Calculus Area between curves
posted by Amy .
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
3y+x=3 , y^2x=1

Why are you switching names ??
find the intersection;
from the first:
x = 3  3y
into the 2nd:
y^2  (33y) = 1
y^2 + 3y  4 = 0
(y+4)(y1) = 0
y = 4 or y = 1
in x = 33y ....
if y = 1, x = 0 >(0,1)
if y = 4 , x = 15 , > (15, 4)
Your sketch should look like this and my answers are confirmed
http://www.wolframalpha.com/input/?i=plot+3y%2Bx%3D3+%2C+y%5E2x%3D1
judging from the graph I would integrate with respect to y
so the effective width of my horizontal slice
= 33y  (y^2  1)
= 4  3y  y^2
area = ∫(4  3y  y^2) dy from 4 to 1
= (4y  (3/2)y^2  (1/3)y^3)  from 4 to 1
= (4  3/2  1/3)  (16 24 + 64/3)
= 13/6  (56/3)
= 125/6