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Help with Volumes by slicing

  1. Feb 18, 2004 #1
    I'm having trouble coming up with the correct solution to a number of these questions.. This is one I've been stumped on..

    Find the Volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis.

    y = sqrt(25-x^2), y = 3

    Since I believe it is a washer method, what I did was:

    V = int pi[f(x)]^2 - [g(x)]^2 dx

    = pi int{-5to5} [25-x^2] - [9] dx

    = pi int{-5to5} [25x-x^3/3] - [9x]

    = pi [(-125) + (125/3) + (45)] - [(125) - (125/3) - (45)]

    = pi (365/3) - (115/3)

    = 250pi/3

    The book says 256pi/3, which part am I doing wrong?

    Any help is greatly appreciated.
  2. jcsd
  3. Feb 18, 2004 #2


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    Y= sqrt(25- x^2) is, of course, the sphere centered at (0,0) with radius 5.

    You are really asking for the volume of that sphere with a cylinder of radius 3 removed (I prefer to think of it as an apple with the core removed. Stuff it with cinnamon and brown sugar and bake!).

    Yes, a cross section is a "washer" and the area of a washer is the difference of the areas of the two circles: exactly, as you give, pi[f(x)]^2 - [g(x)]^2. The volume is the integral of that, dx.

    The only thing you are doing wrong is that the integral is NOT from "- 5 to 5". Drawing a line from the center of the circle to the line where y= 3, you will see a "3-4-5" right triangle. 5 is the length of the hypotenuse. You can't go out to x= 5- that is the edge of the sphere but is beyond the cylinder. You need to take the integral from -4 to 4.
  4. Feb 20, 2004 #3
    Oh okay, I think I'm understanding it better now..
  5. Feb 21, 2004 #4
    I thought I was getting the hang of these but here's another one I'm having trouble with.. I believe again that I have my limits of integration wrong, but I don't understand why.. I also believe this problem is an "up and down" type problem rather than a "right left", but probably wrong..

    Find the volume of the solid that results when the region enclosed by the given curves is revolved about the y-axis.

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

    V = 2pi int x[f(x) - g(x)] dx

    = 2pi int{0to1} x[x^3] - [0]dx

    = 2pi int{0to1} [x^4] dx

    = 2pi int{0to1} [x^5/5]

    = 2pi [1/5] - [0]

    = 2pi/5

    The answer in the book is 3pi/5.

    Any help is greatly appreciated. :smile:
  6. Feb 21, 2004 #5
    Your problem is that you are trying to mix two methods, the method of shells and the method of disks.

    Here is how to do it by the method of DISKS:

    The radius from the y axis to any point is a function of x, not y.

    Therefore, we must solve for x and you get x=y^(1/3)

    Now, the area of a disk of thickness dy is pi*r^2*dy

    so your total area is the intergral of all your infintessimal pieces...which is


    Integrate that and you will get the right answer.

    Now the the way to do it by SHELLS is:

    first you must know that to do a volume problem with shells, you must make your "cylinders" parrallel to the axis of rotation. So we would want to keep our function in terms of y

    However, you have your two functions mixed up! This is not a right to left problem when done as shells (it is when done with disks)....but with shells, since they must be parrallel to the axis or rotation, it is a top to bottom problem.

    So...your f(x) is one (the top line) and your g(x) is x^3 (the bottom)

    Now, I think you can finish it with that little help, since the rest of it you did right


    Notice how the problem is done with dx in shells (since your thickness is an x value) and with dy in the disk method (since your thickness is a y value)

    It seems that you were trying to do shells because your whole method was correct except the functions you choose to be f(x) and g(x)
    Last edited: Feb 21, 2004
  7. Feb 21, 2004 #6
    Okay, I'm gonna work through it again right now. Another part I believe why I'm getting confused is, how do I tell for sure if it's a "top to bottom" or "right to left" type problem? Because I'm having trouble seeing it when just drawing the graphs of f(x) and g(x)?
  8. Feb 21, 2004 #7
    You use "top to bottom" if you are doing disks (or washers) about a horizontal line or if you do shells about a vertical line.

    You use "right to left" if you are doing disks (or washers) about a vertical line or if you do shells about a horizontal line.
  9. Feb 22, 2004 #8
    Okay thanks. I'm understanding it more now! :smile:
  10. Feb 24, 2004 #9
    Okay, here's yet another one that I'm having trouble with.. I don't have the correct answer since it's an even number but I don't think it's right..


    Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis.

    x = sqrt(y) x = y/4

    Okay, I figured since it is rotating along the x-axis I'd rewrite them as: y = x^2 , y = 4x and use the method of washers.

    I also figured when (x=2 y=4) and when (x=2 y=8) so, y = 4x is the top and y = x^2 is the bottom function.

    I then set them equal to themselves: x^2 = 4x which came out to have the limits as (x=0 and x=4).


    V = pi int {0to4} [(4x)]^2 - [(x^2)]^2 dx

    = pi int {0to4} 16x^2 - x^4 dx

    = pi int {0to4} 16x^3/3 - x^5/5

    = pi [(16x^3/3) - (x^5/5)] - [(0)]

    = pi [(1024/3) - (1024/5)]

    = 2048pi / 15

    Any help is greatly appreciated.
  11. Feb 24, 2004 #10
    Looks okay to me.

  12. Feb 25, 2004 #11
    Okay, thanks!! :smile:
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