The area of the region bounded by the curve defined by \( r = \sqrt{\sin(\theta)} \) within the sector \( 0 \leq \theta \leq \frac{\pi}{2} \) is calculated using the formula \( A = \frac{1}{2} \int_a^b f(\theta)^2 d\theta \). Substituting \( f(\theta) = \sqrt{\sin(\theta)} \), the area is computed as \( A = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(\theta) d\theta \), resulting in an area of \( \frac{1}{2} \). However, the correct area should be \( \frac{2\pi}{3} \), highlighting the importance of accurately identifying \( r \) as \( f(\theta) \).
PREREQUISITESStudents and educators in mathematics, particularly those focusing on calculus and polar coordinates, as well as anyone interested in understanding area calculations in polar systems.
The formula to calculate the area is :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?