Questions: Evaluate ∬D x dA, where D is given by x ≥ 0, y ≥ 0, x²+y²=4, x²+y²=2y.

Transcript text: Evaluate $\iint_{D} x d A$, where $D$ is given by \[ x \geqslant 0, y \geqslant 0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=2 y . \]

Solution

Solution Steps

Solution Approach

Identify the region $ D $ bounded by the given curves.
Convert the given equations to polar coordinates for easier integration.
Set up the double integral in polar coordinates.
Evaluate the integral using Python.

Step 1: Identify the Region $ D $

The region $ D $ is defined by the inequalities $ x \geq 0 $, $ y \geq 0 $, and the curves $ x^2 + y^2 = 4 $ and $ x^2 + y^2 = 2y $. The first curve represents a circle of radius 2 centered at the origin, while the second curve can be rewritten as $ x^2 + (y - 1)^2 = 1 $, which is a circle of radius 1 centered at $ (0, 1) $. The region $ D $ is the area in the first quadrant that lies between these two curves.

Step 2: Convert to Polar Coordinates

In polar coordinates, we have:

$ x = r \cos(\theta) $
$ y = r \sin(\theta) $

The equations transform as follows:

$ r^2 = 4 $ gives $ r = 2 $.
$ r^2 = 2r \sin(\theta) $ simplifies to $ r = 2 \sin(\theta) $.

Step 3: Set Up the Integral

The double integral to evaluate $ \iint_{D} x \, dA $ in polar coordinates becomes: \[ \iint_{D} r \cos(\theta) \, r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2 \sin(\theta)} r^2 \cos(\theta) \, dr \, d\theta \]

Step 4: Evaluate the Integral

Evaluating the integral yields: \[ \int_{0}^{\frac{\pi}{2}} \left[ \frac{r^3}{3} \cos(\theta) \right]_{0}^{2 \sin(\theta)} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{(2 \sin(\theta))^3}{3} \cos(\theta) \, d\theta = \frac{8}{3} \int_{0}^{\frac{\pi}{2}} \sin^3(\theta) \cos(\theta) \, d\theta \] Using the identity for $ \sin^3(\theta) $ and integrating gives the final result of $ \frac{2}{3} $.