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Volumes of solids of revolution

  1. Aug 20, 2015 #1
    I have a few questions about finding volumes of solids of revolution (in a typical first year single variable calculus course).
    1) I can rotate any region about any horizontal/vertical axis. How exactly do I rotate a region about a line that is neither horizontal nor vertical (##y = x - 1## for example)? Or is this beyond Calculus I/II?
    2) Every solids of revolution example I have come across so far always involved an axis of rotation that does not pass through the region (which is to be rotated). How can one find the volume of the solid obtained when a region is rotated about an axis passing through the region itself?
    3) The method of "disks/washers" fails in cases where, say, ##y## is given as a function of ##x## and it's impossible to find ##x## as a function of ##y## and we're supposed to rotate a region about a vertical axis, in which case we resort to the "shell" method. Does the shell method ever fail? Or does it always work?
     
  2. jcsd
  3. Aug 20, 2015 #2

    BvU

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    1) a transformation is easiest. So from x, y to u = x+y-1 , v = x-y
    2) Try to imagine what kind of solid that would be. Or do you rotate over ##\pi## instead of full circle ? In the latter case there are two things to integrate.
    3) I think it does. Or can you think of some pathological function that doesn't work ?
     
  4. Aug 20, 2015 #3

    HallsofIvy

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    Do you not understand that a "rotation" is a geometric transformation that does NOT depend on a choice of coordinate system?

    If you rotate a region about a line that passes through that region, you get one part of the solid inside another. Do you treat those as having two different volumes or just the larger volume (in which case you can ignore the smaller part of the region)?

    It is always possible to cut a region into parts on which you can use either the disk or shell method.
     
  5. Aug 20, 2015 #4
    I am fully aware that rotation does not depend on a choice of coordinate system. What I meant was "how do we define a new ##x' y'## coordinate system where ##y = x - 1## is a horizontal line?"

    I know there are no inherent limitations for either method. However, sometimes it's almost (computationally) impossible to use the method of disks. For instance, finding the volume obtained by rotating the region bounded by ##y = (x - 1)(x - 3)^2## and the ##x##-axis about the ##y##-axis. In this case, to use the method of disks, one must solve for ##x## in terms of ##y##, which is not easy to do for this particular cubic polynomial, let alone a degree 7 polynomial. Are there any cases where the converse is true? In other words, does the shell method "fail" in some cases?
     
  6. Aug 20, 2015 #5
    I do not follow. But I have attempted deriving general expressions for ##x'## and ##y'## (new coordinates corresponding to the rotation of the ##xy## coordinate system by an angle ##\theta## anticlockwise). The results I obtained where ##y' = y \cos{\theta} - x\sin{\theta}## and ##x' = y \sin{\theta} + x \cos{\theta}##. Is that correct? Where do I go from here? I know that ##y = x -1## would be horizontal if the system were rotated by ##\frac{\pi}{4}##. Therefore the line ##y=x-1## is equivalent to the line ##y' = -\frac{\sqrt{2}}{2}##, assuming that my transformation is true.
     
  7. Aug 21, 2015 #6

    BvU

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    Excellent. With a little sketch (I'm fond of making little drawings for most of the exercises) I can convince myself you are doing fine.

    Where to go from there ? In the new coordinate system you want to rotate around ##y'= -\sqrt{1\over 2}##. To do that you need to express the function for the region contour in ##x'## and ## y'##. Now that you have the coordinate transformation available, that can be done: a simple rotation over ##-\theta## !
     
  8. Aug 21, 2015 #7
    Say we're given ##y = f(x)## and/or ##x = f(y)##. We can substitute either ##x = f(y)## or ##y = f(x)## into the transformation. We could then express ##y'## and ##x'## in terms of another parameter (namely ##x## or ##y##, depending on the problem). The problem is, expressing ##y'## in terms of ##x'## isn't always easy.
     
  9. Aug 21, 2015 #8

    BvU

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    True
     
  10. Aug 22, 2015 #9
    Which means I have to do some parametric integration, right?
    Anyway, when substituting ##x = g(y)## or ##y = f(x)## into the transformation:
    $$y' = y\cos{\theta} - x\sin{\theta}$$
    $$x' = y\sin{\theta} + x\cos{\theta}$$
    we're basically rotating the coordinate axes while keeping the curve(s) fixed in space, right? If we wanted to rotate the curve itself while preserving the orientation of the coordinate axes we would do something like ##x' = g(y')## or ##y' = f(x')##, is that correct? Or does it make no difference?
     
    Last edited: Aug 22, 2015
  11. Aug 22, 2015 #10

    phion

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    It's easy to think about it as rotating the paper to re-orient the graph and manipulating the function(s) into more familiar or malleable terms with respect to the axis of rotation.
     
  12. Aug 22, 2015 #11
    Update: I realize it won't make a difference in terms of the final answer, but the intermediate steps are vastly different. Rotating the axes while keeping the function "fixed" gives a pair of parametric equations. Rotating the entire region while keeping the axes "fixed" gives an equation in which ##y## is defined implicitly, often impossible to isolate.
    Conclusion: Method I (rotating the axes) is more reliable.
    Is my analysis correct?
     
  13. Aug 24, 2015 #12
    Anyone?
     
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