theta)\), \(y = r \sin(\theta)\). Replacing these into the boundaries of \(y\) in the integral, we obtain \(0 \leq r \leq a\) and \(0 \leq \theta \leq \pi/2\).

Change the Double Integral to Polar Form

The double integral can be rewritten in polar coordinates. The element of area \(dxdy\) becomes \(r dr d\theta\) in polar coordinates. Hence, the given integral can be rewritten to leverage these simpler boundaries as: \[\int_{0}^{\pi/2} \int_{0}^{a} r d r d \theta\]

Evaluate the Integral

We now have a simple double integral that can be separated and evaluated. It is the product of two single integrals that can be solved directly to get:\[\frac{1}{2} \int_{0}^{a} r^{2} d r \int_{0}^{\pi / 2} d \theta = \frac{1}{2} * \frac{a^{3}}{3} * \frac{\pi}{2} = \frac{\pi a^{3}}{12}\]