Problem 6 - APEX Calculus

Question

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. \(f(x, y)=4 ; R\) is the region enclosed by the petal of the rose curve \(r=\sin (2 \theta)\) in the first quadrant.

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} \).

Step 1: Understand the Problem

We need to compute the double integral \( \iint_{R} f(x, y) \, dA \) over a given region \( R \), where \( f(x, y) = 4 \). The region \( R \) is the area enclosed by the petal of the rose curve \( r = \sin(2\theta) \) in the first quadrant.

Step 2: Recognize the Region in Polar Coordinates

The rose curve \( r = \sin(2\theta) \) forms a petal in the first quadrant between \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \). The radial component \( r \), in this case, varies from 0 to \( \sin(2\theta) \) for a given \( \theta \).

Step 3: Set up the Integral in Polar Coordinates

To express the integral in polar coordinates, use \( x = r\cos\theta \) and \( y = r\sin\theta \). The differential area element \( dA \) becomes \( r \, dr \, d\theta \). Thus, the double integral is set up as: \[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin(2\theta)} 4r \, dr \, d\theta \]

Step 4: Evaluate the Inner Integral

First, integrate with respect to \( r \): \[ \int_{0}^{\sin(2\theta)} 4r \, dr = \left[ 2r^2 \right]_{0}^{\sin(2\theta)} = 2(\sin(2\theta))^2 \]

Step 5: Simplify the Expression

Recall that \( (\sin(2\theta))^2 = \frac{1 - \cos(4\theta)}{2} \), so the result from Step 4 becomes: \[ 2 \cdot \frac{1 - \cos(4\theta)}{2} = 1 - \cos(4\theta) \]

Step 6: Evaluate the Outer Integral

Now, integrate with respect to \( \theta \): \[ \int_{0}^{\frac{\pi}{2}} (1 - \cos(4\theta)) \, d\theta \] Breaking this into simpler parts: \[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos(4\theta) \, d\theta \]

Step 7: Compute Each Part

For \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta \), the result is simply \( \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \).For \( \int_{0}^{\frac{\pi}{2}} \cos(4\theta) \, d\theta \), use the substitution \( u = 4\theta \), which gives: \[ \frac{1}{4} \int_{0}^{2\pi} \cos(u) \, du = \frac{1}{4} [\sin(u)]_{0}^{2\pi} = 0 \]

Step 8: Compute the Final Result

Combine the results from Step 7: \[ \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Therefore, the value of the double integral is \( \frac{\pi}{2} \).

Key Concepts

Double Integral

A double integral is a powerful tool in calculus that allows us to compute volumes under surfaces or find areas of more complex shapes. This concept extends the idea of a single integral, which deals only with functions of a single variable, to functions of two variables, say \( f(x, y) \). The double integral of a function over a region \( R \) is represented as \( \iint_R f(x, y) \, dA \).

Here's a breakdown of what this means:

• \( f(x, y) \) is the function you are integrating.
• \( R \) is the specific region in the plane over which you are integrating.
• \( dA \) denotes a small element of area in the plane.

In simpler terms, you're essentially summing up an infinite number of tiny rectangles (or boxes in 3D) over a specified region \( R \). Depending on the function and region, this can compute things like total mass, area, or electric charge, as determined by the specific setup in applied problems.

Polar Coordinates

Polar coordinates are a system of mathematical coordinates that works well for integrating regions with circular symmetry, like circles or petals. In this system, points are determined by a radius \( r \) and an angle \( \theta \) instead of \( x \) and \( y \, \) as in Cartesian coordinates.

When using polar coordinates, you convert the original \( x \) and \( y \) values using the formulas:

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

Also, when setting up the integral in polar coordinates, the area element \( dA \) is replaced by \( r \, dr \, d\theta \). This accounts for the wedge-shaped area sweeping out in polar plots, which is different from the rectangular area sweep in Cartesian coordinates. This conversion is essential when dealing with shapes like circles, ovals, and rose curves.

Rose Curve

A rose curve is a fascinating graph that looks like petals forming a flower. It's described using polar coordinates by the equation \( r = \sin(n\theta) \) or \( r = \cos(n\theta) \), where \( n \) determines the number of petals.

Key insights for rose curves:

• If \( n \) is even, the rose curve will have \( 2n \) petals.
• If \( n \) is odd, the rose curve will have \( n \) petals.

In this exercise, the curve \( r = \sin(2\theta) \) is used, forming a petal in the first quadrant from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \). Creaming these petals using polar coordinates simplifies finding regions underneath them.

Trigonometric Substitution

Trigonometric substitution is a useful technique for simplifying integrals, especially when the integrand involves complex trigonometric expressions. This technique exploits trigonometric identities to transform the integrand into a simpler form.

For example:

• We used the identity \( (\sin(2\theta))^2 = \frac{1 - \cos(4\theta)}{2} \) to simplify expressions involving \( (\sin(2\theta))^2 \).

This reduces the complexity of the integral and makes calculations more manageable. During substitution, ensure you account for all elements, such as updating the limits of integration and substituting back to the original variables after evaluation. It is a powerful strategy in calculus, especially when handling integrals with curves and circles.