# Area between a parabola and line?

## 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

Related Calculus and Beyond Homework Help News on Phys.org
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
Mentor
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.