Definite Integrals: Areas and Volumes Qn 1015b

With origin

{O,}

the curves with equations

{y=\sin x}

and

{y=\cos x,}

where

{0 \leq x \leq \frac{1}{2} \pi,}

meet at the point

{P}

with coordinates

{\left( \frac{1}{4} \pi, \frac{1}{2} \sqrt{2} \right).}

The area of the region bounded by the curves and the ${x\textrm{-axis}}$ is ${A_1}$ and the area of the region bounded by the curves and the ${y\textrm{-axis}}$ is ${A_2.}$

Show that

\frac{A_1}{A_2} = \sqrt{2}.

The region bounded by

{y=\sin x}

between

{O}

and

{P,}

the line

{y=\frac{1}{2} \sqrt{2}}

and the

{y\textrm{-axis}}

is rotated about the

{y\textrm{-axis}}

through

{360^\circ.}

Show that the volume of the solid formed is given by

\pi \int_{0}^{\frac{1}{2} \sqrt{2}} (\sin^{-1} y)^2 \; \mathrm{d}y.

Show that the substitution

{y=\sin u}

transforms the integral in part (ii) to

\pi \int_a^b u^2 \cos u \; \mathrm{d}u,

for limits

{a}

and

{b}

to be determined.

Hence find the exact volume.

Question 1015b

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