Area Bounded By Two Polar Curves Practice Test

$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$

$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$

$\dfrac{1}{2}\displaystyle\int_{0}^{2\pi}\big[(2\cos\theta)^2-(2\sin\theta)^2\big], d\theta$

$\dfrac{1}{2}\displaystyle\int_{\pi/4}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$

$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big], d\theta$