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Derivative with e and Definte integral

  1. Jun 7, 2009 #1
    1. Find y' if xey + 1 = xy

    The attempt at a solution
    I separated the y:
    y= ey + 1/x

    After deriviation:

    y' = e2(dy/dx) -1/(x2)

    Is this correct? If so, how do solve after since this question is m/c which one of choices is (y-ey)/ (xey-1), though it could be none of above.

    2. Find the definite integral that represents the area of the surface formed by revolving the graph of f(x) = x2 on the interval [0, sqrt(2)] about the x axis
    This is another m/c problem

    The attempt at a solution
    The problems asks for the proper shell method formula I beleive.
    one of the choice is 2pi [tex]\int[/tex] [tex]^{sqrt(2)}_{0}[/tex] x2[tex]\sqrt{1+x}[/tex]4dx

    and another 2pi [tex]\int[/tex] [tex]^{2}_{0}[/tex] y[tex]\sqrt{1+1/y}[/tex]dy

    ok my question is where do they get a sqrt for part of the answer

    3. Find the volume of the solid formed by revolving the region bounded by y=2x2 + 4x and y = 0 about the y-axis.

    Since the volume is forming on the y-axis, how do I separate x from the given function
    Last edited: Jun 7, 2009
  2. jcsd
  3. Jun 7, 2009 #2


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    Homework Helper

    For problem (2), where do you get those formulae from? When you rotate x^2 about the x-axis you form a solid with a circular cross section. That is if you slice the solid perpendicular to the yz-plane every slice is a circle with radius r and therefore with area [itex]\pi r^2[/itex]. You want to add all the slices in the interval [itex][0,\sqrt{2}][/itex] together. The r however is different for different values of x so you want to express r in terms of x. Can you see what r(x) is?
    Last edited: Jun 7, 2009
  4. Jun 7, 2009 #3
    Those formulas in problem 2 are some of the answer choices of that multiple choice question
  5. Jun 7, 2009 #4


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    Homework Helper

    Well forget multiple choice for a while. Do you understand my previous post and can you derive an expression for r(x)?
  6. Jun 7, 2009 #5
    Since you still had y on both sides of the equation, it was not really necessary to divide both sides by x, especially since it changed the nature of the possible values for x (in your new equation, it is no longer valid to have x = 0, which was perfectly valid in the old equation). Keep an eye out for losing the scope of solutions like this.
    So going back to the old equation:
    xey + 1 = xy
    all you really have to do is take the derivative of each side of the equation with respect to x, and note that the derivatives of each side should be equal as well. That is to say, if f(x) = g(x), then f'(x) = g'(x). Both sides are still functions of x even with the y present, because in general, we restrict attention to those regions of the graph of the solutions to the equation on which y is an implicit function of x. Then solve for the term (dy/dx) which is the same as y'(x) in the resulting equation.
    Last edited: Jun 7, 2009
  7. Jun 7, 2009 #6
    O yea, I forgot about that. Duh. Thank you so much
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