# Rayanjafar's parametric integral question for Y!Answers

#### CaptainBlack

##### Well-known member
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

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#### CaptainBlack

##### Well-known member
(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 $$x=1$$.

The 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

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