# Area of a region

1. Mar 21, 2007

### Aerosion

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=3(x^1/2)
y=4
2y+4x=7

2. Relevant equations

3. The attempt at a solution

I decided to solve for y, since that would make it easier to graph. This got me y=(3(x^1/2)/2) and y=7/2-2x. I then graphed these, finding that the span on the x-axis would be from 0 to 2.

I integrated the problem, getting 3-2*(2^1/2). This seems wrong. What happened? I'm probably overlooking something small...

(By the way, Latex isn't working, so the x^1/2 things means that that number or variable is square rooted.

Last edited: Mar 21, 2007
2. Mar 22, 2007

### d_leet

Hang on, what function did you integrate exactly?

3. Mar 22, 2007

### Aerosion

Well, when I graphed it, I noticed that 7/2-2x was above 3(x^1/2)/2. So I integrated (7/2-2x) - (3(x^1/2)/2) with an upper limit of 2 and a lower limit of 0, and that got me 3-2*(2^1/2).

$$\int_{0}^{2} (\frac{7}{2} - 2x) - (\frac{3*\sqrt{x}}{2}) dx$$

Last edited: Mar 22, 2007
4. Mar 22, 2007

### Aerosion

Okay, I've sort of figured it out. I have to solve for x in the equation, even though I solved for y in the graph. So here it is, revised...

$$\int_{1.5}^{3} \frac{4*y^2}{9} - \frac{-(2*y - 7)}{4} dy$$

It's still not coming out correct, though (I'm doing an internet-generated problem and it's saying that it's wrong). I'm getting 2.5625, by the way. Am I still doing something wrong?

5. Mar 22, 2007

### d_leet

Are you sure about " 3-2*(2^1/2)" as an answer, I can see where the 3 comes from, but I don't see how you got the square root of two there, what is the integral of the squareroot of x?

6. Mar 23, 2007

### HallsofIvy

Staff Emeritus
Why do you need to solve for x rather than y? That's a crucial part of the problem.