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Homework Help: Integration sqrt(10z - z^2)

  1. Aug 3, 2005 #1
    trying to integrate [tex] \sqrt{10z-z^2} [/tex]

    i have been advised to used trig subsitution wherein i subsitute z = cos theta
    however i end up at a dead end

    [tex] \int \sqrt{10 \cos \theta - \cos^2 \theta} sin \theta d \theta [/tex]

    i could certainly replace the sin

    [tex] \int \sqrt{10 \cos \theta - \cos^2 \theta} \sqrt{1- \cos^2 \theta} d \theta [/tex]

    hwat now do i expand?
    or do i use parts?
     
  2. jcsd
  3. Aug 3, 2005 #2
    alternatively i can use [tex] z = 10 sin^2 \theta [/tex]

    i am currenttly working o ntaht because i made a mistake... z should be cos theta

    is that hter ight way to go??
    [tex] \int \sqrt{100 sin^2 \theta - 100 \sin^4 \theta} (20 sin \theta cos \theta) dz [/tex]
     
    Last edited: Aug 3, 2005
  4. Aug 3, 2005 #3

    GCT

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    try [tex]z=10-w[/tex] substitution
     
  5. Aug 3, 2005 #4

    arildno

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    The simplest choice is to rewrite your root expression as:
    [tex]\sqrt{10z-z^{2}}=5\sqrt{1-(\frac{z-5}{5})^{2}}[/tex]
     
  6. Aug 3, 2005 #5
    [tex]10z-z^2=5^2-(z-5)^2[/tex]
    Try [tex]z-5=5\cos{t}[/tex]
     
  7. Aug 3, 2005 #6

    GCT

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    what's more simple than this?

    [tex] I = \int \sqrt{10z-z^2} dz [/tex]

    [tex]z=10-t,~dz=-dt[/tex]

    [tex]I= - \int (10-t) \sqrt{10-(10-t)} dt [/tex]

    [tex] I= -10 \int t^{ \frac{1}{2}} dt + \int t^{ \frac{3}{2}} dt [/tex]

    arildno, can you explain your solution?
     
  8. Aug 3, 2005 #7
    Last line isn't correct. it should be [tex]I= - \int \sqrt{t (10-t)} dt [/tex]
     
  9. Aug 3, 2005 #8

    arildno

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    It's the same as Yegor's, since we now make the substitution
    [tex]\frac{z-5}{5}=\cos(t)[/tex]
     
  10. Aug 3, 2005 #9

    GCT

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    I see, well with yegor's/arildno proposal, I end up with [tex]25 \int cos^{2} \theta d \theta[/tex]
     
  11. Aug 3, 2005 #10
    .. then just use [itex]\cos 2\theta = 2\cos^{2} \theta - 1[/itex], presumably.
     
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