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Area Between Two Curves

  • Thread starter Slimsta
  • Start date
190
0
1. The problem statement, all variables and given/known data

Find the area of the region R enclosed by the line y=2x−3 and the parabola y2=4x+93.

2. Relevant equations



3. The attempt at a solution
[tex]
\int_{-9}^{11} 2x-3 dx~-~\int_{-9}^{11} (4x+93)^0^.^5 /6 dx
[/tex]

[x2-3x]11-9 - [(4x+93)1.5 / 6 ]11-9
i get: The area of the region R is A = 215.5338875

whats wrong with it?
 
Last edited:
If I have the root of a number, what is the power it is raised to? It is not 1.5 (that is the root of a number, cubed). Also, why dt? Shouldn't you be integrating with respect to x (dx)?
 
32,357
4,142
There are at least a couple things wrong with your integral.
  1. You are integrating with respect to t - there 's not a t anywhere in sight.
  2. There should not be an exponent of 1.5 on anything.
  3. You are apparently using vertical strips, which makes your integrals incorrect, since you have the wrong limits of integration for that. You can do this problem with vertical strips, but that's the harder way to do this.

Slimsta, did you graph the region? If you did, you would see that horizontal strips are the easier way to go. If you do it this way, you need only one integral, not two, as you would need if you use vertical strips. And besides that, the integral is easier to work with.

Using horizontal strips, the typical area element is [itex]\Delta A = [x_{line} - x_{parabola}]\Delta y[/itex].

In your equation for the line, you'll need to solve for x as a function of y. In your parabola equation, you'll also need to solve for x as a different function of y. Your horizontal strips run from y = -9 to y = 11. Note that vertical strips run from x = -93/4 to x = 7, but the lower end of the vertical strips changes from the y values on the parabola to the y values on the line at x = -3, making it necessary to have two integrals with different limits of integration (x = -93/4 to x = -3 and x = -3 to x = 7).

That should get you started.

Draw the graph if you haven't already done so!
 
190
0
There are at least a couple things wrong with your integral.
  1. You are integrating with respect to t - there 's not a t anywhere in sight.
  2. There should not be an exponent of 1.5 on anything.
  3. You are apparently using vertical strips, which makes your integrals incorrect, since you have the wrong limits of integration for that. You can do this problem with vertical strips, but that's the harder way to do this.

Slimsta, did you graph the region? If you did, you would see that horizontal strips are the easier way to go. If you do it this way, you need only one integral, not two, as you would need if you use vertical strips. And besides that, the integral is easier to work with.

Using horizontal strips, the typical area element is [itex]\Delta A = [x_{line} - x_{parabola}]\Delta y[/itex].

In your equation for the line, you'll need to solve for x as a function of y. In your parabola equation, you'll also need to solve for x as a different function of y. Your horizontal strips run from y = -9 to y = 11. Note that vertical strips run from x = -93/4 to x = 7, but the lower end of the vertical strips changes from the y values on the parabola to the y values on the line at x = -3, making it necessary to have two integrals with different limits of integration (x = -93/4 to x = -3 and x = -3 to x = 7).

That should get you started.

Draw the graph if you haven't already done so!
I made some mistakes in my last one when copying it from my paper work
okay i finally got it! my mistake was that i didnt make XR - XL
i used YR - YL lol.. finally! such a small mistake f-ed everything up
 
Last edited:

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