Question

Evaluate the iterated integrals. \(\int_{0}^{\pi} \int_{0}^{2} r \cos \frac{\theta}{4} d r d \theta\)

Short answer

The result of the iterated integrals is \( 4\sqrt{2} \).

Step-by-step solution

Evaluate the Inner Integral

For the inner integral, we focus on \( \int_{0}^{2} r \cos \frac{\theta}{4} \, dr \). Since \( \cos \frac{\theta}{4} \) is a constant with respect to \( r \), we can integrate directly with respect to \( r \). The integral becomes:\[ \int_{0}^{2} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{2}. \]Calculate the definite integral:\[ \frac{(2)^2}{2} - \frac{(0)^2}{2} = 2.\]Thus, the inner integral results in \( 2 \cos \frac{\theta}{4} \).

Evaluate the Outer Integral

Now, substitute the result from the inner integral into the outer integral:\[ \int_{0}^{\pi} 2 \cos \frac{\theta}{4} \, d\theta. \]To solve this, we need to find:\[ 2 \int_{0}^{\pi} \cos \frac{\theta}{4} \, d\theta. \]To evaluate this integral, we use a substitution method. Let \( u = \frac{\theta}{4} \), so \( d\theta = 4 \, du \). Change the limits accordingly: when \( \theta = 0, u = 0 \) and when \( \theta = \pi, u = \frac{\pi}{4} \).

Substitute and Evaluate the Integral

Substitute into the integral:\[ 2 \cdot 4 \int_{0}^{\frac{\pi}{4}} \cos u \, du = 8 \int_{0}^{\frac{\pi}{4}} \cos u \, du. \]Evaluate:\[ 8 \left[ \sin u \right]_{0}^{\frac{\pi}{4}} = 8 \left( \sin \frac{\pi}{4} - \sin 0 \right). \]Since \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) and \( \sin 0 = 0 \), the evaluated integral becomes:\[ 8 \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2}. \]

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify the process of finding integrals, especially when dealing with composite functions. It's analogous to the reverse of the chain rule in differentiation. Here's a basic step-by-step explanation:

• Select a substitution: The goal is to find a substitution that simplifies the expression. In our exercise, we chose to set \( u = \frac{\theta}{4} \) because it reformulates the trigonometric function into something more straightforward to integrate.
• Calculate the differential: Here, if \( u = \frac{\theta}{4} \), then \( d\theta = 4 \, du \). This step helps replace \( d\theta \) in the original integral with \( du \).
• Adjust the limits of integration: This part can be tricky. You need to convert the original limits for \( \theta \) to the new limits for \( u \). In the exercise, with \( \theta = 0 \), \( u = 0 \); with \( \theta = \pi \), \( u = \frac{\pi}{4} \).
• Integrate with respect to \( u \): Once the integral is in terms of \( u \), perform the integration. This usually results in a simpler expression compared to integrating with the original variable.
• Back-substitute if necessary: If the problem requires a solution in the original variable, substitute back to express the answer in terms of the original variable.

Integration by substitution helps break down complex integrals into simpler parts, making them more manageable and paving the way for successful integration.

Definite Integral

A definite integral calculates the "net area" under the curve of a function on a specified interval. Unlike indefinite integrals that lack specific bounds and include a constant of integration \( C \), definite integrals give a value representing the area or accumulation. Let's explore:

• Limits of Integration: The numbers at the top and bottom of the integral sign are the limits of integration, defining the interval over which the integral is calculated. In this case, \( \int_{0}^{\pi} \) and \( \int_{0}^{2} \) were used for the \( \theta \) and \( r \) variables, respectively.
• Evaluating Definite Integrals: Calculate the integral of the function first, then apply these limits by finding the difference of the values at the upper and lower limits. This gives the "net change" in the dependent variable as we vary the variable of integration between its bounds.
• Geometric Interpretation: For functions represented in Cartesian coordinates, a definite integral can be viewed as the area between the curve and the horizontal axis, within the specified bounds. However, with polar coordinates, it might represent a different geometric concept based on the radial distances.

In the context of our problem, integrating with respect to \( r \) first provided an interim result that allowed us to simplify and focus on the calculations concerning \( \theta \) in the second part, precisely using the integral's properties.

Polar Coordinates

Polar coordinates offer a different way to describe locations in a plane, essential when dealing with problems involving circular symmetry. In polar coordinates, each point is defined by a radius \( r \) and angle \( \theta \). Here's how it works and applies:

• Components of Polar Coordinates: The radius \( r \) specifies the distance from the origin to the point, while the angle \( \theta \) indicates the direction from the positive x-axis. Together, they fix any location on the plane.
• Conversion to Cartesian Coordinates: Sometimes, you may need to convert between polar and Cartesian coordinates: \( x = r\cos\theta \) and \( y = r\sin\theta \). This conversion is useful in applying standard integration techniques.
• Multiple Uses and Applications: Polar coordinates are particularly helpful in evaluating integrals with symmetrical circular bounds or when the area of interest is within a circular region like in our exercise.
• Integrating in Polar Coordinates: The integration process often involves multiplying the integrand by \( r \), as shown in our problem, and sometimes requires additional transformations for integration, such as the use of symmetry or trigonometric identities to simplify results further.

Polar coordinates simplify integral evaluation in certain cases, making them crucial in multi-variable calculus, especially in real-world applications involving circular and radial symmetries.