Question

polar coordinates lecture participation day 2: probl(2 points)find the area of the region that is bounded by the curve $r = \sqrt{5\sin(\theta)}$ and lies in the sector $0 \leq \theta \leq \pi$.area =you have attempted this problem 0 times.you have unlimited attempts remaining.page generated december 1, 2025, 10:11:0webwork © 1996-2025 | theme: math4 | ww_version:

Explanation:

Step1: Recall the formula for the area in polar coordinates

The formula for the area \( A \) bounded by a polar curve \( r = f(\theta) \) from \( \theta = a \) to \( \theta = b \) is \( A=\frac{1}{2}\int_{a}^{b}r^{2}d\theta \). Here, \( r = \sqrt{5\sin\theta} \), \( a = 0 \), and \( b=\pi \). So first, we square the radius function: \( r^{2}=5\sin\theta \).

Step2: Set up the integral for the area

Substitute \( r^{2} \) into the area formula: \( A=\frac{1}{2}\int_{0}^{\pi}5\sin\theta d\theta \). We can factor out the constant \( \frac{5}{2} \) from the integral: \( A = \frac{5}{2}\int_{0}^{\pi}\sin\theta d\theta \).

Step3: Integrate \( \sin\theta \)

The integral of \( \sin\theta \) with respect to \( \theta \) is \( -\cos\theta \). So we evaluate \( \frac{5}{2}[-\cos\theta]_{0}^{\pi} \).

Step4: Evaluate the definite integral

First, substitute \( \theta=\pi \): \( -\cos(\pi)=-(-1) = 1 \). Then substitute \( \theta = 0 \): \( -\cos(0)=-1 \). Now subtract the lower limit from the upper limit: \( 1-(-1)=2 \). Then multiply by \( \frac{5}{2} \): \( \frac{5}{2}\times2 = 5 \).

Answer:

\( 5 \)