Rayanjafar's parametric integral question for YAnswers

[CaptainBlack]

"C4 question, please help.?
the curve C has parametric equations x = sint , y = sin2t, 0<t<pi/2
a) find the area of the region bounded by C and the x-axis

and, if this region is revolved through 2pi radians about the x-axis,
b) find the volume of the solid formed

How do you do this question. Can anyone please show me step by step?"

C4 here denotes a question appropriate to the UK Core 4 A-Level Maths Exam

 

[CaptainBlack]

(a) First sketch the curve. It obviously starts with slope \(2\) at \((0,0)\) and rises to a maximum of \(y=1\) at \(x=1/\sqrt(2)\) and then falls to \(y=0\) at \(x=1\).
View attachment 515The area we want is the integral:

\[I = \int_{x=0}^1 y(x) dx\]
Use the substitution \(t=arcsin(x), x=sin(t)\). Then \(dx = cos(t) dt\), and the integral becomes:

\[I = \int_{t=0}^{\pi/2} sin(2t) cos(t) dt\]
Now we replace the \(sin(2t)\) using the double angle formula by \(2 sin(t) cos(t)\) to get:

\[I = \int_{t=0}^{\pi/2} 2sin(t) (cos(t))^2 dt\]
As the integrand is the derivative of \(-(2/3) (cos(t))^3\) we get:

\[I = -(2/3) [0-1] = 2/3\].
The second part proceeds in much the same way once we write down the volume of revolution:

\[V= \int_{x=0}^1 \pi (y(x))^2 dx\]
and proceed in much the same way as before

CB