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Prooving Area of Circle Equation

  1. Oct 12, 2006 #1
    Hi, i am hoping that someone could help me or give me a hint for the follwing question...i will show what I have done so far

    The area of a quarter circle given is

    Integral sqrt(25-x^2) dx

    Use the substitution x = 5 sinu to evaluate this def integral exactly an show that your answer is consistent with area pi*r^2 with radius r

    firstly i did dx/du = 5cos u

    then i said:
    Integral 5sqrt(1-sin^2u).5cosu du
    Integral 5cosu.5cosu
    Integral 25 cos^2u du
    Integral 25(1+cos2u)/2
    Integral 25/2 +25cos2u/2

    and well the problem i am having is that i dont know how to get the u =5sin u to SUB into the intergal so i can solve it (ie i can SUB u into cos2u...it doesnlt work)

    so hopefully someone could point me in a better diretion to solve this??

    thanks heaps!!
     
  2. jcsd
  3. Oct 12, 2006 #2
    it shouldn't be [tex] x = 5\sin u [/tex], it should be [tex] x = 5\sin \theta [/tex].
     
  4. Oct 12, 2006 #3
    And the difference between those two is?
     
  5. Oct 13, 2006 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't understand what your question is. You seemed to be able to substitute and integrate!
    If your question is "How do you know to make that substitution?", it's because of: sin2u+ cos2u= 1 so that
    1- sin2 u= cos2u, a "perfect square".

    If your question is "how do I do that last integral", [itex]\frac{25}{2}\int 1+ cos(2u) du[/itex], you certainly should know that the integral of 1 du is u and the integral of cos(2u) is (1/2)sin(2u). That integral is
    [itex]\frax{25}{2}u+ \frac{1}{2}sin(2u)[/itex]. Since you say you are dealing with a quarter circle and you appear to have set it up with center at (0,0) and radius 5, your original integral is from x= 0 to x= 5. 5 sin u= x= 0 or sin u= 0 when u= 0 and 5 sin u= x= 5, or sin u= 1 when u= [itex]\pi/2[/itex] your limits of integration in the final integral are 0 and [itex]\pi/2[/itex].
     
  6. Oct 14, 2006 #5
    sorry

    the question is to proove using intergation techniques and this quarter circle as an example, that the area of a quarter circle is 1/4*pi*r^2. Ie intergrate the given integral and compare that to the area of a circle formula and see that they match. However, the question stipulates that you must use the substitution which I outlined....which leads to my prob when i get to the part where you substitute this : x = 5 sin u back into the equation which is meant to be easier to integrate.... so i guess you could say

    i dont know how to get u =5sin u (the upper limit) to substitute into the integral:

    25/2 + 25 (cos 2u)/2 (which is as simplified as i can get the intergal)

    thanks for all you responses btw!!! appreciated
     
  7. Oct 14, 2006 #6
    No Its A Half Circle!!!

    SORRY!! just rechecked the Q, its a semi circle, not a quarter!!!
     
  8. Oct 14, 2006 #7
    I understand the intergal you did completely, no probs there!!

    okay heres wat i dont get, how did you change the limits which i had i had, ie 0 and 5 sin u into numbers which I can actaually substitute into the intergal and then solve, it is irrlevant whether a quarter circle or half circle, but i chose quarter circle since the limits 0 and 5 seemed easier.

    hang, i think i found my error, the limits therfore should be what you said, i was doing sumthin completely stupid with my limits, i dont know where 5sinu came from as a limit!!!! sorry...THANKU HEAPS FOR YOU HELP!!!!
     
  9. Oct 14, 2006 #8
    5 = 5sinu, so sin u = 1.



    x^2 + y^2 = r^2, y^2 = sqrt(r^2 - x^2)
     
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