# Area between curves

1. Feb 3, 2010

### hover

1. The problem statement, all variables and given/known data
Find the area of the bounded region enclosed by the curves: 6x+y^2=13, x=2y

2. Relevant equations
Integration

3. The attempt at a solution
Finding the area shouldn't be too much of a problem. In this particular problem, integrating with respect to y is the better choice compared to integrating with x. If you graph out the two lines, you will see that most of the area will lie under the x axis.

My real question is do I treat that intended area as a positive value even though its under the curve? I mean the question states find the area. Would I treat this area as negative are because its under the x axis? I don't know whether to treat it positive or negative .

Thanks for the response!

2. Feb 3, 2010

### Staff: Mentor

Area is never negative. For example,
$$\int_0^{2\pi} sin x dx = 0$$
but the area between the curve y = sin x and the x-axis, between 0 and 2pi, is 4.

For your real question, it all boils down to how you set up your incremental area element, $\Delta A$. For horizontal area elements, the area of each is (xright - xleft)$\Delta y$, which always gives you a positive value.

For vertical area elements, the areaof each is (ytop - ybottom)$\Delta x$, which also always gives you a positive value.

3. Feb 3, 2010

### hover

So any area I want can be considered to be positive all the time? Even in my example? http://www.wolframalpha.com/input/?i=plot+6x%2By^2%3D13%2C+x%3D2y+

4. Feb 3, 2010

### Staff: Mentor

It doesn't matter whether the region is above or below the x-axis or mixed. You calculate area in such a way that it comes out positive or possibly zero.

5. Feb 3, 2010

### hover

Ok. I just want to make sure. I know that integrating under the x axis returns a negative number by nature. I should be able to solve this.

6. Feb 3, 2010

### Staff: Mentor

Not necessarily. If you set up the integral with the right area element, an integral that represents the area will come out positive. See what I said in post #2.

7. Feb 4, 2010

### hover

I did out the problem and I would like someone to confirm my answer please. I got 686/9 for an answer. The bounds I used were b=1 and a=-13. If you want to see my work in more detail that's fine (I'll put it through a scanner) but i just want to make sure i have the right answer.

Thanks!!

8. Feb 5, 2010

### Staff: Mentor

That's correct.

9. Feb 5, 2010

### hover

Thanks for your help Mark44! I thought I may have had it right but as anyone knows, all it takes is a flip in one number or sign before everything comes crashing down!

Thanks!

10. Feb 5, 2010

### Staff: Mentor

Yes, and I found this to be a little tricky to evaluate at -13. Had to go back and check my work a couple of times before I caught an error I had made.