Consider the double integral f(x,y)=(x²+y2) 3/2 over the region R where y is bounded between 0 and (1-x²)¹/2 and x is bounded between -1 and 1. Convert this integral to polar coordinates. (You do not need to integrate)
Consider the double integral f(x,y)=(x²+y2) 3/2 over the region R where y is bounded between 0 and (1-x²)¹/2 and x is bounded between -1 and 1. Convert this integral to polar coordinates. (You do not need to integrate)
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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![### Convert a Double Integral to Polar Coordinates
Consider the double integral:
\[ f(x, y) = (x^2 + y^2)^{3/2} \]
over the region \( R \) where \( y \) is bounded between 0 and \((1 - x^2)^{1/2} \), and \( x \) is bounded between \(-1\) and \(1\).
Convert this integral to polar coordinates. (You do not need to integrate).
### Detailed Explanation
The given function and limits will be converted into polar coordinates. Here’s a step-by-step guide:
1. **Express \( x \) and \(y \) in Polar Coordinates:**
\[
x = r \cos \theta
\]
\[
y = r \sin \theta
\]
2. **Convert the Function to Polar Coordinates:**
\[
(x^2 + y^2)^{3/2} = (r^2)^{3/2} = r^3
\]
3. **Change the Integration Bounds:**
For a given region \( R \):
- \( x \) ranges from \(-1\) to \(1\), and
- \( y \) ranges from \(0\) to \(\sqrt{1 - x^2}\).
In polar coordinates, the bounds change to:
- \( r \) ranges from \(0\) to \(1\), and
- \( \theta \) ranges from \(0\) to \(\pi\).
4. **Change the Area Element \( dA \) in Polar Coordinates:**
The area element \( dx \, dy \) is replaced by \( r \, dr \, d\theta \).
5. **Rewrite the Integral in Polar Coordinates:**
The double integral becomes:
\[
\iint_R (x^2 + y^2)^{3/2} \, dA = \iint_R (r^3) \, r \, dr \, d\theta
\]
Simplify the integrand:
\[
\iint_R r^4 \, dr \, d\theta
\]
Adjusting for the bounds:
\[
\int_0^\pi \int_0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1219e52d-80cb-4922-a983-3e539dd2ff27%2F09315a29-6d8a-421f-8a68-3530a2951ce6%2Fa53uulo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Convert a Double Integral to Polar Coordinates
Consider the double integral:
\[ f(x, y) = (x^2 + y^2)^{3/2} \]
over the region \( R \) where \( y \) is bounded between 0 and \((1 - x^2)^{1/2} \), and \( x \) is bounded between \(-1\) and \(1\).
Convert this integral to polar coordinates. (You do not need to integrate).
### Detailed Explanation
The given function and limits will be converted into polar coordinates. Here’s a step-by-step guide:
1. **Express \( x \) and \(y \) in Polar Coordinates:**
\[
x = r \cos \theta
\]
\[
y = r \sin \theta
\]
2. **Convert the Function to Polar Coordinates:**
\[
(x^2 + y^2)^{3/2} = (r^2)^{3/2} = r^3
\]
3. **Change the Integration Bounds:**
For a given region \( R \):
- \( x \) ranges from \(-1\) to \(1\), and
- \( y \) ranges from \(0\) to \(\sqrt{1 - x^2}\).
In polar coordinates, the bounds change to:
- \( r \) ranges from \(0\) to \(1\), and
- \( \theta \) ranges from \(0\) to \(\pi\).
4. **Change the Area Element \( dA \) in Polar Coordinates:**
The area element \( dx \, dy \) is replaced by \( r \, dr \, d\theta \).
5. **Rewrite the Integral in Polar Coordinates:**
The double integral becomes:
\[
\iint_R (x^2 + y^2)^{3/2} \, dA = \iint_R (r^3) \, r \, dr \, d\theta
\]
Simplify the integrand:
\[
\iint_R r^4 \, dr \, d\theta
\]
Adjusting for the bounds:
\[
\int_0^\pi \int_0
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