Express the following integral as a double integral in polar coordinates 1 Suvaldy x² + y² dy dr
Express the following integral as a double integral in polar coordinates 1 Suvaldy x² + y² dy dr
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Transcription for Educational Website**
**Topic: Converting Cartesian Integrals to Polar Coordinates**
**Instruction:**
Express the following integral as a double integral in polar coordinates.
**Given Integral:**
\[
\int_{1/\sqrt{2}}^{1} \int_{\sqrt{1-x^2}}^{x} \frac{1}{x^2 + y^2} \, dy \, dx
\]
**Explanation:**
The provided integral needs to be converted from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). Polar coordinates are often used when the region of integration is circular or when the integrand involves expressions like \(x^2 + y^2\).
In polar coordinates:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(x^2 + y^2 = r^2\)
- The differential area element \(dy \, dx\) becomes \(r \, dr \, d\theta\).
**Step-by-step Conversion:**
The limits of integration need to be translated in terms of \(r\) and \(\theta\). Sketching the region of integration will help identify these limits.
1. **Convert Integrand:**
\[
\frac{1}{x^2 + y^2} \to \frac{1}{r^2}
\]
2. **Determine New Limits:**
- The range of \(x\) and \(y\) defines a region in the first quadrant.
- For \(x\), the limits are \(1/\sqrt{2}\) to 1.
- The inner integral's limit \(y = \sqrt{1-x^2}\) suggests a transformation involving a circle of radius 1.
3. **Polar Coordinates Setup:**
- Radius \(r\) ranges from \(/sqrt{2}\) to 1 based on the outer \(x\)-limits.
- Angle \(\theta\) ranges based on the region, often determined visually or by substitution.
4. **Transform the Integral:**
Replace \(x\) and \(y\) with their polar equivalents and include \(r\) from the Jacobian determinant:
\[
\int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5563205c-2551-4780-a535-119e8d9f322e%2F145a93ca-8cce-4fab-a7cc-34e741188d9f%2Fhklc03g_processed.png&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
**Topic: Converting Cartesian Integrals to Polar Coordinates**
**Instruction:**
Express the following integral as a double integral in polar coordinates.
**Given Integral:**
\[
\int_{1/\sqrt{2}}^{1} \int_{\sqrt{1-x^2}}^{x} \frac{1}{x^2 + y^2} \, dy \, dx
\]
**Explanation:**
The provided integral needs to be converted from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). Polar coordinates are often used when the region of integration is circular or when the integrand involves expressions like \(x^2 + y^2\).
In polar coordinates:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(x^2 + y^2 = r^2\)
- The differential area element \(dy \, dx\) becomes \(r \, dr \, d\theta\).
**Step-by-step Conversion:**
The limits of integration need to be translated in terms of \(r\) and \(\theta\). Sketching the region of integration will help identify these limits.
1. **Convert Integrand:**
\[
\frac{1}{x^2 + y^2} \to \frac{1}{r^2}
\]
2. **Determine New Limits:**
- The range of \(x\) and \(y\) defines a region in the first quadrant.
- For \(x\), the limits are \(1/\sqrt{2}\) to 1.
- The inner integral's limit \(y = \sqrt{1-x^2}\) suggests a transformation involving a circle of radius 1.
3. **Polar Coordinates Setup:**
- Radius \(r\) ranges from \(/sqrt{2}\) to 1 based on the outer \(x\)-limits.
- Angle \(\theta\) ranges based on the region, often determined visually or by substitution.
4. **Transform the Integral:**
Replace \(x\) and \(y\) with their polar equivalents and include \(r\) from the Jacobian determinant:
\[
\int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} \frac
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