Suppose R is the shaded region in the figure. As an iterated integral in polar coordinates, B D f f(x, y) dA= [₁ fº R with limits of integration A = ? B = ? C = 0 D = 3sqrt(2) ← - ▶ 2 -3 [ f( cos(9), 7 sin(0)) r dr de -2 2 3

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Suppose \( R \) is the shaded region in the figure. As an iterated integral in polar coordinates,

\[
\iint_R f(x, y) \, dA = \int_A^B \int_C^D f(r \cos(\theta), r \sin(\theta)) \, r \, dr \, d\theta
\]

with limits of integration

- \( A = ? \)
- \( B = ? \)
- \( C = 0 \)
- \( D = 3\sqrt{2} \)

**Diagram Explanation:**

The graph is a polar coordinate plot showing a sector of a circle. The sector is bounded by two lines forming an angle at the origin and a curved arc at the top. The bounding lines create an angle with the x-axis, forming the sector shown in the shaded area.

The shaded area forms a sector that is symmetric about the y-axis, extending from \(-\pi/4\) to \(\pi/4\) in terms of angle \(\theta\). The arc at the top of the shaded region is part of a circle with radius \(3\sqrt{2}\). 

To complete the integration limits:

- \( A = -\frac{\pi}{4} \)
- \( B = \frac{\pi}{4} \)

This corresponds to complete coverage of the shaded sector.
Transcribed Image Text:Suppose \( R \) is the shaded region in the figure. As an iterated integral in polar coordinates, \[ \iint_R f(x, y) \, dA = \int_A^B \int_C^D f(r \cos(\theta), r \sin(\theta)) \, r \, dr \, d\theta \] with limits of integration - \( A = ? \) - \( B = ? \) - \( C = 0 \) - \( D = 3\sqrt{2} \) **Diagram Explanation:** The graph is a polar coordinate plot showing a sector of a circle. The sector is bounded by two lines forming an angle at the origin and a curved arc at the top. The bounding lines create an angle with the x-axis, forming the sector shown in the shaded area. The shaded area forms a sector that is symmetric about the y-axis, extending from \(-\pi/4\) to \(\pi/4\) in terms of angle \(\theta\). The arc at the top of the shaded region is part of a circle with radius \(3\sqrt{2}\). To complete the integration limits: - \( A = -\frac{\pi}{4} \) - \( B = \frac{\pi}{4} \) This corresponds to complete coverage of the shaded sector.
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