Change the Cartesian integral A B 0 S* Sº 0 *1/2 •2cos(e) [²/² [ 20x(®) [cos(0) + sin(0) ] drdo 0 -2cos(@) Ⓒ ~2cos © √²/2 √200x(10) 0 0 *2cos (6) •√1-(x-1)² x+y x² + y² 0 0 S²S√²- 00 [cose + sine] rdrd0 [cose + sine] rdrde [cos (0) + sin (0)] drdo dy dr into an equivalent polar integral.
Change the Cartesian integral A B 0 S* Sº 0 *1/2 •2cos(e) [²/² [ 20x(®) [cos(0) + sin(0) ] drdo 0 -2cos(@) Ⓒ ~2cos © √²/2 √200x(10) 0 0 *2cos (6) •√1-(x-1)² x+y x² + y² 0 0 S²S√²- 00 [cose + sine] rdrd0 [cose + sine] rdrde [cos (0) + sin (0)] drdo dy dr into an equivalent polar integral.
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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![### Converting Cartesian to Polar Coordinates
**Objective:**
Change the Cartesian integral
\[
\int_{0}^{2} \int_{0}^{\sqrt{1-(x-1)^2}} \frac{x+y}{x^2+y^2} \, dy \, dx
\]
into an equivalent polar integral.
**Problem:**
Given the Cartesian integral mentioned above, convert it to the polar coordinate system and evaluate the correct equivalent integral from the options listed below.
---
#### Options:
(A)
\[
\int_{0}^{\pi/2} \int_{0}^{2 \cos(\theta)} [\cos(\theta) + \sin(\theta)] \, r \, dr \, d\theta
\]
(B)
\[
\int_{0}^{\pi} \int_{-2 \cos(\theta)}^{0} [\cos(\theta) + \sin(\theta)] \, r \, dr \, d\theta
\]
(C)
\[
\int_{0}^{\pi/2} \int_{0}^{2 \cos(\theta)} [\cos(\theta) + \sin(\theta)] \, r \, dr \, d\theta
\]
(D)
\[
\int_{0}^{\pi} \int_{0}^{2 \cos(\theta)} [\cos(\theta) + \sin(\theta)] \, dr \, d\theta
\]
---
#### Analysis:
To determine the correct equivalent integral in polar coordinates, we proceed with the following steps:
1. **Convert Cartesian to Polar Coordinates:**
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
- \( dx \, dy = r \, dr \, d\theta \)
- \( (x-1)^2 + y^2 = r^2 \) leads to converting the limits properly
2. **Transformation and Limits:**
- For \( x \) going from 0 to 2 and \( y \) going from 0 to \( \sqrt{1-(x-1)^2} \), the region described transforms in polar coordinates to \( r \) from 0 to \( 2 \cos(\theta) \) and \( \theta \) from 0 to \( \frac{\pi}{2} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f7f7e91-311f-40b2-8ef2-058480eb16a0%2F34a3a50d-f388-4cf7-a12c-1d142d4cceda%2F7vwhsvl_processed.png&w=3840&q=75)
Transcribed Image Text:### Converting Cartesian to Polar Coordinates
**Objective:**
Change the Cartesian integral
\[
\int_{0}^{2} \int_{0}^{\sqrt{1-(x-1)^2}} \frac{x+y}{x^2+y^2} \, dy \, dx
\]
into an equivalent polar integral.
**Problem:**
Given the Cartesian integral mentioned above, convert it to the polar coordinate system and evaluate the correct equivalent integral from the options listed below.
---
#### Options:
(A)
\[
\int_{0}^{\pi/2} \int_{0}^{2 \cos(\theta)} [\cos(\theta) + \sin(\theta)] \, r \, dr \, d\theta
\]
(B)
\[
\int_{0}^{\pi} \int_{-2 \cos(\theta)}^{0} [\cos(\theta) + \sin(\theta)] \, r \, dr \, d\theta
\]
(C)
\[
\int_{0}^{\pi/2} \int_{0}^{2 \cos(\theta)} [\cos(\theta) + \sin(\theta)] \, r \, dr \, d\theta
\]
(D)
\[
\int_{0}^{\pi} \int_{0}^{2 \cos(\theta)} [\cos(\theta) + \sin(\theta)] \, dr \, d\theta
\]
---
#### Analysis:
To determine the correct equivalent integral in polar coordinates, we proceed with the following steps:
1. **Convert Cartesian to Polar Coordinates:**
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
- \( dx \, dy = r \, dr \, d\theta \)
- \( (x-1)^2 + y^2 = r^2 \) leads to converting the limits properly
2. **Transformation and Limits:**
- For \( x \) going from 0 to 2 and \( y \) going from 0 to \( \sqrt{1-(x-1)^2} \), the region described transforms in polar coordinates to \( r \) from 0 to \( 2 \cos(\theta) \) and \( \theta \) from 0 to \( \frac{\pi}{2} \).
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