91Ó°ÊÓ

Integral conversion is a powerful technique used in calculus, especially when dealing with complex regions of integration. When solving a double integral, such as \[ \int_{0}^{3} \int_{0}^{\sqrt{9-y^{2}}}(x^{2}+y^{2}) dx \, dy \]converting from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) can significantly simplify the integration process.In polar coordinates, every point in space is represented as a distance from the origin, \(r\), and an angle, \(\theta\). The formulas used for conversion are:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• Differential area element: \(dx \, dy = r \, dr \, d\theta\)

When these transformations are applied, they often turn complex regions into simpler shapes—like circles centered at the origin—which are easily represented in polar coordinates.

The Cartesian to polar transformation helps translate a region of integration from Cartesian axes to polar coordinates. This process becomes indispensable when the region of interest forms a circle or part of it, as in our given problem.Originally, the region of integration is described by the inequalities: \[0 \leq y \leq \sqrt{9-y^2}\] However, inspecting this inequality reveals it forms a semi-circle of radius 3. When shifting to polar coordinates, the task is to redefine these boundaries in terms of \(r\) and \(\theta\).For the current example:

• The range for \(r\) goes from 0 to 3 because \(x^2 + y^2 \leq 9\), forming a circle with radius 3.
• \(\theta\) varies from 0 to \(\frac{\pi}{2}\) since the region for integration is only a quarter of the circle (first quadrant).

This conversion defines a much simpler domain for integration and leads to a more manageable integral setup.

Evaluating double integrals requires integrating a function of two variables over a specified region. Once the integral is adjusted using polar coordinates, the process entails evaluating it step-by-step with respect to \(r\) and then \(\theta\).In our specific integral, after converting:\[ \int_{0}^{\pi/2} \int_{0}^{3} r^3 \, dr \, d\theta \]First, we integrate with respect to \(r\). The inner integral results in:\[ \int_{0}^{3} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{3} = \frac{81}{4} \]Substituting this value into the outer integral, we proceed to integrate with respect to \(\theta\):\[ \int_{0}^{\pi/2} \frac{81}{4} \, d\theta = \frac{81}{4} \left[ \theta \right]_{0}^{\pi/2} = \frac{81}{4} \cdot \frac{\pi}{2} \]This yields the final result:\[ \frac{81 \pi}{8} \]The step-by-step integration clearly reflects how progressing from inner to outer integrals leads to evaluating double integrals efficiently.