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Solids of revolution

  1. Dec 2, 2004 #1
    Ok. I just need someone to explain the methods for me, if it's not too much trouble.

    I know you have the shell method, disk method, and washer method. I am taking AP Caclulus BC in high school. I want to know if all of these methods are useful of if one of them is more important.

    Thanks,
    Jameson
     
  2. jcsd
  3. Dec 2, 2004 #2

    WaR

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    Well, they are all useful because they are used in different situations. The volume of a solid is different than the volume of the same solid with a hole in it :)
     
  4. Dec 3, 2004 #3

    HallsofIvy

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    I am inclined to consider the "disk" method as more fundamental than the others (not necessarily "more important"!). You can do any "washer" problem by doing two disk calculations and subtracting. A volume by "shells" you can (theoretically!) do by swapping variables.
     
  5. Dec 4, 2004 #4

    mathwonk

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    they are all based on a sort of "cavalieri's" principle. i.e. if you let a certain surface move along a line perpendicular to the surface, and if the moving surface sweeps out the whole solid, then the volume of the solid is the integral of the surface's area, along the moving line segment.

    The various methods you mention are all of this type, but they just choose different dierctions along which to move the surface. thus the answer to your question is: the best method depends on the shape of the givenm solid, and what you know about it.

    For example, given a sphere, we can move a plane perpendicular to the north south axis, and sweep out the volume this way. the advantage to this choice is that each plane cuts the sphere in a circle, which is an area formula that we already know. so the volume of the sphere is obtained by integrating the area of a moving circle, along the north south axis.

    On the other hand a sphere may be swept out by cylinders expanding radially from the north south axis. then each cylinder cuts the (solid) sphere in a cylinder, and we can get the volume by integrating the area of these cylinders along a radius of the sphere.

    in this case neither method is more natural than the other, although the planes cut discs, which seem easier to picture in the mind's eye. the only difference in this case is in which integral, i.e. which area formula is easier to integrate afterwards.

    now the reaslly natiral way to sweep out spherical volume is by spheres of smaller radii, moving outward along a radius of the sphere. then the volume of the sphere is seen to be the integral of the formula for the area of the sphere.

    as a method of getting the volume this does not work, because we do not know the area formula for a sphere in advance. but behold, after using one of the other methods to get the volume formula, this method tells us how to get the area formula. just differentiate the now known volume formula!

    consider now a doughnut obtained by revolving a circle lying in the x,z plane around the z axis. you can sweep out the volume in several ways here. by a moving family of parallel planes, all perpendicular to the z axis, and moving along the z axis. these each cut the doughnut in a "washer". i.e. the difference of two discs, as Halls of Ivy noted. so since you know th4e formula for the area of a washer, you can get the volume this way, provided you can do the square root integral it yields.

    the shell method sweeps out the volume by a family of cylinders with center the z axis, expanding radially outward from the z axis. eacxh of these cuts the doughnut in a cylinder and again the only advantage to one of these is which one gives an easier integral.

    there is another way however, anmalogous to thew third approach above to the sphere namely sweep out the volume of the doughnut by an expanding family of doiughnuts, as if you were building the doughnut by dippeing a circular wire in wax, over and over and letting it thicken around the wire.

    this method finds the volume of a doughnut as the integral of the area of the doughnut. Again it does not help us at first since we do not know the area of siughnut, but used backwards, aftern we know the volume, it tells us we can ghet the area by differentiating the volume formula with respect to this radius.

    finally one should know that pappus knew the volume and area of a soiughnut long bwfore calculus was invented, so there must be a more naturla, easier way and indeed there is.

    namely the most natural simple way to sweep out the volume of a doughnut is by the method sued to describe the doughnut, i.e. by revolving the circle around a circular path around the z axis. so the volume of the doughnut should be obtained by integrating the area of a circle around a circular interval. but which radius to choose? intuition and pappus theorem says that it should be the path travelled by the center of the circle, so the volume of a circle, whiuch is a difficult integral by any of the other methi=ods, becomes trivial here, equal merely to the product of the area of the circle of constant radius being revolved, times the distance traveled by the center of that circle.

    the area is likewise the product of the length of that circle times the distance traveled by its center. this mof cousre is the same as the derivative of the volume wrt the radius of the circle being revolved, (since the length of the circle being revolved is the derivative of its area).


    so there is really only one method of compouting volumes, integrating a moving area formula. the choice of which area to integrate is dictated by the shape of the region and the difficulty of determining and then integrating the subsequent area formula.


    for instance if you revolve the graph of a cubic equation y = x^3 + x, around the x axis, then theoretically you could use either discs or shells, but to do both, you need to be able to solve both for y in terms of x, and for x in terms of y. now unless you know Cardano's "cubic formula", (at least published by him), you are going to have trouble solving for x in terms of y, so you are advised to use discs, and integrate area of circles with respect to x.


    Exercise: find the volume of the region inside two perpendicular cylinders both of radius one. one cylinder has central axis equal to the x axis, and the other has central axis equal to the y axis.

    hint: try moving planes perpendicular to the z axis.

    Remark: in this rpoblem your figure is not generated by revolution, so none of the cookbook methods above works. but the idea they are based on still works.


    here is an easier one: find the volume of a pyramid whose base is a square of side a, and whose height is given as h.


    another one: find the volume generated by revolving an equilateral triangle of base length one, lying inm the x,y plane, and centered at the point (3,0), around the y axis. hint: reread all the solution methods above used on the doughnut.
     
    Last edited: Dec 4, 2004
  6. Dec 5, 2004 #5

    mathwonk

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    hey mr jameson, did this explanation make any useful impression on you?
     
  7. Dec 5, 2004 #6
    Yes, it helped. I am still confused about when to use what equation though. But I think I am getting better at it. I don't believe I can do a problem with a "z axis" though.

    Jameson
     
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