Finding volume given by curves using integral

[Moonstar]

Homework Statement

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=5 sqrt(x), y=3 and 2y+4x=9

Homework Equations

Integral of right function - left function if integrating with respect to y
F(b)-F(a)

The Attempt at a Solution

I decided to integrate with respect to y since after I got the graph, the area is kind of split up into two different parts if you decide to do it with respect x. So if I do it with respect to y the right function would be 2y=5 sqrt(x) and to make it with respect to y it would be (4/5)y^2. The left function is 2y+4x=9 or with respect to y it would be -(2y-9)/4.

To find the bounds I found where the line y=3 intersected with 2y=5 sqrt(x) and where 2y=5 sqrt(x) intersected with 2y+4x=9. I found my "y bounds" to be 2 to 3.So I plug everything in.

Integral from 2 to 3 of: (4/25)*y^2 - [(-2y-9)/4]
So then I took the anti-derivative and got

4/75y^3 - 1/4y^2 + 9/4y

Then I plugged 3 into the above equation to get my F(b) and then plugged in 2 into the equation to get my F(a). I did F(b)-F(a)

(4/75*(3)^3+1/4*(3)^2+9/4*(3)) - (4/75*(2)^3+1/4*(2)^2+9/4*(2)) = 4.513333

... which is not the answer :( Can someone please help me? I'm sorry if this is hard to read, I tried using the symbols but it kept messing everything up; I'm new to this forum and I don't really know how to use the symbols.

 

[murmillo]

I think you got the lower bound of y wrong. Check your algebra. Hint: It might be easier to find the value of x that satisfies 2y=5sqrt (x) and 2y+4x=9, then plug x into the equation to get the y-value.

 

[Moonstar]

You're right, the lower bound was my problem... I think I estimated it on a graph instead of actually getting the correct decimals and the online homework I am doing is very specific with answers. Thank you, I have completed this problem finally :) Now on to more...