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Can you help me , please?

  1. Apr 30, 2008 #1
    Can you help me , please??

    Can you solve this problem by using Calculus l "" Volumes by Slicing and Rotation About an Axis"" ??
    Develop a formula for the volume of an ellipsoid of the form

    ( x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1, a, b, c > 0.

    I want the steps , because I know the final answer .
  2. jcsd
  3. Apr 30, 2008 #2


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    Welcome to PF!

    Hi vip89! Welcome to PF! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help you! :smile:
  4. Apr 30, 2008 #3
    I know that I must sketch,find the equation of the cross section , find the area from A=piab and x^2/a^2+y^2/b^2=1
    Finally,I must find the definit integral of the area
    the final answer will be 4pi/3 abc
  5. Apr 30, 2008 #4
    I dont know how to do this , and how to apply my steps?!
  6. Apr 30, 2008 #5


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    Hi vip89! :smile:
    ok … we usually take horizontal cross-sections (don't have to) …

    in other words, the intersection with a plane z = constant.

    So what is the equation (in x and y) if you put z = w (a constant) … ? :smile:

    … and then what is the area? :smile:
  7. May 2, 2008 #6
    thnkx very much
  8. May 6, 2008 #7
    I trid to solve it by this way,but strang things will appear,the equation be more complicated!!
  9. May 6, 2008 #8
    Show us your work. You can either take a picture, scan it, and upload it, or you can try using LaTeX in between [noparse][tex] and [/tex][/noparse] tags.
  10. May 7, 2008 #9


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    Hi vip89! :smile:

    Show us how far you got …

    What equation (in x and y) did you get for the horizontal cross-sections when you put z = w (a constant)? :smile:
  11. May 7, 2008 #10

    lol ... funny kid we're not here to do your hw
  12. May 7, 2008 #11


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    Welcome to this thread, rocomath!

    oi … rocomath! … that was vip89's very first post (seven days ago)!

    … but he's getting the hang of it now! :smile:

    If you can't help, just say "welcome!" :smile:
  13. May 7, 2008 #12
    sorry :( i've grown less tolerant towards ppl just wanting us to do their hw.
  14. May 9, 2008 #13
    These are my trials
    I have the idea but cant get the right answer
    pls reply

    Attached Files:

  15. May 9, 2008 #14

    Attached Files:

  16. May 10, 2008 #15


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    In your first sheet, you start with [tex]\int A(x)dx[/tex] and then immediately switch to [tex]\int x y[/itex] without a "dx" at all. How did that happen? I also cannot see any good reason for writing x as a function of y and z and writing y as a function of x and z.
    You are correct that [tex]y= b\sqrt{1- x^/a^2- z^2/c^2}[/tex] and that [tex]z= c\sqrt{1- x^2/a^2- y^2/b^2[/tex]. The first tells you that when z= 0, [tex]y= b\sqrt{1- x^2/a^2}[/tex] and the second tells you that when y= 0, [tex]z= c\sqrt{1- x^2/a^2}[/tex]. In other words, at each x, the cross section is an ellipse with semi-axes [tex]b\sqrt{1-x^2/a^2}[/tex] and [tex]c\sqrt{1- x^2/a^2}[/tex]. Do you know that the area of an ellipse with semi-axes a and b is [tex]\pi ab[/tex]?
  17. May 11, 2008 #16


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    hmm … let's start at the top of your page 3 …
    which I take it is following my suggestion …
    And then you try to use the formula A = πab for the area of an ellipse.

    BUT … that formula only applies if the equation for the ellipse is in the standard form, with nothing but x² and y² on the left and "= 1" on the right.

    You must put w² = (1 - x²/a² - y²/b²)c² into that form first.

    Then you will get a "new a and b" that are not the same as the original a and b.

    ok … rearrange w² = (1 - x²/a² - y²/b²)c² into the standard form … and then find the area. :smile:
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