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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    06 Feb '09 04:30
    The question:

    Find the volume of a solid, who's base is bounded by the graph of

    y=x^3 , y=0, x=1

    and whose cross section is that of a semiellipse(a hieght, twice that of it base) taken perpendicular to the y axis.

    solving the problem isn't a problem, Im intentionally making it more difficult that it has to be in hopes of gaining a little insight.

    here is how i want to approach it!

    find the quadratic equation that has the ordered pair solutions (0,0),(b,0), (1/2*b, 2b)

    the equation in what would be the z direction would be

    z = -x^2 + 8x

    now back to the x-direction, the base of this elipse in terms of "y" must be

    base = (1-y^(1/3))

    now here is where im not sure i can do what im doing

    the hieght in terms of y = 2-2y^(1/3)

    so setting

    -x^2 + 8x = 2-2y^(1/3)

    I arrive at

    x = sqrt( 2y^(1/3) + 14 ) + 4

    before I go and try to evaluate this integral which i believe would be

    INT(0_1) (sqrt( 2y^(1/3) + 14 ) + 4 )*(1-y^(1/3)) dy

    I dont even relistically think i can?

    does this seem logical, or is it horse crap?
  2. 06 Feb '09 06:14 / 1 edit
    Originally posted by joe shmo
    The question:

    Find the volume of a solid, who's base is bounded by the graph of

    y=x^3 , y=0, x=1

    and whose cross section is that of a semiellipse(a hieght, twice that of it base) taken perpendicular to the y axis.

    solving the problem isn't a problem, Im intentionally making it more difficult that it has to be in hopes of gaining a little insigh dy

    I dont even relistically think i can?

    does this seem logical, or is it horse crap?
    a cursory examination of your computation seems like you're being accurate... however, what is your goal in solving the problem this way?

    are you trying to take a multivariable problem down to a single parameter? because as you can probably tell it makes the problem a lot harder than it needs to be. this is like finding the volume of a sphere in rectangular coordinates vs. using spherical coordinates, or like stubbornly sticking to cartesian coordinates in a problem better suited to polar.

    that being said, i don't think anything you've done is "wrong" except it seems you've arrived at an antiderivative that can't be computed algebraically (as far as i can tell). this doesn't mean it doesn't exist - in fact the integral is definitely possible. you just might have to find the limit of a nasty riemann sum, rather than using algebraic methods of finding an antiderivative and subsequently computing F(1) - F(0).

    all in all, i'd rather just use a double integral with respect to x and y. much easier

    EDIT: in a look back, i'm not sure why you searched for the quadratic at the beginning of the problem. isn't the problem saying that if you take a cross-section perpendicular to the y-axis (and thus see the relationship between z and x) that the height is twice the length of the base? i.e. wouldn't z = 2x rather than z = -x^2 + 8x? or am i too tired to think straight right now? lol
  3. Subscriber joe shmo On Vacation
    Strange Egg
    06 Feb '09 20:58
    Originally posted by Aetherael
    a cursory examination of your computation seems like you're being accurate... however, what is your goal in solving the problem this way?

    are you trying to take a multivariable problem down to a single parameter? because as you can probably tell it makes the problem a lot harder than it needs to be. this is like finding the volume of a sphere in rectan ...[text shortened]... 2x rather than z = -x^2 + 8x? or am i too tired to think straight right now? lol
    I think I may be getting in over my head, The reason I was searching for the quadratic, was because I was trying to model the area of the semiellipse...I thought that if I integrated that function I could derive the area for the semiellipse..Then from there use the formula to integrate the cubic function.....waste of time I know.......so basically I was trying to find the formula for area of a semiellipse, and integrate it into the integral...if that make any sense, I not sure it does?
  4. 07 Feb '09 14:25
    Originally posted by joe shmo
    I think I may be getting in over my head, The reason I was searching for the quadratic, was because I was trying to model the area of the semiellipse...I thought that if I integrated that function I could derive the area for the semiellipse..Then from there use the formula to integrate the cubic function.....waste of time I know.......so basically I was try ...[text shortened]... a semiellipse, and integrate it into the integral...if that make any sense, I not sure it does?
    hmm... i think you may have been a bit off base here? perhaps if this was your intent, you could solve the problem normally, compute an anti-derivative of f(x,y) and then differentiate with respect to y? this would give you an area function for the semi-ellipse in terms of x, i think (provided you plug in x = y^(1/3) wherever there is a y in your resulting expression). then you could consider this a "formula" for the area of the semi-ellipse, and use it for whatever purposes you want... but this is like doing the problem in question twice, and isn't really worth it? i'm not really sure i'm just throwing out ideas here, but it seems like you could use the conventional method of solving to derive the area formula, but not derive it on the way to the solution. and since you need the solution to get to the formula for the area, does this method cease to be useful in terms of your overall plan? depends on what you are getting out of the exercise i think.