Area Bounded By Two Polar Curves Practice Test

$\dfrac12\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$

$\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)-(2\sin\theta)\big]d\theta$

$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$

$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(4\sin\theta)^2\big],d\theta$

$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$