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Math Help - Cylinder and Sphere

  1. #1
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    Cylinder and Sphere

    A cylinder of length 2x is inscribed in a sphere of radius a. Between one end of this cylinder and the sphere, another cylinder of length 2y is inscribed with one end on an end of the first cylinder so that the axes of the cylinder are collinear.

    Show that the sum of the volume of the two cylinder is

    2pi(x+y)(a^2 - x^2 -4y^2)

    how can i solve this?
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  2. #2
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    Quote Originally Posted by MilK View Post
    A cylinder of length 2x is inscribed in a sphere of radius a. Between one end of this cylinder and the sphere, another cylinder of length 2y is inscribed with one end on an end of the first cylinder so that the axes of the cylinder are collinear.

    Show that the sum of the volume of the two cylinder is

    2pi(x+y)(a^2 - x^2 -4y^2)

    how can i solve this?
    1. Make a sketch. (I've made a little mistake: The red line should be labeled as 2y)

    2. Calculate the radius of the greater cylinder:

    a^2-x^2=r_g^2

    3. Calculate the volume of the greater cylinder:

    V_g=\pi \cdot r_g^2 \cdot 2x = \pi (a^2-x^2) \cdot 2x

    4. Calculate the radius of the smaller cylinder:

    a^2-(x+2y)^2 = r_s^2

    5. Calculate the volume of the smaller cylinder:

    V_s=\pi \cdot r_s^2 \cdot 2y = \pi (a^2-(x+2y)^2) \cdot 2y

    6. Add the volumina, expand the brackets and factor again. You'll get the result given in the problem.
    Attached Thumbnails Attached Thumbnails Cylinder and Sphere-zweizyl_inkugl.gif  
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  3. #3
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    'volumina'. Cool word. I looked it up and it wasn't in the dictionary.
    How erudite.
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  4. #4
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    Got it! Thanks

    After i've shown the eqn, i need to find its max value,

    i procced to find partial x and y.

    Now the eqn have a unknown a, should i treat this as constant? or find partial a?
    Last edited by MilK; February 16th 2008 at 05:45 PM.
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    Quote Originally Posted by MilK View Post
    Got it! Thanks

    After i've shown the eqn, i need to find its max value,

    i procced to find partial x and y.

    Now the eqn have a unknown a, should i treat this as constant? or find partial a?
    Keep \, a\, as a constant. Then you'll get x as a function of \, a\, and y as a function of \, a\, .
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  6. #6
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    Quote Originally Posted by galactus View Post
    'volumina'. Cool word. I looked it up and it wasn't in the dictionary.
    How erudite.
    I'm guessing it's German for the plural of volumes ....

    Edit: Confirmed.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Keep \, a\, as a constant. Then you'll get x as a function of \, a\, and y as a function of \, a\, .
    do you mean treat a as a constant when finding partial x and y OR do u mean keep treat a as a constant for the whole equation? it is not stated in the question than a is a constant. if a is not a constant in the eqn, do i need to find partial a?
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  8. #8
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    Quote Originally Posted by MilK View Post
    do you mean treat a as a constant when finding partial x and y OR do u mean keep treat a as a constant for the whole equation? it is not stated in the question than a is a constant. if a is not a constant in the eqn, do i need to find partial a?
    a is the radius of the sphere. There is absolutely nothing in the question (that I can see) that suggests that this radius can change. It is clearly an implied constant for the problem and obviously should be treated a such.

    If it worries you that much, give it a value like a = 1. Then forget about it and get on with solving the problem.

