91Ó°ÊÓ

In mathematics, converting between coordinate systems can make integrating certain functions easier. Polar coordinates represent a point in a plane using an angle and a radius from a reference point, usually the origin. Instead of using Cartesian coordinates \( x \) and \( y \), we use \( r \) (the distance from the origin) and \( \theta \) (the angle from the positive x-axis). This conversion is especially useful when integrating functions over circular regions.
For example, the conversion formulas are:

• \[ x = r \cos(\theta) \]
• \[ y = r \sin(\theta) \]

These relationships help modify the area element \[ dx \, dy \] to \[ r \, dr \, d\theta \]. This change is crucial when dealing with circular regions, like in the given problem where \[ x^2 + y^2 \leq 1 \] becomes \[ 0 \leq r \leq 1 \] and \[ 0 \leq \theta \leq 2\pi \].
Utilizing this transformation simplifies double integral calculations over circular areas, leading us directly to the next concept.

A double integral extends single variable integration to functions of two variables. Here, we integrate over an area in the plane rather than over an interval. For a function \( f(x, y) \), the double integral \[ \iint_D f(x, y) \, dx \, dy \] sums the values of \( f \) over the region \( D \).
In the exercise, the goal is to integrate \( f(x, y) = k\sqrt{1 - x^2 - y^2} \) over the disk defined by \( x^2 + y^2 \leq 1 \):

• Set up the integral: \[ \iint_{x^2 + y^2 \leq 1} k\sqrt{1 - x^2 - y^2} \, dx \, dy = 1 \]
• Convert it to polar coordinates: \[ k \int_{0}^{2\pi} d\theta \int_{0}^{1} r \sqrt{1 - r^2} \, dr = 1 \]

After converting to polar coordinates, the double integral simplifies and separates into two single integrals which are easier to evaluate. The solution involves computing these single integrals and solving for \( k \), as previously detailed.

In probability theory, the marginal density function gives the probability distribution of a subset of a collection of stochastic variables. For example, if \(f(x, y) \) is a joint density function, the marginal density of \( X \) is obtained by integrating out \( y \):
\[ f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy \]
In the problem, one must find the marginal density of \( X \) using the calculated value of \( k \):

• Set up the integral: \[ f_X(x) = \int_{x^2 + y^2 \leq 1} f(x, y) \, dy = \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \frac{3}{2\pi} \sqrt{1 - x^2 - y^2} \, dy \]
• Evaluate the integral carefully, considering symmetry and appropriate substitutions.

This operation condenses the joint distribution \( f(x, y) \) into a format that describes just \( X \), revealing how likely different values of \( X \) are.

Substitution is a powerful technique in calculus, used to simplify the process of integration. When faced with difficult integrals, a substitution can make the integral more manageable.
In the exercise, to solve the radial part of the integral, we use the substitution \( u = 1 - r^2 \). This changes the differential and simplifies the integral:

• Change of variables: \[ u = 1 - r^2, \, du = -2r \, dr \Rightarrow -\frac{1}{2} \, du = r \, dr \]
• New bounds: As \( r \) goes from 0 to 1, \( u \) goes from 1 to 0:

Substituting and solving the integral becomes more straightforward: \[ \frac{1}{2} \int_{0}^{1} \sqrt{u} \, du = \frac{1}{2} \left[ \frac{2u^{3/2}}{3} \right]_{0}^{1}= \frac{1}{3} \]
This technique reduces the complexity of integrals by transforming them into more familiar forms, aiding in quicker and more accurate solutions.