Find the area of the specified region.
-Inside one leaf of the four-leaved rose $r = 3 \sin 2 \theta$
A) $\frac { 3 \pi } { 2 }$
B) $\frac { 9 \pi } { 8 }$
C) $\frac { 9 \pi } { 2 }$
D) $\frac { 9 \pi } { 8 } - \frac { 9 } { 16 }$

Question

Quizplus · Accepted Answer

The area inside one leaf of the four-leaved rose $r = 3 \sin 2\theta$ can be found using the formula for the area in polar coordinates, $A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$. For one leaf of $r = 3 \sin 2\theta$, the limits of integration are from $0$ to $\frac{\pi}{4}$ because the rose has four leaves and spans $2\pi$ radians, so each leaf spans $\frac{\pi}{2}$, but since the equation gives us twice the angle, the effective range for one leaf is halved. Thus, the area is $A = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (3 \sin 2\theta)^2 d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} 9\sin^2 2\theta d\theta$. Solving this integral gives $A = \frac{9\pi}{8}$.

Find the Area of the Specified Region r=3sin⁡2θr = 3 \sin 2 \thetar=3sin2θ

Find the Area of the Specified Region $r = 3 \sin 2 \theta$