Question

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region \(R\). \(\int_{R} \int\left(x^{2}+y^{2}\right) d A\) \(R\) : semicircle bounded by \(y=\sqrt{4-x^{2}}, y=0\)

Short answer

The integral over the region \(R\) is \(4\pi\).

Step-by-step solution

Sketch the region of integration

The region of integration \(R\) is bounded by the semicircle \(y=\sqrt{4-x^2}\) and the x-axis (y=0). This semicircle is centered at the origin with a radius of 2 and occupies the upper half of the xy-plane. So after sketching the region, we will see that the \(\theta\) in polar coordinates will range from 0 to \(\pi\) and the r will range from 0 to 2.

Rewrite the integral in polar coordinates

When we want to integrate over a circular or semi-circular region, polar coordinates are often the best choice. In polar coordinates \(x = r \cos \theta\) and \(y = r \sin \theta\). The double integral can be rewritten as: \(\int_{0}^{\pi} \int_{0}^{2} ((r \cos\theta)^2+(r \sin\theta)^2)r dr d\theta\). We applied \(r\) as the Jacobian determinant when converting from Cartesian to polar coordinates. Since \((r cos(θ))^2 + (r sin(θ))^2\) can be simplified to \(r^2\), the integral becomes: \(\int_{0}^{\pi} \int_{0}^{2} r^3 dr d\theta\).

Evaluate the integral

This is a separable integral, meaning the integration with respect to \(r\) does not depend on \(\theta\), and vice versa. We can compute it as the product of two single integrals. Integrate first with respect to \(r\): \(\int_{0}^{2} r^3 dr = [0.25 * r^4]_{0}^{2} = 4\). Then, integrate with respect to \(\theta\): \(\int_{0}^{\pi} d\theta = [\theta]_{0}^{\pi} = \pi\). Multiplying these results together, we find the final answer is \(4\pi\).