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Magazine Chapter 4: Problem 39
In the following exercises, evaluate the double integral \(\iint_{R} f(x, y) d A\) over the polar rectangular region \(D\). \(\iint_{D} e^{x^{2}+y^{2}}\left[1+2 \arctan \left(\frac{y}{x}\right)\right] d A, D=\left\\{(r, \theta) \mid 1 \leq r \leq 2, \frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\right\\}\)
Short Answer
Expert verified
The value of the double integral is \( \frac{\pi + \pi^2}{12}(e^4 - e) \).
Step by step solution
01
Understanding the Region D
The region \( D \) given is in polar coordinates, defined by \( 1 \leq r \leq 2 \) and \( \frac{\pi}{6} \leq \theta \leq \frac{\pi}{3} \). This describes a sector of an annulus (a ring) centered at the origin, with an inner radius of 1, an outer radius of 2, and angles ranging between \( \frac{\pi}{6} \) and \( \frac{\pi}{3} \).
02
Convert the Function to Polar Coordinates
The integrand includes \( e^{x^2 + y^2} \) and \( \arctan \left( \frac{y}{x} \right) \). In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \). Thus, \( x^2+y^2 = r^2 \) and \( \arctan\left(\frac{y}{x}\right) = \theta \). So the integrand becomes \( e^{r^2} (1 + 2\theta) \).
03
Set Up the Double Integral in Polar Coordinates
In polar coordinates, the area element \( dA \) is \( r \, dr \, d\theta \). Therefore, the integral becomes:\[\int_{\theta=\frac{\pi}{6}}^{\frac{\pi}{3}} \int_{r=1}^{2} e^{r^2}(1 + 2\theta) \cdot r \, dr \, d\theta.\]
04
Evaluate the Inner Integral (with respect to r)
For the inner integral, fix \( \theta \) and integrate with respect to \( r \):\[\int_{1}^{2} e^{r^2}(1 + 2\theta) \, r \, dr.\]Use the substitution \( u = r^2 \), thus \( du = 2r \, dr \) or \( r \, dr = \frac{1}{2}du \). The limits of integration change from 1 to 4 (since \( u = r^2 \)). The integral becomes:\[\frac{1}{2} \int_{1}^{4} e^u(1 + 2\theta) \, du.\]This simplifies to:\[\frac{1}{2}(1 + 2\theta) \left[ e^u \right]_{1}^{4} = \frac{1}{2}(1 + 2\theta)(e^4 - e^1).\]
05
Evaluate the Outer Integral (with respect to 胃)
Now integrate the result from the inner integral with respect to \( \theta \):\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{2}(1 + 2\theta)(e^4 - e^1) \, d\theta.\]This can be broken down into two integrals:\[\frac{1}{2} (e^4 - e) \left[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, d\theta + 2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \theta \, d\theta \right].\]The first integral evaluates to \( \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \), and the second integral evaluates to:\[\left. \theta^2 \right|_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \left( \frac{\pi}{3} \right)^2 - \left( \frac{\pi}{6} \right)^2 = \frac{\pi^2}{9} - \frac{\pi^2}{36} = \frac{3\pi^2}{36} = \frac{\pi^2}{12}.\]
06
Compute the Final Value
Combine the results of the outer integral:\[\frac{1}{2}(e^4 - e) \left[ \frac{\pi}{6} + 2 \cdot \frac{\pi^2}{12} \right] = \frac{1}{2}(e^4 - e) \left[ \frac{\pi}{6} + \frac{\pi^2}{6} \right].\]This simplifies to:\[\frac{1}{2}(e^4-e) \cdot \frac{\pi + \pi^2}{6} = \frac{\pi + \pi^2}{12}(e^4 - e).\] This is the value of the double integral over the given region.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates Simplified
Polar coordinates offer a refreshing way to handle integrals, especially when dealing with circular or cylindrical regions. Instead of using the conventional Cartesian coordinates \(x, y\), polar coordinates represent positions in a plane using the distance from the origin, \(r\), and the angle from a fixed direction, \(\theta\).
