N2015 P1 Q10

Do not use a calculator in answering this question.

With origin \(O\), the curves with equations \(y=\sin x\) and \(y=\cos x\), where \(0\le x\le\frac12\pi\), meet at the point \(P\) with coordinates \((\frac14\pi,\frac12\sqrt2)\). The area of the region bounded by the curves and the \(x\)-axis is \(A_1\) and the area of the region bounded by the curves and the \(y\)-axis is \(A_2\) (see diagram).

1. Show that \(\dfrac{A_1}{A_2}=\sqrt2\).[4]
2. The region bounded by \(y=\sin x\) between \(O\) and \(P\), the line \(y=\frac12\sqrt2\) and the \(y\)-axis is rotated about the \(y\)-axis through \(360^\circ\). Show that the volume of the solid formed is given by \[\pi\int_0^{\frac12\sqrt2}(\sin^{-1}y)^2\,\mathrm dy.\][2]
3. Show that the substitution \(y=\sin u\) transforms the integral in part (ii) to \(\pi\displaystyle\int_a^b u^2\cos u\,\mathrm du\), for limits \(a\) and \(b\) to be determined. Hence find the exact volume.[6]