N2003 P1 Q14

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 \({{A}_{2}}=\left( \sqrt{2} \right){{A}_{1}}\).[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

\(C = \pi \int_{0}^{\frac{1}{\sqrt{2}}} \left(\sin^{-1} y\right)^2 \mathrm{d}y.\)

1. By substituting \(u={{\sin }^{-1}}y\), show that \(V = \pi \int_{0}^{\frac{1}{4}\pi} u^2 \cos u \, \mathrm{d}u\).

   [2]
2. Show that \(\frac{\mathrm{d}}{\mathrm{d}u}\left( {{u}^{2}}\sin u+2u\cos u-2\sin u \right)={{u}^{2}}\cos u\).[2]
3. Hence find the exact value of \(V\).[2]