(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

Integral of right function - left function if integrating with respect to y

F(b)-F(a)

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Finding volume given by curves using integral

**Physics Forums | Science Articles, Homework Help, Discussion**