Evaluate the given integral by changing to polar coordinates. sin(x² + y²) da, where R is the region in the first quadrant between the circles with center the origin and radii 3 and 4
Evaluate the given integral by changing to polar coordinates. sin(x² + y²) da, where R is the region in the first quadrant between the circles with center the origin and radii 3 and 4
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Integral Evaluation in Polar Coordinates
Evaluate the given integral by changing to polar coordinates.
\[ \iint_R \sin(x^2 + y^2) \, dA \]
where \( R \) is the region in the first quadrant between the circles with center the origin and radii 3 and 4.
---
**Explanation**:
To solve this integral, we need to convert from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\).
1. **Conversion**:
- \( x = r\cos\theta \)
- \( y = r\sin\theta \)
- \( x^2 + y^2 = r^2 \)
- The differential area element \( dA \) in polar coordinates is \( r \, dr \, d\theta \).
2. **Integral Setup**:
\[ \iint_R \sin(x^2 + y^2) \, dA \]
in polar coordinates transforms to
\[ \int_{\theta=\frac{\pi}{2}}^{\theta=0} \int_{r=3}^{r=4} \sin(r^2) \cdot r \, dr \, d\theta \]
3. **Integral Boundaries**:
- \(r\) ranges from 3 to 4.
- \(\theta\) ranges from 0 to \(\frac{\pi}{2}\) since we are in the first quadrant.
४. **Integrating**:
- First integrate with respect to \( r \):
\[ \int_{r=3}^{r=4} r\sin(r^2) \, dr \]
- Then integrate with respect to \(\theta\):
\[ \int_{\theta=0}^{\theta=\frac{\pi}{2}} d\theta \]
This converts the original integral into a more manageable form suited for evaluation using polar coordinates.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab6b79a9-c663-4f87-993a-883a678be91b%2F5a142d20-3517-4423-bef1-271d72119b13%2Fqy4meta_processed.png&w=3840&q=75)
Transcribed Image Text:### Integral Evaluation in Polar Coordinates
Evaluate the given integral by changing to polar coordinates.
\[ \iint_R \sin(x^2 + y^2) \, dA \]
where \( R \) is the region in the first quadrant between the circles with center the origin and radii 3 and 4.
---
**Explanation**:
To solve this integral, we need to convert from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\).
1. **Conversion**:
- \( x = r\cos\theta \)
- \( y = r\sin\theta \)
- \( x^2 + y^2 = r^2 \)
- The differential area element \( dA \) in polar coordinates is \( r \, dr \, d\theta \).
2. **Integral Setup**:
\[ \iint_R \sin(x^2 + y^2) \, dA \]
in polar coordinates transforms to
\[ \int_{\theta=\frac{\pi}{2}}^{\theta=0} \int_{r=3}^{r=4} \sin(r^2) \cdot r \, dr \, d\theta \]
3. **Integral Boundaries**:
- \(r\) ranges from 3 to 4.
- \(\theta\) ranges from 0 to \(\frac{\pi}{2}\) since we are in the first quadrant.
४. **Integrating**:
- First integrate with respect to \( r \):
\[ \int_{r=3}^{r=4} r\sin(r^2) \, dr \]
- Then integrate with respect to \(\theta\):
\[ \int_{\theta=0}^{\theta=\frac{\pi}{2}} d\theta \]
This converts the original integral into a more manageable form suited for evaluation using polar coordinates.
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