Results 1 to 9 of 9

Math Help - Finding the area of a region bounded by THREE curves

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    242

    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
    Attached Thumbnails Attached Thumbnails Finding the area of a region bounded by THREE curves-untitled.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by fattydq View Post
    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
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by mr fantastic View Post
    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.
    Last edited by mr fantastic; February 20th 2009 at 03:48 PM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by fattydq View Post
    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 ....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by mr fantastic View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,792
    Thanks
    1532
    Quote Originally Posted by fattydq View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by HallsofIvy View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2008
    Posts
    242
    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by fattydq View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 9th 2011, 05:15 AM
  2. Replies: 4
    Last Post: August 13th 2009, 05:27 PM
  3. Replies: 3
    Last Post: February 20th 2009, 01:02 PM
  4. Replies: 3
    Last Post: February 20th 2009, 04:46 AM
  5. Area of region bounded by these curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 19th 2009, 06:19 PM

Search Tags


/mathhelpforum @mathhelpforum