Trig Substitution with circle

Question

Answer 1

function semicircle with positive y values is y=(r²-x²)^1/2

so area circle A=2∫(r²-x²)^1/2dx, where x is from -r to r.

set x=rcos(t) where r is constant. If t ranges from π to 0, x ranges from -r to r.

dx--rsin(t)dt

Now A=2∫r(1-cos²(t))^1/2(-rsin(t))dt=-r²∫sin²(t)dt ,,,,where t is from π to 0

sin²(t)=(1/2)(1-cos(2t))

A=-2r²∫(1/2)(1-cos(2t))dt ,,,t from π to 0

A=-r²(t-sin(2t)/2) ,,,, from π to 0

A=-r²((0-0)-(π-0))=πr²

Answer 2

The integral you want in rectangular co-ordinates is 4 times the integral from 0 to r of sqrt(r²-x²) dx.

The substitution you want is x=r sin u, dx=r cos u and the integral becomes 4 times the integral from 0 to pi/2 of r²cos²u du. Use the identity cos²u = (1/2)(1+cos 2u).