Question

Evaluate the double integral \(\int_{R} \int f(r, \theta) d A,\) and sketch the region \(R\). $$ \int_{0}^{\pi / 2} \int_{0}^{1+\sin \theta} \theta r d r d \theta $$

Short answer

The detailed solution steps consist of first evaluating the inner integral with respect to \(r\), then evaluating the outer integral with respect to \(\theta\), and at last the sketching of the region R that is bounded by the polar coordinates provided. The final answer of the integral computation is obtained through detailed computation of \( \int_{0}^{\pi / 2} \frac{1}{2}\theta (1 + \sin(\theta))^2 d\theta \).

Step-by-step solution

Compute the inner integral

For the given integral \(\int_{0}^{\pi / 2} \int_{0}^{1+\sin \theta} \theta r d r d \theta\), evaluate the integral with respect to \(r\) first. Using the power rule for integration, which is \(\int x^n dx = \frac{1}{n+1}x^{n+1}\), the integral becomes \(\int_{0}^{\pi / 2} [\frac{1}{2}\theta r^2]_{0}^{1+\sin(\theta)} d\theta\). Replace the limits of \(r\) to get \(\int_{0}^{\pi / 2} \frac{1}{2}\theta (1 + \sin(\theta))^2 d\theta\).

Compute the outer integral

Now, to compute the final integral \(\int_{0}^{\pi / 2} \frac{1}{2}\theta (1 + \sin(\theta))^2 d\theta\), use the formula for integral of functions of the form \(\sin^n(x)\) or \(\cos^n(x)\). Break down the integral into smaller components if needed and compute it carefully.

Sketch the region R

The region R is defined by the limits of the polar coordinates. For \(\theta\), the region ranges from 0 to \(\pi / 2\); meaning it is the first quadrant. For \(r\), the region ranges from 0 to \(1 + \sin(\theta)\). This means it will start from the origin and end on the circle \(r = 1 + \sin(\theta)\) in the first quadrant.

Key concepts

Polar Coordinates Integration

To fully understand double integral calculus, particularly when integrating over a region using polar coordinates, one must grasp the concept of polar coordinates integration. This method is incredibly useful when the region of integration, denoted by 'R', has a circular or sector shape, as is commonly the case in problems involving circular symmetry.

In polar coordinates, a point in the plane is described by an ordered pair \( (r, \theta) \) where \( r \) represents the radius—distance from the origin—and \( \theta \) denotes the angle formed with the positive x-axis. The element of area in polar coordinates, \( dA \) is not as simple as \( dx dy \) in Cartesian coordinates; it takes the form \( r dr d\theta \), emphasizing how the 'size' of a piece of the region depends on how far out from the origin the piece is.

The given exercise requires the evaluation of the double integral \( \int_{0}^{\pi / 2} \int_{0}^{1+\sin \theta} \theta r dr d\theta \), which naturally fits polar coordinates due to the presence of trigonometric function \( \sin \theta \) in the limit of integration for \( r \) and the angle \( \theta \) varying between 0 and \( \pi/2 \)—the first quadrant. \( \theta \) represents the angular sweep while \( r \) varies from the pole to the curve \( r=1+\sin(\theta) \).

Understanding the area element and the limits of integration is crucial for setting up and evaluating a double integral in polar coordinates. To sketch region R, one must locate the region in the polar plane bounded by the given radial and angular constraints. For the given integral, this region lies within the first quadrant and extends radially from the origin to \( r=1+\sin(\theta) \).

Power Rule for Integration

One of the foundational tools in calculus is the power rule for integration. It simplifies the process of finding antiderivatives of polynomials and is expressed as \( \int x^n dx = \frac{1}{n+1}x^{n+1} + C \) where \( n \) is a real number other than \( -1 \) and \( C \) represents the constant of integration. This rule is indispensable when dealing with integrals involving powers of the variable, and it's elegantly used in the inner integral of the given problem.

Consider the first step of the problem: evaluating the integral with respect to \( r \) first within the given limits. Since the integrand is \( \theta r \) and \( \theta \) acts as a constant with respect to \( r \), the power rule is applied to \( r \) alone, stated as \( \int \theta r dr \). This application turns \( r \) into \( r^2 \) upon integration and brings out a coefficient \( \frac{1}{2} \) as per the power rule, resulting in \( \frac{1}{2}\theta r^2 \).

Students often make mistakes by not recognizing when to apply the power rule or by applying it incorrectly to coefficients that are not part of the power of \( x \) (or in this case, \( r \)). It is key to remember that any coefficient not related to the variable being integrated remains unchanged during the application of the power rule.

Integrating Trigonometric Functions

Integrating trigonometric functions is an essential skill in calculus. These functions often appear in problems involving periodic phenomena or geometric contexts like the one presented in the exercise. The process of integrating trigonometric functions may involve applying identities or breaking down complex expressions into simpler, integrable parts.

For example, after completing Step 1 in the given solution, the integral becomes \( \int_{0}^{\pi / 2} \frac{1}{2}\theta (1 + \sin(\theta))^2 d\theta \). To tackle this, one might expand the square and then integrate term by term. However, integrating products of trigonometric functions and polynomials, as in this case, often requires the use of trigonometric identities or integration by parts.

Let's focus on the squared sine term, \( (1 + \sin(\theta))^2 \). When such terms are expanded, they generate a polynomial in \( \sin(\theta) \) which in turn can be integrated using known antiderivatives of sin and cos functions and possibly other trigonometric identities. This approach can be seen as breaking down the integral into smaller, more digestible components which can then be integrated separately.

Understanding the role of trigonometric identities and having the skill to manipulate expressions are invaluable when performing integrations involving trigonometric functions. These techniques are not just mechanical; they often require creativity and a deep understanding of trigonometry and calculus.