Area between a parabola and line?

[smandphysics]

Homework Statement

Find the area of the region enclosed by the curves:
2y=sqrt(4x)
y=5 and
2y+3x=7

Homework Equations

A = integral from a to b of f(x)-g(x) dx

The Attempt at a Solution

Tried to integrate this with respect to y.
Found the intersection points to be y=2, -14/3
Then did the integral from -14/3 to 2 of (7/3 - 2y/3) - (1/4y^2)
My answer was 1000/81 but this is incorrect.
Also tried integral from -14/3 to 5 of (7/3 - 2y/3) - (1/4y^2)./
My answer was 841/324 but this was also incorrect.

Thanks to anyone who can help explain this to me, I'm so lost :s

 

[armolinasf]

I would recommend you graph all those functions to give you an idea of where all the intersections are. For example y=(7-3x)/2 intersects y=5 in one place. Similarly y=(sqrt4x)/2 intersects y=5 in one place. And y=(7-3x)/2 and y=(sqrt4x)/2 intersect each other in one place. Your job is to sort out which integral to subtract from which to get at the area between all three functions. Hope this helps.

 

[Mark44]

In addition to what armolinasf said, the equation 2y = sqrt(4x) can be simplified to y = sqrt(x).

If you break up the region into vertical strips, you will need to use two integrals. If you break the region up into horizontal strips, you'll need only one integral.