1. Not finding help here? Sign up for a free 30min 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!

Area of shaded object

  1. Nov 28, 2013 #1
    DLOq3MC.jpg

    For question 16, I am having trouble assessing the question. I set the sides equal to x and put the image in the middle of a coordinate axes hoping it would be easier to assess. I also set r2 = [(1/2)x]2 + y2 and tried to set up an integration but cannot come up with a relation for the inner square's area. Any help with this question would be great!
     
  2. jcsd
  3. Nov 28, 2013 #2
    The question says that the shaded region is the set of all points inside of the square that are closer to the origin than the edges.

    Can you write down an inequality for this (Hint: Pythagorean theorem)? What if you were only to pick the outermost edges of the region, what would the equation for that be?
     
  4. Nov 29, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Further to Sentin3l's reply...
    Simplify the shape by cutting it into a number of identical regions, each with a much simpler boundary.
    Pick one of those pieces. If (x, y) is a point in that piece, how far is it from the origin, and how far is it from the square?
     
  5. Dec 2, 2013 #4
    Ok, well my solution was:

    Split the square into 4 quadrants and integrate 1 of them and then multiply that area by 4 to find the answer. So, I did:

    x2 + y2 < 0.25 since it must not be closer to the outer edge of the square

    So I rearranged and got:

    y2 < 0.25 - x3/3

    Now I integrated:

    0<0.5∫y2dy < A

    (1/3)y3l0<0.5 < A

    (0.53)/3 > A since using 0.5 means it is the same distance b/w the square's edge and the actual shape

    0.125/3 > A and so I multiply this by 4 and find 0.5/3 > A....

    this isn't an exact answer and my textbook doesn't have the answer for this question, so any help would be great!

    Also, is it mathematically improper to put inequalities as the upper or lower limits if you're just trying to assess the maximum and minimum possible area?
     
  6. Dec 2, 2013 #5
    Also, I found this solution. http://mymathforum.com/viewtopic.php?f=15&t=33895

    The problem I have is that he assumes that (x2 + y2) = (y-1)2 since the origin and square are equidistant from the edge of the shaded shape. Why can you assume they're equidistant? Doesn't the questions say that it is always closer to the origin?
     
  7. Dec 2, 2013 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Do you mean the usual four quadrants, as bounded by the x and y axes? That won't help much because of the awkward bends in the boundary at NE, SE, SW and NW positions. Cut the shape so that those bends are at the boundaries of the cuts.
    That is the arc of a circle centred at the origin, radius 1/2. It is not composed of arcs of circles, and it extends beyond such a circle.
    You need an algebraic expression for the boundary of the shape (or rather, the boundary as it is within one of the regions you cut it into). Write an equation representing points which are equidistant from the centre and the square.
    Where did the cube come from?
    Why are you integrating y2? Area is ∫y.dx
     
  8. Dec 2, 2013 #7
    Ok, I initially thought the radius was constant for some absurd reason. I realize the solution I gave is incorrect. Why can we assume the area in each of the 8 quadrants is circular? Also, why do we consider the lines equidistant when the question says they square is farther from the shape compared to the origin at all points?
     
  9. Dec 2, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It isn't circular, and I didn't say it was. A one eighth sector is an octant.
    If a region is defined by f(x,y) < c then its boundary is given by f(x,y)=c.
     
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: Area of shaded object
Loading...