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Am I doing this correctly?

  1. Aug 28, 2004 #1
    Hello everyone,

    I was wondering if someone could take a look at two equations to see if I am solving them properly. I'm currently in Calculus II, but is been a year since I took Calculus I (budget cuts cancelled the Calc II class last spring,) so I want to make sure I still haven't forgotten anything.

    These questions came from my Calc II professor's final exam she gave last year to her Calc I class. She gave us the exam for homework to see where our skills were. I believe these are correct, but I wanted to make sure my work was correct. Anyway, here they are (my apologies for a long post):
    Question 1: Consider the region bounded by the graph [tex] y = 2x - x^2[/tex] and [tex] y = x [/tex]

    Part 1: Set up the integral for the volume of the solid formed by revolving the region around the y-axis

    Part 2: Find the volume of the solid.

    [tex] V = \int_{a}^{b} 2 \pi x f(x)dx [/tex]

    [tex] 2x - x^2 = x [/tex]

    [tex] x - x^2 = 0 [/tex]

    [tex] x (1 - x) = 0 [/tex]

    [tex] x = 0 [/tex] and [tex] x = 1 [/tex]

    [tex] V = \int_{0}^{1} 2 \pi x (x) dx [/tex]

    [tex] V = 2 \pi \int_{0}^{1} x (x) dx [/tex]

    [tex] V = 2 \pi \int_{0}^{1} x^2 dx [/tex]

    [tex] V = 2 \pi \int_{0}^{1} \frac{1}{3}x^3 dx [/tex]

    [tex] V = 2 \pi \left[ \frac{1}{3}x^3 \right]_{0}^{1} [/tex]

    [tex] V = 2 \pi \left[ \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 \right] [/tex]

    [tex] V = \frac{2 \pi}{3} [/tex]

    Question 2: Evaluate [tex] \int \frac{\cos \left[ \frac{1}{t} \right]}{t^2} dt [/tex]

    [tex] \int \cos \left[ \frac{1}{t} \right] t^{-2} dt [/tex]

    let [tex] u = \frac{1}{t} [/tex]

    [tex] du = - t^{-2} dx [/tex]

    [tex] - \int \cos (u) du[/tex]

    [tex] = - sin (u) + C [/tex]

    [tex] = - sin \left[ \frac{1}{t} \right] + C [/tex]

    Again, my apologies for such a long post.
  2. jcsd
  3. Aug 28, 2004 #2
    #1 what you are doing is rotating the area between the two curves

    To find the limits of integration you need to figure out where they are equal to each other (which you did correctly)

    Then you need to subtract the rotated area of the lower curve (y=x) from the rotated area of the higher curve (y=2x-x^2)

    Which would give you the integral:
    [tex] 2 * \pi * \int_0^1 2x^2 - x^3 -x^2 dx [/tex]

    The last x^2 comes from f(x)*x where f(x) is y=x. It is minus because we are subtracting that area.

    The 2x^2 – x^3 comes from f(x)*x where f(x) is y = 2x-x^2
    The 2 pi comes from the formula to do cylindrical shells. (basically it is the out side of a shell)

    So your new integral is
    [tex] 2 * \pi * \int_0^1 x^2 - x^3 dx [/tex]

    I get an volume of pi/6

    #2 looks good to me.
    Last edited: Aug 28, 2004
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