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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![**Topic: Evaluating Iterated Integrals by Converting to Polar Coordinates**
To solve this problem, we must evaluate the given iterated integral by converting to polar coordinates.
**Given Integral:**
\[
\int_0^a \int_0^{\sqrt{a^2 - y^2}} y \, dx \, dy
\]
**Conversion to Polar Coordinates:**
In polar coordinates, the variables \( x \) and \( y \) are expressed as:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
The differential area element \( dx \, dy \) becomes \( r \, dr \, d\theta \).
**New Integral Limits:**
Convert the Cartesian limits:
- The region is defined within a circle of radius \( a \), centered at the origin.
The polar coordinate limits for \( r \) range from \( 0 \) to \( a \), and \( \theta \) varies from \( 0 \) to \( \frac{\pi}{2} \).
**Transformed Integral:**
\[
\int_0^{\frac{\pi}{2}} \int_0^a y \cdot r \, dr \, d\theta
\]
Hence, the integrand and the limits are expressed in terms of polar coordinates.
**Conclusion:**
The problem transforms the original iterated integral into polar coordinates for easier evaluation. The exact solution will involve performing the integration in the new coordinate system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ec57636-08bb-40ad-b0ca-9d615e386c7c%2Fc2f61fed-8265-4ead-bf38-b3161b6b3f24%2Fj9uo8yq_processed.png&w=3840&q=75)
Transcribed Image Text:**Topic: Evaluating Iterated Integrals by Converting to Polar Coordinates**
To solve this problem, we must evaluate the given iterated integral by converting to polar coordinates.
**Given Integral:**
\[
\int_0^a \int_0^{\sqrt{a^2 - y^2}} y \, dx \, dy
\]
**Conversion to Polar Coordinates:**
In polar coordinates, the variables \( x \) and \( y \) are expressed as:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
The differential area element \( dx \, dy \) becomes \( r \, dr \, d\theta \).
**New Integral Limits:**
Convert the Cartesian limits:
- The region is defined within a circle of radius \( a \), centered at the origin.
The polar coordinate limits for \( r \) range from \( 0 \) to \( a \), and \( \theta \) varies from \( 0 \) to \( \frac{\pi}{2} \).
**Transformed Integral:**
\[
\int_0^{\frac{\pi}{2}} \int_0^a y \cdot r \, dr \, d\theta
\]
Hence, the integrand and the limits are expressed in terms of polar coordinates.
**Conclusion:**
The problem transforms the original iterated integral into polar coordinates for easier evaluation. The exact solution will involve performing the integration in the new coordinate system.
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