# Finding the area of a region bounded by THREE curves

• Feb 20th 2009, 12:14 PM
fattydq
Finding the area of a region bounded by THREE curves
I've attached the problem I'm working on. I didn't think doing these types of problems when there were only two curves was too bad but where would I even begin now that there's three? How do I know what to subtract from what to subtract from what? I've sketched it and I can figure it out from that but...then what becomes the bounds of the integral. I'm so confused (Speechless)
• Feb 20th 2009, 12:20 PM
mr fantastic
Quote:

Originally Posted by fattydq
I've attached the problem I'm working on. I didn't think doing these types of problems when there were only two curves was too bad but where would I even begin now that there's three? How do I know what to subtract from what to subtract from what? I've sketched it and I can figure it out from that but...then what becomes the bounds of the integral. I'm so confused (Speechless)

Draw a graph of the the three curves and shade the required regions. Find the x-coordinates of the intersection points of the curves. Break up the required region into subregions that have a clear upper and lower curve. Find the area of each subregion.
• Feb 20th 2009, 12:31 PM
fattydq
Quote:

Originally Posted by mr fantastic
Draw a graph of the the three curves and shade the required regions. Find the x-coordinates of the intersection points of the curves. Break up the required region into subregions that have a clear upper and lower curve. Find the area of each subregion.

Ok I found it to be 1/2 between 4x^2 and-5x+6, 1 between x^2 and -5x+6 and the obvious 0. Now I'll have two integrals, one from 0 to 1/2 and the other from 1/2 to 1, but what values do I subtract from what? Just whatever values were used to find those intersection points? Like between 4x^2 and -5x+6 I'd use 4x^2 as the upper and -5x+6 as the lower with the integral with bounds 1/2 to 1? Which ones would I use from 0 to 1/2? I'll be doing this twice and just adding them?

Edit: I thought of a better way to word what I'm asking. I understand that I'll end up breaking it into two subintervals and adding them, but I don't understand what function would be associated with each curve...and how to find out this.
• Feb 20th 2009, 03:47 PM
mr fantastic
Quote:

Originally Posted by fattydq
I thought of a better way to word what I'm asking. I understand that I'll end up breaking it into two subintervals and adding them, but I don't understand what function would be associated with each curve...and how to find out this.

When you drew "a graph of the the three curves and shade the required regions" did you label each graph with its equation? So just read the equation associated with each curve ....
• Feb 20th 2009, 04:17 PM
fattydq
Quote:

Originally Posted by mr fantastic
When you drew "a graph of the the three curves and shade the required regions" did you label each graph with its equation? So just read the equation associated with each curve ....

Yeah but I can't even tell which is the bottom and which is the top curve on some because they cross at a seemingly even point.
• Feb 20th 2009, 05:10 PM
HallsofIvy
Quote:

Originally Posted by fattydq
Yeah but I can't even tell which is the bottom and which is the top curve on some because they cross at a seemingly even point.

I have no idea what a "seemingly even point" is.

It looks to me easy to see that the "top" curve is $y= 4x^2$ up until the line 5x+ y= 6 intersects it and that line is the "top" curve untill it intersects $y= x^2$.
• Feb 23rd 2009, 07:36 AM
fattydq
Quote:

Originally Posted by HallsofIvy
I have no idea what a "seemingly even point" is.

It looks to me easy to see that the "top" curve is $y= 4x^2$ up until the line 5x+ y= 6 intersects it and that line is the "top" curve untill it intersects $y= x^2$.

so it'd be the integral of 4x^2-(-5x+6) added to the integral of -5x+6-(x^2), and the bounds for each integral would just be the intersection points between the two that were subtracted, correct?
• Feb 23rd 2009, 01:57 PM
fattydq
4x^2=-5x+6, how would I find this intersecting point? Because it could simplify out to 4x^2+5x=6 but I can't think of any number that would satisfy that? And then same with -5x+6=x^2, it doesn't simplify out evenly other than using 1, but what would the other value be?
• Feb 23rd 2009, 04:40 PM
mr fantastic
Quote:

Originally Posted by fattydq
4x^2=-5x+6, how would I find this intersecting point? Because it could simplify out to 4x^2+5x=6 but I can't think of any number that would satisfy that? And then same with -5x+6=x^2, it doesn't simplify out evenly other than using 1, but what would the other value be?

I suggest re-arranging the equations into the form $y = ax^2 + bx + c = 0$ and then using the quadratic formula.