1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral calculus -- regions

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Find all values of a such that this set is a type II region (i.e. the bounds of x can be represented as functions of y, while the bounds of y are constant valued)
    -1<y<0, Y<x<-y (union) 0<y<1, -y+a<x<y+a

    2. Relevant equations


    3. The attempt at a solution
    I arrived at a being any value in the bound (-inf, 0}, since if a is a value greater than 0, then the y bounds (0<y<1) start to exist outside of the region bounded by -y+a and y+a. In other words, a must be negative or 0.
     
  2. jcsd
  3. Feb 4, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Check a = - 1/2.
     
  4. Feb 5, 2015 #3
    I've tried this, and it appears to work. Maybe I'm missing something. What's your reasoning for this?
     
  5. Feb 7, 2015 #4
  6. Feb 7, 2015 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I don't understand that statement.
    Call the regions A = {-1<=y<=0 & y<=x<=-y}, B = {0<=y<=1 & -y+a<=x<=y+a}. (Allowing equality won't affect integration.)
    What do A and B look like, in words?
    Where do they meet?
    What happens if you flip the signs of a and x consistently throughout?
    However, I'm not sure how to interpret the question. What sort of function of x is allowed? I originally was thinking only of 'nice' functions, so drew attention to a = -1/2. If anything is allowed, check a = +1/2.
     
  7. Feb 7, 2015 #6
    So Region A looks like a triangle with its top vertex at the origin (the entire triangle is below the horizontal x-axis and its y coordinates range from -1 to 0); this region is bounded by x=y to the left and x=-y to the right. I am asking what values of a are allowed such that the union of regions A and B is a type II region (i.e. the x bounds can be represented as functions of y, while the y bounds are constant values).
    If a is >0, the region does not fall in the 0<=y<=1 bound because the linear functions shift up away from the origin, but it works for any negative a values, I think?
     
  8. Feb 7, 2015 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Check what solutions there are for y=0 in region B.
    Anyway, even if there appeared a gap in the range for y, there would still be a way to treat it as a contiguous range for integration purposes. If the lower bound is expressed as y=f(x) and the upper bound as y=g(x) then through the gap region specify f=g. If that's not allowed then this categorisation strikes me as somewhat arbitrary and pointless.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integral calculus -- regions
  1. Calculus (integration) (Replies: 3)

  2. Calculus Integral (Replies: 4)

  3. Integral Calculus (Replies: 15)

  4. Integral Calculus (Replies: 3)

Loading...