/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 An iterated integral in polar co... [FREE SOLUTION] | 91Ó°ÊÓ

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An iterated integral in polar coordinates is given. Sketch the region whose area is given by the iterated integral and evaluate the integral, thereby finding the area of the region. \(\int_{0}^{\pi / 4} \int_{0}^{2} r d r d \theta\)

Short Answer

Expert verified
The area is \( \frac{\pi}{2} \).

Step by step solution

01

Understand the integral in polar coordinates

The iterated integral is given as \( \int_{0}^{\pi / 4} \int_{0}^{2} r \, dr \, d\theta \). This represents the area of a region in the polar coordinate system, with \( r \) being the radius ranging from 0 to 2, and \( \theta \) being the angle from 0 to \( \frac{\pi}{4} \).
02

Sketch the region

In polar coordinates, the region defined by the given integral is bounded by the radius values between 0 and 2, and the angle \( \theta \) from 0 to \( \frac{\pi}{4} \). This forms a sector of a circle with radius 2, covering an angle of \( \frac{\pi}{4} \) radians.
03

Integrate with respect to \( r \) first

To evaluate the integral, begin by computing the inner integral with respect to \( r \):\[ \int_{0}^{2} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = 2. \]
04

Integrate with respect to \( \theta \)

Now, substitute the result from the initial integral into the outer integral:\[ \int_{0}^{\pi/4} 2 \, d\theta = 2 \cdot \left[ \theta \right]_{0}^{\pi/4} = 2 \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{2}. \]
05

Conclusion

The area of the region defined by the given iterated integral is \( \frac{\pi}{2} \) square units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iterated Integral
An iterated integral involves the sequential integration of a function of more than one variable. In the case of the polar coordinates provided in the exercise, the iterated integral is performed over two variables: the radius \( r \) and the angle \( \theta \).
In this scenario, the integral processes are carried out starting from the innermost integral, which is with respect to \( r \). This calculates the area of infinitely small slices defined by the radius.
Once the first integral is completed, its result becomes part of the outer integral with respect to \( \theta \). This part handles the broader path, or arc, that these slices are forming while sweeping from the angle \( 0 \) to \( \frac{\pi}{4} \). By integrating iteratively, we effectively cover the entire area of a sector which is the focus of this problem.
Area of a Region
Calculating the area of a region using polar coordinates is a powerful method, particularly effective for circular or angular regions. The given iterated integral represents such a calculation resulting in a sector area.
First, with the region defined by \( 0 \leq r \leq 2 \) and \( 0 \leq \theta \leq \frac{\pi}{4} \), you're essentially identifying a slice of a circle.
This slice—or sector—has a radius of 2 and spans an angle of \( \frac{\pi}{4} \) radians (equivalent to 45 degrees). Sketching this can visually aid in understanding how \( r \) and \( \theta \) parameters define a distinct part of the plane.
Ultimately, the area arises from integrating the function \( r \) over the specified limits, resulting in the piece of the circle subtended by the angle \( \frac{\pi}{4} \).
Integration in Polar Coordinates
Integration in polar coordinates is a transformative approach for evaluating integrals over regions better suited for circular measurements.
Converting to polar coordinates involves using \( r \) for the radial distance from the origin and \( \theta \) for the angular position. This method is especially useful when dealing with regions or boundaries that are inherently circular in nature.
The typical differential element in polar coordinates, \( r dr d\theta \), accounts for both the radial and angular expansion of each infinitesimal area element, taking into account that area depends on both radius and angle.
Applying integration in this system, especially through an iterated integral, enables solving for areas, volumes, and other properties of figures shaped around a circular geometry, efficiently and intuitively. The given exercise perfectly exemplifies how switching to polar coordinates streamlines the process of evaluating the complex region formed by specific radius and angle limits.

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