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. 2y=3sqrt(x) , y=3 and 2y+3x=6.

Answer 1

It is relatively easy to find the intersections

y = 3 and 2y+3x = 6 intersect at (0,3)
y = (3/2)√x and y = 3 intersect at (4,3)
y = (3/2)√x and the line 2y+3x = 6 intersect at (1,3/2)

If you take vertical slices you must find the area in two parts, but if we take horizontal slices we can do it in one calculation , letting y run from 3/2 to 3

we have to solve 2y = 3√x for x ---> x = (4/9)y^2
and solve 2y+3x=6 for x ---> x = 2 - (2/3)y
width of region = (4/9)y^2 - 2 + (2/3)y

area = [integral] (4/9)y^2 - 2 + (2/3)y dy from 3/2 to 3
= [(4/27)y^3 - 2y + (1/3)y^2] from 3/2 to 3
= (4 - 6 + 3) - (1/3 - 3 + 3/4)
= 35/12

check my arithmetic