The region bounded by the axes and the curve $y=\cos x$ from $x=0$ to $x=\frac{1}{2}\pi $ is divided

Question

The region bounded by the axes and the curve $y=\cos x$ from $x=0$ to $x=\frac{1}{2}\pi $ is divided into two parts, of areas ${{A}_{1}}$ and ${{A}_{2}}$, by the curve $y=\sin x$ (see diagram). Prove that ${{A}_{2}}=\left( \sqrt{2} ight){{A}_{1}}$. [6] The two curves meet at $P$. The line through $P$parallel to the $x$-axis meets the $y$-axis at $Q$. The region $OPQ$, bounded by the arc $OP$ and the lines $PQ$ and $QO$, is rotated through $4$ right angles about the $y$-axis to form a solid of revolution of volume $V$. It is given that $C=\pi \displaystyle\int
olimits_{0}^{\frac{1}{\sqrt{2}}}{{{\left( {{\sin }^{-1}}y ight)}^{2}}}\mathrm{d}y$. By substituting $u={{\sin }^{-1}}y$, show that $V=\pi \displaystyle\int
olimits_{0}^{\frac{1}{4}\pi }{{{u}^{2}}\cos u\,\mathrm{d}u}$. [2] Show that $\frac{\mathrm{d}}{\mathrm{d}u}\left( {{u}^{2}}\sin u+2u\cos u-2\sin u ight)={{u}^{2}}\cos u$. [2] Hence find the exact value of $V$. [2]

Accepted Answer

(iii) $\frac{\pi }{\sqrt{2}}\left( \frac{{{\pi }^{2}}}{16}+\frac{\pi }{2}-2 \right)$ units$^{3}$

N2003 P1 Q14

N2003 P1 Q14

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