3y = -3y + 9 ???
Fix that.
Then, find where the two graphs intersect. Since the parabola opens to the right, it's easier to integrate along dy. The area is just a bunch of thin strips of height dy and length the distance between the two curves. Express that and integrate from bottom to top.
Calculate the area of the bounded region between the curves y^2=x and 3y = -3y + 9 ?
3 answers
3y = -3x + 9
the curves intersect at y = (-1-√13)/2 and (-1+√13)/2
area is thus
∫[(-1-√13)/2,(-1+√13)/2] (-y+3)-(y^2) dy
area is thus
∫[(-1-√13)/2,(-1+√13)/2] (-y+3)-(y^2) dy