### Problem StatementLet $ R = \{(r, \theta) \mid 1 \leq r \leq 3, 0 \leq \theta \leq \pi/2 \} $. Sketch the region of integration $ R $ and evaluate the following integral over $ R $ using polar coordinates:\[\iint\limits_{R} 2xy \, dA \quad \text{(remember that } dA = rdrd\theta \text{)}.\]### Explanation of RegionThe region $ R $ is defined in polar coordinates ($r, \theta$) where:- $ r $ (the radius) ranges from 1 to 3.- $ \theta $ (the angle) ranges from 0 to $\pi/2$ (which is the first quadrant).In Cartesian coordinates, this region would correspond to a sector of a circle located in the first quadrant, with inner and outer radii of 1 and 3, respectively, and spanning from the positive x-axis (0 radians) to the positive y-axis ($\pi/2$ radians). ### Evaluating the IntegralTo solve the integral, convert the integral into polar coordinates and use the given transformation $dA = rdrd\theta$. Therefore, the integral becomes:\[\iint\limits_{R} 2xy \, dA = \iint\limits_{R} 2 \cdot (r\cos\theta) \cdot (r\sin\theta) \cdot r d\theta dr\]Substituting $x$ and $y$ in polar coordinates ($x = r\cos\theta, y = r\sin\theta$):\[= \iint\limits_{R} 2r^3 \cos\theta \sin\theta \, dr d\theta\]Now, calculate the double integral:\[\int_{0}^{\pi/2} \int_{1}^{3} 2r^3 \cos\theta \sin\theta \, dr d\theta\]Proceed by evaluating the inner integral with respect to $ r $:\[= \int_{0}^{\pi/2} \cos\theta \sin\theta \left[ \frac{r^4

Answered: Image /qna-images/answer/3eb0a9e1-53e0-4f60-b3e9-cd2a1b747cf0.jpg

Answered: (2) Let R = evaluate the following…

(2) Let R = evaluate the following integral over R using polar coordinates: {(r, 0) | 1 < r < 3,0 < 0 < T/2}. Sketch the region of integration R and 2.xy dA (remember that dA = rdrde).

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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Question

### Problem Statement
Let $ R = \{(r, \theta) \mid 1 \leq r \leq 3, 0 \leq \theta \leq \pi/2 \} $. Sketch the region of integration $ R $ and evaluate the following integral over $ R $ using polar coordinates:

\[
\iint\limits_{R} 2xy \, dA \quad \text{(remember that } dA = rdrd\theta \text{)}.
\]

### Explanation of Region
The region $ R $ is defined in polar coordinates ($r, \theta$) where:

- $ r $ (the radius) ranges from 1 to 3.
- $ \theta $ (the angle) ranges from 0 to $\pi/2$ (which is the first quadrant).

In Cartesian coordinates, this region would correspond to a sector of a circle located in the first quadrant, with inner and outer radii of 1 and 3, respectively, and spanning from the positive x-axis (0 radians) to the positive y-axis ($\pi/2$ radians).

### Evaluating the Integral
To solve the integral, convert the integral into polar coordinates and use the given transformation $dA = rdrd\theta$. Therefore, the integral becomes:

\[
\iint\limits_{R} 2xy \, dA = \iint\limits_{R} 2 \cdot (r\cos\theta) \cdot (r\sin\theta) \cdot r d\theta dr
\]

Substituting $x$ and $y$ in polar coordinates ($x = r\cos\theta, y = r\sin\theta$):

\[
= \iint\limits_{R} 2r^3 \cos\theta \sin\theta \, dr d\theta
\]

Now, calculate the double integral:

\[
\int_{0}^{\pi/2} \int_{1}^{3} 2r^3 \cos\theta \sin\theta \, dr d\theta
\]

Proceed by evaluating the inner integral with respect to $ r $:

\[
= \int_{0}^{\pi/2} \cos\theta \sin\theta \left[ \frac{r^4

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