Question 1029007
{{{y^2 = x(2-x)}}} has domain the closed interval [0,2].  (Outside it, y^2 = x(2-x)is negative, which is impossible. Verify!!)

==>Area = {{{int((sqrt(x(2-x))-(-sqrt(x(2-x)))), dx, 0,2) = 2*int(sqrt(x(2-x)), dx, 0,2)}}}

={{{2*int(sqrt(1-(x-1)^2), dx, 0,2) = 2*int(sqrt(1-(x-1)^2), d(x-1), 0,2)}}}.

Now let {{{x - 1 = sin(theta)}}}.
If x = 2, {{{sin(theta) = 1}}},==> {{{theta = pi/2}}}
If x = 0, {{{sin(theta) = -1}}},==> {{{theta = -pi/2}}}.
Also, {{{d(x-1) = dx = cos(theta)d(theta)}}}

==> Area = {{{2*int(cos(theta)* cos(theta),d(theta), -pi/2,pi/2)}}}

={{{2*int(cos^2(theta),d(theta), -pi/2,pi/2)}}}

={{{2*(1/2)int(1+ cos(2theta),d(theta), -pi/2,pi/2)}}}

={{{int(1+ cos(2theta),d(theta), -pi/2,pi/2) =(theta + (1/2)sin(2theta))[-pi/2]^(pi/2)}}}

= {{{pi/2+(1/2)sin(pi)-(-pi/2 +sin(-pi))}}} 

= {{{pi/2 +pi/2 = pi}}} sq. units.