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Negative volume using washer method

  1. Dec 26, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the volume of a solid formed by the area trapped between y= -x^2 and y= -2x rotated 360° around x-axis?

    2. Relevant equations
    V = ∫A(x)dx

    3. The attempt at a solution
    y=y
    -x^2 = -2x
    x^2 -2x = 0
    x(x-2) = 0

    This means that the two functions cross at x = 0 and x = 2
    From x = 0 to x = 2 , y = -x^2 will be the upper bound

    So, the volume is
    ## V = \pi \int_{0}^{2}((-x^2)^2-(-2x)^2))dx = \pi \int_{0}^{2}(x^4 - 4x^2)dx = \pi \left [ \frac{1}{5}x^5-\frac{4}{3}x^3 \right ]^2_0 = \pi (\frac{32}{5}-\frac{32}{3})=-4\frac{4}{15}\pi ##

    Why do I get negative sign?? What's wrong??
     
  2. jcsd
  3. Dec 26, 2015 #2

    Orodruin

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    Because x(x-2) is negative in your specified interval. x^2 is the lower curve!
     
  4. Dec 26, 2015 #3
    So, to determine which one is upper, we need to see which one is at the top at y positive area (by visualizing rotating it) ?
    In other words,
    if y < 0 then the upper bound is actually the bottom curve and the lower bound is the top one ??
    if y > 0, we just see which one at the top/bottom to decide which one is upper/lower bound..

    Is it right?
     
  5. Dec 26, 2015 #4
    I don't think -x^2 is the bottom function, but make sure you understand what solid is being formed when you rotate R around the x-axis. Look at one arbitrary x-value and see what happens when you rotate it around the x-axis, and first see what kind of solid you get.
     
  6. Dec 26, 2015 #5

    Mark44

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    This is more complicated than it needs to be. Because you're squaring the y-values, the squared y-value on the line is larger than the squared y-value on the parabola, so your integral should be ##\int_0^2 (-2x)^2 - (-x^2)^2 dx## to produce a reasonable (i.e., nonnegative) volume.

    The region being revolved is equal in area to the region defined by the curves ##y = x^2## and y = 2x, and would make for a simpler integral.
     
  7. Dec 26, 2015 #6
    I looked at this incorrectly earlier but I think the OP's mistake was that you had y =-2x be the bottom function instead of the top function.
    The easiest way to tell which function is on top of which for problems like these, since that's probably the only challenging part of these problems for most people when it comes to graphing and seeing how the area looks like, is to plug in values between the two points that the curves intersect. Be especially wary about how the area between two curves looks when you have some function like sin or cosine, because you can end up with two integrals if the two curves switch in their orientation to each other (i.e. from 0 to π/4, cos(x) is above sin(x) but from π/4 to π/2, sin(x) is above cos(x)).
     
  8. Dec 26, 2015 #7

    Mark44

    Staff: Mentor

    y = -2x is the "bottom" function relative to ##y = -x^2##. However, once you square both of these, then ##(-x^2)^2 \le (-2x)^2## on the interval in question.
    The simplest thing to do, IMO, for problems such as the one in this thread, is to work with the functions reflected across the x-axis. The area of the region being revolved is the same, and you don't have the complication of figuring out which function is on the bottom.
     
  9. Dec 26, 2015 #8

    Orodruin

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    I agree with this, I did so without even mentioning it, which is why x^2 is my bottom function. The easier way of looking at this is to consider the volume as the difference between two rotational volumes. It then becomes very clear which is which.
     
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