How Do You Find the Area of a Region Enclosed by Given Curves?

[Aerosion]

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

Homework Equations

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.

 

[d_leet]

Hang on, what function did you integrate exactly?

 

[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).

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

 

[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?

 

[d_leet]

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).

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

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?

 

[HallsofIvy]

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