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Magazine Chapter 5: Problem 142
In the following exercises, evaluate the double integral \(\int_{R} f(x, y) d A\) over the polar rectangular region \(D\). $$ \iint e^{x^{2}+y^{2}}\left[1+2 \arctan \left(\frac{y}{x}\right)\right] d A, D=\left\\{(r, \theta) \mid 1 \leq r \leq 2, \frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\right\\} $$
Short Answer
Expert verified
The value of the double integral is \(\frac{1}{2}(e^4 - e)\left(\frac{\pi}{3} + \left(\frac{\pi}{3}\right)^2 - \frac{\pi}{6} - \left(\frac{\pi}{6}\right)^2\right)\).
Step by step solution
01
Convert the Integral to Polar Coordinates
First, we need to rewrite the double integral in terms of polar coordinates. In polar coordinates, the transformation is given by the relationships: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). The region \(D\) becomes \(1 \leq r \leq 2 \) and \(\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\). Additionally, the differential area element \(dA\) is replaced by \(r \, dr \, d\theta\).
02
Express the Function in Polar Form
The given function \(f(x, y) = e^{x^2 + y^2} \left[ 1 + 2 \arctan \left( \frac{y}{x} \right) \right]\) can be expressed using polar coordinates. In polar coordinates, \(x^2 + y^2 = r^2\), and \( \frac{y}{x} = \tan(\theta)\) (since \(y = r \sin(\theta)\) and \(x = r \cos(\theta)\)). The function becomes: \(f(r, \theta) = e^{r^2} \left[1 + 2 \arctan(\tan(\theta))\right]\). But, \(\arctan(\tan(\theta)) = \theta\) when \(\theta\) is in \([0, \pi)\). So, the function simplifies to \(e^{r^2} \left[1 + 2\theta\right]\).
03
Set Up the Double Integral
Now, we can set up the double integral: \[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \int_{1}^{2} e^{r^2} \left[1 + 2\theta\right] r \, dr \, d\theta\]
04
Integrate with Respect to r
First, integrate with respect to \(r\): \[\int_{1}^{2} e^{r^2} \left[1 + 2\theta\right] r \, dr\] Use the substitution \(u = r^2\), so \(du = 2r \, dr\) or \(r \, dr = \frac{1}{2} \, du\). The limits of integration for \(u\) are 1 to 4: \[\frac{1}{2}\int_{1}^{4} e^u \left[1 + 2\theta\right] \, du\] This evaluates to: \[\frac{1}{2} \left[ (e^4 - e) (1 + 2\theta) \right]\].
05
Integrate with Respect to 胃
Now, integrate the result from Step 4 with respect to \(\theta\): \[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{2} (e^4 - e) (1 + 2\theta) \, d\theta\] This becomes: \[\frac{1}{2} (e^4 - e) \left[ \theta + \theta^2 \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\] Calculate the definite integral: This results in: \[\frac{1}{2} (e^4 - e) \left[ \left(\frac{\pi}{3} + \left(\frac{\pi}{3}\right)^2\right) - \left(\frac{\pi}{6} + \left(\frac{\pi}{6}\right)^2\right) \right]\] Compute the specific values and the subtraction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
When dealing with double integrals, it's often useful to convert from Cartesian coordinates (spanning the x and y axes) to polar coordinates (using variables r and \( \theta \)). In polar coordinates, a point on the plane is represented by the distance \( r \) from the origin and an angle \( \theta \) from the positive x-axis. This approach is beneficial especially when the region of integration has circular symmetry.
- Relationships: Transforming from Cartesian to polar coordinates is done using the equations: x = r \cos(\theta) and \( y = r \sin(\theta) \).
- Region Description: In the given problem, the region \( D \) is a sector of a circle with radii 1 to 2 and angles from \( \pi/6 \) to \( \pi/3 \). This translates straightforwardly into constraints on \( r \) and \( \theta \).
- Differential Element: The infinitesimal area element \( dA \), in polar coordinates, converts to \( r \, dr \, d\theta \). The presence of \( r \) acknowledges the circular structure of the area under consideration.
Change of Variables
The change of variables technique is a powerful tool often used to simplify the process of integration. In this context, it primarily involves transforming the original variables (x and y) into a new set (r and \( \theta \)). This not only reformulates the function but also the limits of integration.
- Function Transformation: The original function \( f(x, y) = e^{x^2 + y^2} \left[ 1 + 2 \arctan \left( \frac{y}{x} \right) \right] \)is rewritten in terms of polar coordinates using the identities: \( x^2 + y^2 = r^2 \) and \( \frac{y}{x} = \tan(\theta) \). As a result, the function becomes \( f(r, \theta) = e^{r^2} \left[1 + 2\theta\right] \).
- Integration Limits Conversion: The rectangular region described in polar coordinates allows direct substitution of the new limits of integration: \(1 \leq r \leq 2 \) and \( \pi/6 \leq \theta \leq \pi/3 \).
Definite Integral
A definite integral computes the accumulation of quantities, giving a total value over a specified interval. With double integrals, this concept extends into calculating the volume beneath a surface over a defined region in two variables.
- Structure: The specified double integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \int_{1}^{2} e^{r^2} \left[1 + 2\theta\right] r \, dr \, d\theta \) allows us to find the precise volume under the surface defined by the function \( f(r, \theta) \).
- Evaluation Strategy: Since this is a definite integral, the process involves calculating two successive integrations, one with respect to \( r \) and then \( \theta \). This involves systematically applying the fundamental theorem of calculus for the given interval bounds.
Integration with Respect to Variables
Integrating with respect to a single variable is an essential procedure in solving multi-variable integrals. Here, it involves focusing first on one dimension (integrate with \( r \) first) while treating other variables as constants, followed by the subsequent variables.
- First Dimension: Integration with respect to \( r \) requires solving the inner integral \( \int_{1}^{2} e^{r^2} \left[1 + 2\theta\right] r \ dr \). Using substitution simplifies this step significantly; we let \( u = r^2 \) with \( du = 2r \, dr \), leading to an expression that can be directly integrated.
- Second Dimension: Resulting from the first dimension, you use the resultant expression to integrate with respect to \( \theta \) (\( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{2} (e^4 - e) (1 + 2\theta) \ d\theta \)). Here, \( \theta \) is treated singularly, ensuring the full double integral process is completed.
