Change the Cartesian integral -₁¹* dy dx into an equivalent polar integral. Then evaluate the polar integral. Change the Cartesian integral into an equivalent polar integral. S²-₁ S0¹-* dy dx = =APor dr de (Type exact answers, using as needed.)
Change the Cartesian integral -₁¹* dy dx into an equivalent polar integral. Then evaluate the polar integral. Change the Cartesian integral into an equivalent polar integral. S²-₁ S0¹-* dy dx = =APor dr de (Type exact answers, using as needed.)
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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### Cartesian to Polar Integral Conversion
#### Problem Statement:
Change the Cartesian integral \(\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} dy \, dx\) into an equivalent polar integral. Then evaluate the polar integral.
#### Solution Steps:
1. **Convert the Cartesian Integral to a Polar Integral:**
The given Cartesian integral is:
\[
\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} dy \, dx
\]
2. **Express the integral limits in polar coordinates:**
- In polar coordinates (\(r, \theta\)), the region \(x^2 + y^2 \le 1\) and \(0 \le y \le \sqrt{1 - x^2}\) translates to:
- \(0 \le r \le 1\) (since \(r\) is the radius from the origin)
- \(0 \le \theta \le \pi\) (since \(\theta\) ranges from \(0\) to \(\pi\) to cover the integration over the upper semicircle)
3. **Set up and evaluate the polar integral:**
\[
\int_{0}^{\pi} \int_{0}^{1} r \, dr \, d\theta
\]
4. **Evaluate the integrals:**
The step-wise evaluation of the integrals is:
\[
\left[ \int_{0}^{\pi} d\theta \right] \times \left[ \int_{0}^{1} r \, dr \right]
\]
- Evaluate the angular part:
\[
\int_{0}^{\pi} d\theta = \theta \Big|_{0}^{\pi} = \pi - 0 = \pi
\]
- Evaluate the radial part:
\[
\int_{0}^{1} r \, dr = \frac{r^2}{2} \Big|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
\]
5. **Combine the results:**
Multiply the results of the angular and radial integrals:
\[
\pi \times \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1840ebe-b0df-4fe8-9210-d8e3dcfa32cc%2F55d48bba-a95d-4b97-a4c3-d252794362ed%2Fewexe6_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Cartesian to Polar Integral Conversion
#### Problem Statement:
Change the Cartesian integral \(\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} dy \, dx\) into an equivalent polar integral. Then evaluate the polar integral.
#### Solution Steps:
1. **Convert the Cartesian Integral to a Polar Integral:**
The given Cartesian integral is:
\[
\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} dy \, dx
\]
2. **Express the integral limits in polar coordinates:**
- In polar coordinates (\(r, \theta\)), the region \(x^2 + y^2 \le 1\) and \(0 \le y \le \sqrt{1 - x^2}\) translates to:
- \(0 \le r \le 1\) (since \(r\) is the radius from the origin)
- \(0 \le \theta \le \pi\) (since \(\theta\) ranges from \(0\) to \(\pi\) to cover the integration over the upper semicircle)
3. **Set up and evaluate the polar integral:**
\[
\int_{0}^{\pi} \int_{0}^{1} r \, dr \, d\theta
\]
4. **Evaluate the integrals:**
The step-wise evaluation of the integrals is:
\[
\left[ \int_{0}^{\pi} d\theta \right] \times \left[ \int_{0}^{1} r \, dr \right]
\]
- Evaluate the angular part:
\[
\int_{0}^{\pi} d\theta = \theta \Big|_{0}^{\pi} = \pi - 0 = \pi
\]
- Evaluate the radial part:
\[
\int_{0}^{1} r \, dr = \frac{r^2}{2} \Big|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
\]
5. **Combine the results:**
Multiply the results of the angular and radial integrals:
\[
\pi \times \frac{
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