Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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Question
![**Convert the integral to polar coordinates and evaluate.**
\[ \int_{0}^2 \int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}} (x^2 + y^2)^2 \, dx \, dy \]
When converting from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\), use the transformations:
\[ x = r \cos \theta \]
\[ y = r \sin \theta \]
\[ dx \, dy = r \, dr \, d\theta \]
The given integral can be rewritten in polar coordinates. Evaluate the new integral by following these steps:
1. **Set up the new boundaries:**
The region of integration is a circle of radius 2 centered at the origin. So, the bounds for \(r\) and \(\theta\) are:
\[ 0 \le r \le 2 \]
\[ 0 \le \theta \le 2\pi \]
2. **Rewrite the integrand in polar coordinates:**
Since \(x^2 + y^2 = r^2\), the integrand \((x^2 + y^2)^2\) becomes \(r^4\).
3. **Update the integral in polar coordinates:**
\[ \int_{0}^{2\pi} \int_{0}^{2} (r^4) \cdot r \, dr \, d\theta \]
Simplify \(r^4 \cdot r\) to get \(r^5\):
\[ \int_{0}^{2\pi} \int_{0}^{2} r^5 \, dr \, d\theta \]
4. **Evaluate the inner integral:**
\[ \int_{0}^{2} r^5 \, dr \]
\[ = \left[ \frac{r^6}{6} \right]_{0}^{2} \]
\[ = \frac{2^6}{6} - \frac{0^6}{6} \]
\[ = \frac{64}{6} \]
\[ = \frac{32}{3} \]
5. **Evaluate the outer integral:**
\[ \int_{0}^{2\pi} \frac{32}{3} \, d\theta \]
\[ = \frac{32}{3} \left[ \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad35cd49-88f1-4759-9e4c-232c0c792d3e%2F57b8095d-8812-48ce-8598-18a421f2b7c6%2F88s6rdb.jpeg&w=3840&q=75)
Transcribed Image Text:**Convert the integral to polar coordinates and evaluate.**
\[ \int_{0}^2 \int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}} (x^2 + y^2)^2 \, dx \, dy \]
When converting from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\), use the transformations:
\[ x = r \cos \theta \]
\[ y = r \sin \theta \]
\[ dx \, dy = r \, dr \, d\theta \]
The given integral can be rewritten in polar coordinates. Evaluate the new integral by following these steps:
1. **Set up the new boundaries:**
The region of integration is a circle of radius 2 centered at the origin. So, the bounds for \(r\) and \(\theta\) are:
\[ 0 \le r \le 2 \]
\[ 0 \le \theta \le 2\pi \]
2. **Rewrite the integrand in polar coordinates:**
Since \(x^2 + y^2 = r^2\), the integrand \((x^2 + y^2)^2\) becomes \(r^4\).
3. **Update the integral in polar coordinates:**
\[ \int_{0}^{2\pi} \int_{0}^{2} (r^4) \cdot r \, dr \, d\theta \]
Simplify \(r^4 \cdot r\) to get \(r^5\):
\[ \int_{0}^{2\pi} \int_{0}^{2} r^5 \, dr \, d\theta \]
4. **Evaluate the inner integral:**
\[ \int_{0}^{2} r^5 \, dr \]
\[ = \left[ \frac{r^6}{6} \right]_{0}^{2} \]
\[ = \frac{2^6}{6} - \frac{0^6}{6} \]
\[ = \frac{64}{6} \]
\[ = \frac{32}{3} \]
5. **Evaluate the outer integral:**
\[ \int_{0}^{2\pi} \frac{32}{3} \, d\theta \]
\[ = \frac{32}{3} \left[ \
![**Converting and Evaluating an Integral in Polar Coordinates**
Consider the integral:
\[ \int_{0}^{3} \int_{-\sqrt{9 - x^2}}^{\sqrt{9 - x^2}} \sin(x^2 + y^2) \, dy \, dx \]
To convert this integral to polar coordinates and subsequently evaluate it, follow these steps:
1. **Convert the Limits of Integration to Polar Coordinates**:
- The given limits suggest a region bounded by a circle \( x^2 + y^2 = 9 \) and the vertical line \( x = 3 \).
- The bounds for \( r \) (the radius in polar coordinates) will range from 0 to 3, as per the circle's radius within the first quarter circle and bounded by \( x = 3 \).
- The bounds for \( \theta \) (the angle in polar coordinates) will range from 0 to \( \pi/2 \) to cover the first quadrant.
2. **Apply the Conversion**:
- In polar coordinates, \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).
- \( x^2 + y^2 = r^2 \).
- The integral transforms to:
\[
\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \sin(r^2) r \, dr \, d\theta
\]
- Note: The extra \( r \) term comes from the Jacobian determinant when converting to polar coordinates.
3. **Evaluate the Integral**:
- Integrate with respect to \( r \):
\[
\int_{0}^{3} \sin(r^2) r \, dr
\]
- Use the substitution \( u = r^2 \), hence \( du = 2r \, dr \) or \( \frac{1}{2} du = r \, dr \):
\[
\int_{0}^{9} \frac{1}{2} \sin(u) \, du = \frac{1}{2} \left[-\cos(u)\right]_{0}^{9} = \frac{1}{2} \left[-\cos(9) + 1 \right]
\]
- Evaluate the definite integral](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad35cd49-88f1-4759-9e4c-232c0c792d3e%2F57b8095d-8812-48ce-8598-18a421f2b7c6%2Fz7b8582.jpeg&w=3840&q=75)
Transcribed Image Text:**Converting and Evaluating an Integral in Polar Coordinates**
Consider the integral:
\[ \int_{0}^{3} \int_{-\sqrt{9 - x^2}}^{\sqrt{9 - x^2}} \sin(x^2 + y^2) \, dy \, dx \]
To convert this integral to polar coordinates and subsequently evaluate it, follow these steps:
1. **Convert the Limits of Integration to Polar Coordinates**:
- The given limits suggest a region bounded by a circle \( x^2 + y^2 = 9 \) and the vertical line \( x = 3 \).
- The bounds for \( r \) (the radius in polar coordinates) will range from 0 to 3, as per the circle's radius within the first quarter circle and bounded by \( x = 3 \).
- The bounds for \( \theta \) (the angle in polar coordinates) will range from 0 to \( \pi/2 \) to cover the first quadrant.
2. **Apply the Conversion**:
- In polar coordinates, \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).
- \( x^2 + y^2 = r^2 \).
- The integral transforms to:
\[
\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \sin(r^2) r \, dr \, d\theta
\]
- Note: The extra \( r \) term comes from the Jacobian determinant when converting to polar coordinates.
3. **Evaluate the Integral**:
- Integrate with respect to \( r \):
\[
\int_{0}^{3} \sin(r^2) r \, dr
\]
- Use the substitution \( u = r^2 \), hence \( du = 2r \, dr \) or \( \frac{1}{2} du = r \, dr \):
\[
\int_{0}^{9} \frac{1}{2} \sin(u) \, du = \frac{1}{2} \left[-\cos(u)\right]_{0}^{9} = \frac{1}{2} \left[-\cos(9) + 1 \right]
\]
- Evaluate the definite integral
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