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

[ineedhelpnow]

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?

 

[evinda]

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 :
$$A=\frac{1}{2} \int_a^b f(\theta)^2 d \theta$$

 

[evinda]

In this case, $f(\theta)=r= \sqrt{ \sin(\theta)}$

Therefore:

$$A=\frac{1}{2} \int_0^{\frac{\pi}{2}} (\sqrt{ \sin(\theta)})^2 d \theta =\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(\theta) d \theta= \frac{1}{2}\left [- \cos(\theta) \right ]_0^{\frac{\pi}{2}}=\frac{1}{2} \left(-\cos \frac{\pi}{2}+ \cos 0 \right )=\frac{1}{2}$$

 

[ineedhelpnow]

oops it was supposed to be $2 \pi/3$ thanks though. i didnt realize that r was $f(\theta)$