This transformation is particularly useful for regions that naturally align with circles or parts of circles, such as arcs or sectors. Here, you'll often encounter conversions like \(x = r \cos \theta\) and \(y = r \sin \theta\).
These expressions can simplify the integration process, transforming complicated functions into more manageable forms. It's crucial to remember that the area element in polar coordinates is \(dA = r \, dr \, d\theta\). Always make sure to incorporate \(r\) in the integrand while setting up the double integrals. This integral set-up in polar coordinates can drastically simplify the problem-solving process, providing a clearer geometric interpretation.
This transformation is particularly useful for regions that naturally align with circles or parts of circles, such as arcs or sectors. Here, you'll often encounter conversions like \(x = r \cos \theta\) and \(y = r \sin \theta\).
These expressions can simplify the integration process, transforming complicated functions into more manageable forms. It's crucial to remember that the area element in polar coordinates is \(dA = r \, dr \, d\theta\). Always make sure to incorporate \(r\) in the integrand while setting up the double integrals. This integral set-up in polar coordinates can drastically simplify the problem-solving process, providing a clearer geometric interpretation.
Exploring the Annulus Region
An annulus is like a donut-shaped region, characterized by having two concentric circles: an inner and an outer one. In the context of double integrals, and particularly in polar coordinates, this region can be incredibly handy for evaluating integrals.
The given annulus in the exercise is defined by the radii \(1 \leq r \leq 2\), suggesting that the region spans from a circle of radius 1 to a circle of radius 2. Meanwhile, the angular restriction \(\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\) describes a slice of this annulus, similar to taking a slice of pie.
This configuration of the annulus makes it easier to visualize the area over which we're integrating. Rather than dealing with complicated boundaries, the annular region in polar coordinates provides a straightforward and structured method for evaluating double integrals, removing the ambiguity of complex shapes.
The given annulus in the exercise is defined by the radii \(1 \leq r \leq 2\), suggesting that the region spans from a circle of radius 1 to a circle of radius 2. Meanwhile, the angular restriction \(\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\) describes a slice of this annulus, similar to taking a slice of pie.
This configuration of the annulus makes it easier to visualize the area over which we're integrating. Rather than dealing with complicated boundaries, the annular region in polar coordinates provides a straightforward and structured method for evaluating double integrals, removing the ambiguity of complex shapes.
Integral Transformation Techniques
Integral transformation is a powerful technique that allows us to simplify integrals by changing the coordinate system. The conversion from Cartesian to polar coordinates is a classic example. This method involves translating variables from one system to another, often turning a complex integral into a more manageable one.
The transformation encompasses converting the variables \(x, y\) to \(r, \theta\). For instance, replacing \(x^2 + y^2\) with \(r^2\) elegantly simplifies exponential expressions.
Moreover, expressions like \( an^{-1}\left(\frac{y}{x}\right)\) are represented directly by \(\theta\) in polar coordinates, streamlining trigonometric functions.
One of the critical steps in transforming the integral is substituting \(dA = r \, dr \, d\theta\) into the integral, which adapts the area element to polar coordinates. This step ensures the captivated expression correctly represents the polar region's area and leads to more straightforward calculations. Ultimately, integral transformation bridges the gap between complex geometries and the elegance of calculus.
The transformation encompasses converting the variables \(x, y\) to \(r, \theta\). For instance, replacing \(x^2 + y^2\) with \(r^2\) elegantly simplifies exponential expressions.
Moreover, expressions like \( an^{-1}\left(\frac{y}{x}\right)\) are represented directly by \(\theta\) in polar coordinates, streamlining trigonometric functions.
One of the critical steps in transforming the integral is substituting \(dA = r \, dr \, d\theta\) into the integral, which adapts the area element to polar coordinates. This step ensures the captivated expression correctly represents the polar region's area and leads to more straightforward calculations. Ultimately, integral transformation bridges the gap between complex geometries and the elegance of calculus.
