Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric Argument

  1. Mar 14, 2008 #1
    I was given an ap problem in class, specifically:

    Problem 2:http://www.collegeboard.com/prod_downloads/ap/students/calculus/b_calculus_bc_frq_03.pdf

    I was able to do it just fine, but then I had the idea to try and solve it geometrically/algebraically. However, I haven't been able to come up with a solution this way and I was wondering if anyone could see a way that it could indeed be solved using some sort of a geometric argument.
  2. jcsd
  3. Mar 14, 2008 #2
    Wait I know that they intersect at (1,1) so then I can find the angle, which is pi/4. Then I can find the area of the sector of the circle centered at the origin and subtract 1/2 (to get the portion of the shaded region bounded by the larger circle. The rest of the area is just 1/4 the are of the smaller circle, so...

    [tex]A_{sector} = \frac{1}{2}(\frac{\pi}{4})\sqrt{2}^2[/tex]

    [tex]A_{sector} = \frac{\pi}{4}[/tex]

    [tex]A_{total} = \frac{1}{4}\pi(1)^2 + [\frac{\pi}{4} - \frac{1}{2}][/tex]

    [tex]A_{total} = \frac{1}{2}(\pi -1)[/tex]

    Look right?
    Last edited: Mar 14, 2008
  4. Mar 17, 2008 #3

    Gib Z

    User Avatar
    Homework Helper

    I don't understand what you meant after "and subtract....".

    When you connect (1,1) and the origin, its basically the sum of a sector and a minor segment.
  5. Mar 17, 2008 #4
    I subtracted 1/2 because I was just trying to find the area of the shaded region past x=1.

    To do so I found the area of the sector then subtracted the area of the triangle that was part of the sector with vertices's at (0,0) (1,1) and (1,0).

    The area of this triangle is 1/2 so that is why I subtracted it.
  6. Mar 18, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    yes it is very easy to subdivide the region into triangles whose areas are easily found and sectors of circles which are equally easy.

    in fact no one in his right mind would do this area problem by calculus.
  7. Mar 18, 2008 #6
    Haha yeah, think you could speak to CollegeBoard on my behalf? :rofl:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Geometric Argument
  1. Geometric arguments (Replies: 5)

  2. Geometric progressions (Replies: 1)

  3. Geometric softwares (Replies: 3)

  4. Geometric absolute (Replies: 1)

  5. Complex arguments (Replies: 2)