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Homework Help: Find volume of solid produced

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data
    y = x³ y= 0 and x = 1

    and its revolved around the line x = 2

    okay i have drawn the graph of y = x³ and other paramaters, but when i get ther area being rotated it produces a hollow center. how do i go about finding the volume?

    would it be a washers i don't understand the gap between the line x = 2 and x = 1?
     
  2. jcsd
  3. Jul 7, 2008 #2

    Defennder

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    Yes there will be a hollow centre. The solid of revolution generated looks like a ring. You can use the washer method of calculating volume. First find the volume generated rotating f(y) (express y=f(x) in terms of x first) around x=2 for the region defined within the range of y=0 and y=1. When that is done, subtract the unwanted volume which corresponds to the hollow centre beteen the lines x=2 and x=1. You don't have to use calculus to do this one, just use the formula for volume of cylinder.
     
  4. Jul 7, 2008 #3
    thanks!
    using washer and shell

    V=∫2pi(2-x)x³dx

    i got 3pi / 5
     
  5. Jul 8, 2008 #4

    dynamicsolo

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    So you chose the shell method. I agree with your result.

    To use the disk/washer method, the infinitesimal slices are taken perpendicular to the rotation axis, so we would have to integrate in the y-direction. The limits of integration would now be y = 0 to y = 1. The "washers" would be disks with an outer radius following the y = x^3 curve and an inner radius for the hole following the x = 1 curve. So the washers have holes of constant radius 2 - 1 unit. The outer radius curve will have to be inverted into x = y^(1/3); since it is to the left of x = 2, the outer radius will be 2 - y^(1/3).

    This makes the infinitesimal volume of a washer

    [tex]dV = [\pi(r_{outer})^2 - \pi(r_{inner})^2] dy
    = [\pi(2 -y^{1/3})^2 - \pi(1)^2] dy [/tex]

    The volume integral is then

    [tex]V = \pi \int_{0}^{1} 4 - 4y^{1/3}+ y^{2/3} - 1 dy[/tex] ,

    which also gets you [tex]V = \frac{3\pi}{5}[/tex]. But shells is definitely the easier method for this one...
     
  6. Jul 8, 2008 #5

    HallsofIvy

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    Another way to think about it (but exactly the same as the washer method) is this:

    1. Suppose you rotate the area above the line y= 0, from x= 0 to x= 1, about the line y= 2. What volume would it have? (That's simply a cylinder- area= [itex]\pi r^2 h[/itex].)

    2. Suppose you rotate the area above the curve y= x3, from x= 0 to x= 1, about the line y= 2. What volume would that have?

    Now subtract the two.
     
  7. Jul 10, 2008 #6
    so do disk and washer rotate perpendicular to rotation of axis and shell rotates paralell
     
  8. Jul 10, 2008 #7

    dynamicsolo

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    That's correct, and that is why the variable of integration depends on which method is used.
     
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