Which set up below finds the area of the region inside the first curve and outside the second curve in the third quadrant? 5/12 de r = 4 sin(20), r2 = 2 A = 2 12 T/12 5/12 (Sam 12 (r - r) do T/12 /12 r de - 2 5/12 /12 r de A = ( do- do) 2 T/12 i de 0