Find the Area bound by the curves

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=3sqrtx
y=5
and
2y+4x=7

Help Please!!

Help please!!

I'm have prepared a graph for you. The variable will be y. The work will be on the graph. I couldn't stand all of the fractions so I got lazy and used a calculator. Feel free to do it with fractions if you want to.

http://i369.photobucket.com/albums/oo133…

The answer is 17.5810

.

Refer to the figure in the link as under.
http://www.flickr.com/photos/52771834@N0…

Solving y = 5 with 2y + 4x = 7
=> x = - 3/4
=> A = (-3/4, 5) and D = (-3/4, 0)

Solving 2y = 3√x with 2y = 7 - 4x
=> 4x + 3√x - 7 = 0
=> √x = (1/8) [-3 + √(121)] = 1
=> x = 1
=> E = (1, 3/2) and F = (1, 0)

Solving y = 5 with 2y = 3√x,
B = (100/9, 5) and C = (100/9, 0)

Required area
= area of rectangle ABCD - area of trapezium AEFD - area under the parabola between EF and BC
= AB * AD - (1/2 (AD + EF) * DF - ∫ (x = 1 to 100) (1/2) √x dx
= (100/9 + 3/4) * 5 - (1/2) (5 + 3/2) * (1 + 3/4) - (1/3) x^(3/2) [x=1 to 100/9]
= (427/36)*5 - 91/16 - (1000/27 - 1)/3
= 59.306 - 5.688 - 37
= 16.618.

Edit:
Arithmetical error corrected.