Area between Curves: Finding the Solution

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Homework Statement

Find the area of the bounded region enclosed by the curves: 6x+y^2=13, x=2y

Homework Equations

Integration

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!

 

[Mark44]

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 (x_right - x_left)\Delta y, which always gives you a positive value.

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

 

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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 (x_right - x_left)\Delta y, which always gives you a positive value.

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

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+

 

[Mark44]

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.

 

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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.

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.

 

[Mark44]

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.

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.

 

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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!

 

[Mark44]

That's correct.

 

[hover]

That's correct.

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!

 

[Mark44]

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.