Problem 43
Question
Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the \(x y\) -plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for \(r\) and \(\theta .\) c. Using the results in part (b), plot the polar region of integration in the \(r \theta\) -plane. d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the limits of integration from your plot in part (c) utility. \(\int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x\)
Step-by-Step Solution
Verified Answer
The evaluated polar integral is \(\frac{\sqrt{2}}{2}\).
1Step 1: Plot the Cartesian Region
The given Cartesian integral is \( \int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x \). The limits of integration for \(y\) are from \(x\) to \(1\), and for \(x\) are from \(0\) to \(1\). This represents a triangular region in the \(xy\)-plane with vertices at \((0,0)\), \((1,1)\), and \((1,0)\). Plot this region to visualize the area over which the integration occurs.
2Step 2: Convert to Polar Coordinates
In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). The curves \(y = x\) translates to \(r\sin\theta = r\cos\theta\), resulting in \(\theta = \frac{\pi}{4}\). The line \(y = 1\) becomes \(r\sin\theta = 1\), and the line \(x = 1\) becomes \(r\cos\theta = 1\). Solve these equations for \(r\): 1. \(r = \frac{1}{\sin\theta}\) for \(y = 1\). 2. \(r = \frac{1}{\cos\theta}\) for \(x = 1\).
3Step 3: Plot the Polar Region
Using the polar boundaries obtained in step 2, plot the region of integration. The region is bounded by \(\theta\) from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\) and \(r\) from \(0\) to \(\min{\left(\frac{1}{\sin\theta}, \frac{1}{\cos\theta}\right)}\). This corresponds to a sector of a circle in the \(r\theta\)-plane.
4Step 4: Change the Integrand to Polar Form
In polar coordinates, the integrand \(\frac{y}{x^{2}+y^{2}}\) becomes \(\frac{r\sin\theta}{r^2}\). Simplified, this is \(\frac{\sin\theta}{r}\). The differential area element \(dy\,dx\) changes to \(r\,dr\,d\theta\). The integral becomes: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\max(\sin\theta,\cos\theta)}} \sin\theta \; dr \; d\theta. \]The inner integral for \(r\) is straightforward, yielding the result \(\sin\theta \cdot \left(\frac{1}{\sin\theta}-0\right)\).
5Step 5: Evaluate the Polar Integral
Now evaluate the integral: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin\theta \, d\theta. \]Since \(\sin\theta\) is a common function, this evaluates to \(-\cos\theta\) from \(\theta = \frac{\pi}{4}\) to \(\frac{\pi}{2}\). The calculation is as follows: \[ [-\cos(\frac{\pi}{2}) - (-\cos(\frac{\pi}{4}))] = [0 - (-\frac{\sqrt{2}}{2})] = \frac{\sqrt{2}}{2}. \]
Key Concepts
Cartesian to polar conversionintegrals in polar coordinatesvisualizing integration regions
Cartesian to polar conversion
Converting from Cartesian to polar coordinates can simplify integration over regions that have circular symmetry. This conversion uses the relationships:
Similarly, a horizontal line like \( y = 1 \) becomes \( r\sin\theta = 1 \), while a vertical line like \( x = 1 \) translates to \( r\cos\theta = 1 \). By solving these equations, we determine boundaries for \( r \):
- For the x-coordinate: \( x = r\cos\theta \)
- For the y-coordinate: \( y = r\sin\theta \)
Similarly, a horizontal line like \( y = 1 \) becomes \( r\sin\theta = 1 \), while a vertical line like \( x = 1 \) translates to \( r\cos\theta = 1 \). By solving these equations, we determine boundaries for \( r \):
- \( r = \frac{1}{\sin\theta} \) for the line \( y = 1 \)
- \( r = \frac{1}{\cos\theta} \) for the line \( x = 1 \)
integrals in polar coordinates
Integrals in polar coordinates often make complex areas much easier to manage. When converting the given Cartesian integral into its polar equivalent, we utilize the substitution \( dy\,dx = r\,dr\,d\theta \) as the differential area element. This is due to the circular representation in polar coordinates. As a result, the integrand \( \frac{y}{x^{2}+y^{2}} \) transforms into \( \frac{r\sin\theta}{r^2} \) in polar form, which simplifies further into \( \frac{\sin\theta}{r} \).
To solve this integral in polar coordinates, you need to determine the bounds for \( r \) and \( \theta \). From the polar plot in the exercise, we see that \( \theta \) varies from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \), while \( r \) ranges from \( 0 \) to \( \min \left( \frac{1}{\sin\theta}, \frac{1}{\cos\theta} \right) \).
Substituting these into the polar integral, it becomes:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\max(\sin\theta,\cos\theta)}} \sin\theta \; dr \; d\theta. \] Solving this step-by-step simplifies the integration process and provides a more straightforward solution.
To solve this integral in polar coordinates, you need to determine the bounds for \( r \) and \( \theta \). From the polar plot in the exercise, we see that \( \theta \) varies from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \), while \( r \) ranges from \( 0 \) to \( \min \left( \frac{1}{\sin\theta}, \frac{1}{\cos\theta} \right) \).
Substituting these into the polar integral, it becomes:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\max(\sin\theta,\cos\theta)}} \sin\theta \; dr \; d\theta. \] Solving this step-by-step simplifies the integration process and provides a more straightforward solution.
visualizing integration regions
Visualizing integration regions is crucial in understanding and solving integrals, especially when transitioning from Cartesian to polar coordinates. The exercise begins with a visual representation of the Cartesian integration region—a triangle with vertices at (0,0), (1,1), and (1,0).
This visualization helps identify the limits of integration by showing the intersections and boundaries of the region. When the region is converted to polar coordinates, the triangle translates into a sector of a circle. This sector is defined by \( \theta \) from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \) and \( r \) varying from \( 0 \) to \( \min \left( \frac{1}{\sin\theta}, \frac{1}{\cos\theta} \right) \).
Being able to sketch or visualize this region in the \( r\theta \)-plane makes it easier to set up the integral, determine the limits accurately, and eventually evaluate the integral with confidence.
This visualization helps identify the limits of integration by showing the intersections and boundaries of the region. When the region is converted to polar coordinates, the triangle translates into a sector of a circle. This sector is defined by \( \theta \) from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \) and \( r \) varying from \( 0 \) to \( \min \left( \frac{1}{\sin\theta}, \frac{1}{\cos\theta} \right) \).
Being able to sketch or visualize this region in the \( r\theta \)-plane makes it easier to set up the integral, determine the limits accurately, and eventually evaluate the integral with confidence.
- Sketching reveals where boundaries overlap,
- Identifies critical points for integration,
- Allows intuitive understanding of the space being integrated,
Other exercises in this chapter
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Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a.
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