21 Comments

  • TheDaekle

    I'm sorry, I may be confused but, If "H" is a length (lets say measured in metres for ease), then "H^2" is an area in metres squared, meaning the units of the final figure suggest we've calculated an area and not a volume. Have I misunderstood something? genuine question, not a criticism, I'm just curious how this works…

  • Doggo Willinks

    Isn't this just the shell method, in terms of y? So you could've just went right to the integral right? This is just the entire idea behind the cylindrical shell method right? Except that would have a 2pi, so maybe not.

  • Sudhakar Medepalli

    It does look confusing! However, if you do this with a more general equation such as: y = a* x^2, where the coefficient 'a' has dimensions of (1/length), the volume comes out to be: V = (Pi/2) * (H^2 / a). Now the second term has dimensions of (length)^3 which is a volume dimension. In this example, a = 1, so the dimension of the constant '1' is hidden.

  • Word Sailor ADD INFUNITEMS

    Volumes of information here, nicely presented by this directrix who displays excellent focus. When you reflect on these concepts, you become a parabolic reflector, which allows you to see really far out places. It's refreshing to have calculus freely available on the web. May the math be with us!

  • ebadullah ahmadzai

    It does look confusing! However, if you do this with a more general equation such as: y = a* x^2, where the coefficient 'a' has dimensions of (1/length), the volume comes out to be: V = (Pi/2) * (H^2 / a). Now the second term has dimensions of (length)^3 which is a volume dimension. In this example, a = 1, so the dimension of the constant '1' is hidden

  • Tareq

    Amazingly, the volume of a paraboloid was computed by the Muslim scientist, Thābit ibn Qurra, in the 9th century, i.e., 8 centuries before Newton invented (or reinvented) calculus

  • shashi shekhar azad

    This is not satisfactory answer mam ……becoz last answer is not a volume according to dimensions or unit……

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