The sketch displays the curve given by $y = \sqrt{\sin 2x + \sin^2 2x}$ for $0 \leq x \leq \frac{1}{6}\pi$. The shaded area lies between the curve, the vertical line $x = \frac{1}{6}\pi$ and the horizontal line $y = 0$.
(a)[3]
Apply the trapezium rule with two intervals to estimate the area of the shaded region. Give your answer correct to $2$ significant figures.
(b)[6]
The shaded region is turned completely about the $x$-axis. Find the exact volume of the solid formed.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Take $y$-values $(0)$, $\sqrt{\sin \frac{\pi}{6} + \sin^2 \frac{\pi}{6}}$, $\sqrt{\sin \frac{\pi}{3} + \sin^2 \frac{\pi}{3}}$ or decimal equivalents” …