91Ó°ÊÓ

In mathematics, converting Cartesian coordinates to polar coordinates is often useful when working with geometric figures or solving integrals. The Cartesian system represents a spatial location using two values, \(x\) and \(y\), corresponding to horizontal and vertical distances from the origin. In contrast, the polar coordinate system uses \(r\) (the distance from the origin) and \(\theta\) (the angle from the positive x-axis) to describe a location.

The conversion formulas are:

• \(r = \sqrt{x^2 + y^2}\)
• \(\theta = \tan^{-1}(y/x)\)

In the given exercise, the region \(R\) is bounded by \(x^2 + y^2 \leq 9\) and \(y \geq 0\). Converting this to polar form involves substituting and simplifying as follows:

• \(x^2 + y^2 = r^2\), which means \(r^2 \leq 9\), or \(r \leq 3\).
• Since \(y \geq 0\), we have \(r\sin{\theta} \geq 0\), leading to \(0 \leq \theta \leq \pi\).

Thus, the polar representation of the region is \(R = \{(r, \theta): 0 \leq r \leq 3, 0 \leq \theta \leq \pi\}\).

A double integral allows us to calculate the volume under a surface in a two-dimensional region. When expressed in Cartesian coordinates, the integrals are typically in the form \(\iint f(x, y)\,dx\,dy\). Converting to polar coordinates transforms area elements from \(dxdy\) to \(r\,dr\,d\theta\).

In our case, we need to evaluate the integral \(\iint_R 2xydA\). By converting to polar coordinates and using the relationship \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\), the expression \(2xy\) becomes \(2r^2\cos{\theta}\sin{\theta}\).

The transformed integral is:

• \(\int_{0}^{\pi} \int_{0}^{3} 2r^3\cos{\theta}\sin{\theta}\,dr\,d\theta\)

This expresses how the function varies over the circular region \(R\).

The region of integration in a double integral is essential because it defines the area over which we compute the integral. For Cartesian to polar transformations, visualizing and sketching \(R\) is helpful. In this exercise, \(R\) is defined as the upper semicircle with radius 3 centered at the origin.

### Characteristics of the Region:

• The inequality \(x^2 + y^2 \leq 9\) describes a circle with radius 3.
• The condition \(y \geq 0\) restricts us to the upper half.
• In polar form, \(r \leq 3\) and \(\theta\) ranges from \(0\) to \(\pi\), simplifying volume calculations.

This transformed region \(R\) allows using polar coordinates to make evaluating the integral more tractable.

Once the integral is set up in polar coordinates, it's time to evaluate it. The evaluation involves handling both the \(r\) and \(\theta\) variables separately.

#### Evaluating the \(r\)-IntegrationFirst, focus on the integral with respect to \(r\):

• \(\int_{0}^{3} 2r^3\cos{\theta}\sin{\theta}\,dr\)
• The antiderivative is \(\frac{2}{3}r^4\cos{\theta}\sin{\theta}\).
• Evaluating from 0 to 3, we get \(54\cos{\theta}\sin{\theta}\).

#### Evaluating the \(\theta\)-IntegrationNext, handle the integral with respect to \(\theta\):

• \(\int_{0}^{\pi} 54\cos{\theta}\sin{\theta}\,d\theta\)
• This simplifies to \( \int_{0}^{\pi} 27\sin{2\theta} \), using \( \sin{2\theta} = 2\sin{\theta}\cos{\theta}\).
• The antiderivative is \(-\frac{27}{2}\cos{2\theta}\), leading to a result of 27 after evaluation over the interval.

Thus, the integral over the given region \(R\) evaluates to \(27\). This final result reflects the total value of the function within the half-circle. Understanding these computations helps in other problems involving polar coordinate integration.