What is the Area of the Region Bounded by r=sinθ in the Sector 0≤θ≤π/2?

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Discussion Overview

The discussion revolves around finding the area of the region bounded by the curve defined by the polar equation \( r = \sqrt{\sin \theta} \) within the sector defined by \( 0 \leq \theta \leq \frac{\pi}{2} \). The focus is on the application of the area formula in polar coordinates.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks how to find the area of the region bounded by the curve \( r = \sqrt{\sin \theta} \) for the specified sector.
  • Another participant provides the formula for calculating the area in polar coordinates, stating \( A = \frac{1}{2} \int_a^b f(\theta)^2 d\theta \).
  • A third participant specifies that in this case, \( f(\theta) = r = \sqrt{\sin(\theta)} \) and proceeds to calculate the area, arriving at \( A = \frac{1}{2} \).
  • A later reply corrects the earlier area calculation, indicating that the area should actually be \( \frac{2\pi}{3} \) and expresses surprise at the realization that \( r \) corresponds to \( f(\theta) \).

Areas of Agreement / Disagreement

There is no consensus on the correct area calculation, as one participant initially arrives at \( \frac{1}{2} \), while another later suggests it should be \( \frac{2\pi}{3} \). The discussion remains unresolved regarding the correct area value.

Contextual Notes

The discussion includes potential confusion regarding the interpretation of the polar function and the application of the area formula, with different participants arriving at different results without resolving the discrepancies.

ineedhelpnow
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find the area of the region that is bounded by the given curve and lies in the specified sector. $r=\sqrt{sin \theta}$, $0 \le \theta \le \pi/2$how do i do this?
 
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ineedhelpnow said:
find the area of the region that is bounded by the given curve and lies in the specified sector. $r=\sqrt{sin \theta}$, $0 \le \theta \le \pi/2$how do i do this?
The formula to calculate the area is :
$$A=\frac{1}{2} \int_a^b f(\theta)^2 d \theta$$
 
In this case, $f(\theta)=r= \sqrt{ \sin(\theta)}$

Therefore:

$$A=\frac{1}{2} \int_0^{\frac{\pi}{2}} (\sqrt{ \sin(\theta)})^2 d \theta =\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(\theta) d \theta= \frac{1}{2}\left [- \cos(\theta) \right ]_0^{\frac{\pi}{2}}=\frac{1}{2} \left(-\cos \frac{\pi}{2}+ \cos 0 \right )=\frac{1}{2}$$
 
oops it was supposed to be $2 \pi/3$ thanks though. i didnt realize that r was $f(\theta)$
 

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