91Ó°ÊÓ

In polar coordinates, the area of a region can be calculated using a specific formula. This formula is:

• \( A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \)

You will need to identify two key elements:

• The limits of integration, \( \theta_1 \) and \( \theta_2 \), which define the interval over which you're calculating the area. These limits correspond to the angles bounding the region.
• The radial coordinate \( r \), expressed as a function of \( \theta \).

For the exercise, we're interested in the area in the first quadrant of a polar graph, which limits \( \theta \) from 0 to \( \frac{\pi}{2} \). This fact becomes important when setting up the integral to find the area. The region defined by the given polar equation, \(r=2(2-\sin 2 \theta)^{1/2}\), requires squaring \( r \) before substituting it back into the area formula. By applying these values and bounds, you can calculate the area using integration.

Definite integrals are a fundamental concept in calculus, used to compute values over an interval. When solving problems involving the area under a curve, the definite integral is employed to accumulate small areas to find a total. The structure of a definite integral looks like this:

• \( \int_{a}^{b} f(x) \, dx \)

where \( a \) and \( b \) are the bounds of integration, and \( f(x) \) is the function integrated over that domain.
In this exercise, we set up a definite integral in polar coordinates. After determining the range \( \theta \), we integrate \( 4(2-\sin 2\theta) \). The integral is then simplified by breaking it into two separate parts:

• \( \int_0^{\frac{\pi}{2}} 2 \, d\theta \) which results in \( \pi \).
• \( -\int_0^{\frac{\pi}{2}} \sin 2\theta \, d\theta \), which needs further manipulation.

Once calculated, we found that the second integral contributes "1" to the solution, leading to the area calculation with ease.

Trigonometric substitution is a technique used to simplify integrals, particularly those involving square roots and other trigonometric functions. This method replaces a portion of the integrand with a trigonometric identity, easing the process of integration.
In our problem, for the integral involving \( \sin 2\theta \), it is beneficial to adjust the integral using a substitution. We let \( u = 2\theta \), which gives \( du = 2 \, d\theta \). This changes the bounds of integration and simplifies complex trigonometric forms to regular trigonometric identities. After substitution, the integral can be expressed and solved as:

• \( -\frac{1}{2} \int_0^{\pi} \sin u \, du \).

This expression becomes more familiar, leading to a direct integration resulting in \( 0 \). By correctly applying this trigonometric substitution, complex integrals become more manageable, allowing easier computation of the area.