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Polar coordinates provide a unique way to represent points in a plane using a distance from a reference point and an angle from a reference direction. This system is particularly useful in cases involving circular regions or angular segments. Instead of the traditional Cartesian coordinates which use

• - x for horizontal displacements
• - y for vertical displacements,

polar coordinates use:

• - \( r \) for the radial distance from the origin,
• - \( \theta \) for the angle measured counterclockwise from the positive x-axis.

These transformations make it easier to handle integrations over circular regions. In our exercise, we were provided a region that forms a quarter of a circle in the third quadrant of the Cartesian plane, which is adeptly represented using polar coordinates.

Cartesian integration is the process of integrating with respect to Cartesian coordinates

• - x and
• - y.

When dealing with a problem set in rectangular domains, Cartesian integrals are straightforward. However, when the area of integration forms a curve, such as part of a circle, the Cartesian method becomes complex. The original exercise involved a quarter-circle defined by

• - x bound between \(-1\) and \(0\)
• - y bound between \(-\sqrt{1 - x^2}\) and \(0\).

This pair of bounds denotes a semi-circular top and vertical linear boundary aligned with the y-axis. That's why transforming into polar coordinates becomes advantageous.

Double integrals are used to compute the volume under a surface in multi-variable calculus. They involve integrating over two different variables, traditionally x and y when working in a Cartesian plane or r and \(\theta\) in a polar plane.
In the exercise, the original integral was in Cartesian form but formed a complex region to integrate over. By converting to polar coordinates, we evaluated using

• \(\int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx\)

which becomes

• \(\int_{\theta_{1}}^{\theta_{2}} \int_{0}^{r} f(r, \theta) \, r \, dr \, d\theta\)

for polar coordinates. This change simplifies the integral because the limits directly match the geometry of the area being integrated. Double integrals in polar form utilize the formula

• \(dA = r \, dr \, d\theta\)

to account for the area differential.

Integral evaluation is key to finding the exact area or volume in calculus problems. After transforming the original Cartesian integral into a polar integral, we proceeded to evaluate it through a series of steps.
First, we handle the inside integral (with respect to r). In our exercise:

• The inner integral of \(\frac{2r}{1+r}\) was transformed using substitution, \( u = 1 + r \).

This transformation made it possible to compute the definite integral smoothly across the bounds we established (from 1 to 2). The result \(2 - 2\ln(2)\) gave us a simplified expression that was then used in the outer integral.
The outer integral concerning \(\theta\) was straightforward since it evaluated the constant over the angular range from \(-\pi\) to \(-\frac{\pi}{2}\). This produced the final result

• \( \frac{(2 - 2\ln(2))\pi}{2} \).

The meticulous transformation and integration allow complex regions of integration to be handled with ease and accuracy.