    Then do the problem again but this time keep the \, a\, as \, a\, . Then check that your answer agrees with what you got for a = 1 ......
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  9. #9
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    i got these 2 eqn for partial x and y

    for partial x = 2pi( -2x(x+y) +a^2 -x^2 -4y^2)
    for partial y = 2pi( -8y(x+y) +a^2 -x^2 -4y^2)

    subtract these two eqn,

    i get 2pi(-6x)(x+y) = 0

    therefore x+y = 0 and -12pi(x) =0
    solving i get x=0, y=0

    the question stated : find the max value of this eqn :

    2pi(x+y)(a^2 - x^2 -4y^2)

    something is wrong somewhere, after finding fxx=0, fyy=0 and fxy=0, (0,0) is not the max point.

    what when wrong?
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  10. #10
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    Quote Originally Posted by MilK View Post
    i got these 2 eqn for partial x and y

    for partial x = 2pi( -2x(x+y) +a^2 -x^2 -4y^2)
    for partial y = 2pi( -8y(x+y) +a^2 -x^2 -4y^2)

    subtract these two eqn,

    i get 2pi(-6x)(x+y) = 0 Mr F says: This is wrong.

    The correct result is 2\pi [-8y(x + y) - -2x(x + y)] = 0 \Rightarrow (x + y)(2x - 8y) = 0 \Rightarrow (x + y) (x - 4y) = 0 ..........

    Therefore x = -y OR x = 4y. x = 4y is the one you want since x = -y is impossible since x> 0 and y > 0. So sub x = 4y into either of your original equations and solve for y. once you have y, get x. Then test the nature of the solutions.


    therefore x+y = 0 and -12pi(x) =0
    solving i get x=0, y=0

    the question stated : find the max value of this eqn :

    2pi(x+y)(a^2 - x^2 -4y^2)

    something is wrong somewhere, after finding fxx=0, fyy=0 and fxy=0, (0,0) is not the max point.

    what when wrong?
    I haven't checked the details but I think it should be blue sky .....
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  11. #11
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    after sub x=4y into 2pi( -8y(x+y) +a^2 -x^2 -4y^2)=0

    i got y = a/60 and x = 4a/60.

    how do i get rid of the a?
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  12. #12
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    Quote Originally Posted by MilK View Post
    after sub x=4y into 2pi( -8y(x+y) +a^2 -x^2 -4y^2)=0

    i got y = a/60 and x = 4a/60.

    how do i get rid of the a?
    equation 1

    y=a/60

    a = 60y

    equation 2

    x = 4a / 60

    4a = 60x

    a = 15x

    so 60y=15x

    but that's not what you want to do ....i think?

    you could solve the equation for a in terms of y?

    2pi(-8y(x+y) + a^2 - x^2 - 4y^2)=0

    2pi(-8y((4y)+y) + a^2 - (4y)^2 - 4y^2)=0

    2pi(-8y(5y) + a^2 - 16^2 - 4y^2)=0

    -60y^2 + a^2 = 0

    a = (+)root(60y^2) and (-)root(60y^2)

    a = + 2y(root15) and - 2y(root15)

    but god knows what the question was ?

    AHHHH LOL page 2/2 >.<
    Last edited by Charbel; February 23rd 2008 at 12:25 AM. Reason: :P opps
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  13. #13
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    Quote Originally Posted by MilK View Post
    after sub x=4y into 2pi( -8y(x+y) +a^2 -x^2 -4y^2)=0

    i got y = a/60 and x = 4a/60. Mr F says: And the value of x simplifies to a/15.

    how do i get rid of the a?
    You don't! It is a constant!

    Quote Originally Posted by mr fantastic View Post
    a is the radius of the sphere. There is absolutely nothing in the question (that I can see) that suggests that this radius can change. It is clearly an implied constant for the problem and obviously should be treated a such.

    [snip]
    Why are you so determined to get rid of a in your answer??
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  14. #14
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    so my final ans is it terms of a?

    sub y = a/60 and x =a/15,

    fxx = -26(pi)a/30
    fyy = -28(pi)a/15
    fxy = 64(pi^2)(a^2) /225

    there its max point since fxy^2-fxxfyy is -ve and fxx is -ve

    and to find its max value, i sub y=a/60 and x=a/15 into fxy which gives max value as -8(pi)a / 15

    correct?
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  15. #15
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    Quote Originally Posted by mr fantastic View Post
    You don't! It is a constant!


    Why are you so determined to get rid of a in your answer??
    i thought we should always get a numerical ans?
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