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Math Help - Volume with Parametric Equations

  1. #1
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    Volume with Parametric Equations

    I've been banging my head against this for a couple days now. The first question is to find the area and the second question (where I am really getting stuck) is to take that area and find a volume. This question is from school but I only have one example in my notes and as much as I look at it...it still makes no sense to me.

    The Parametric Equations are as follows:

    x=t^3-3t+1 and y=\frac{3}{2}t^2+1

    Question 1: Find the area between the curve and the y-axis from t=-\sqrt{3} to t=0

    So I set this up as Area = \int_{t1}^{t2} x dy
    When dy=3tdt
    Area = \int_{-\sqrt{3}}^{0} (t^3-3t+1)(3t)dt and I worked this out to be Area = \frac{18\sqrt{3}}{5}+\frac{9}{2}

    Question 2: What is the volume if the area (in question 1) is revolved about the x-axis?

    This is where I am stuck. The first question may be wrong but atleast I can begin to work it. I don't know where to start setting up the integral to find the volume. I am thinking that the set up will have two integrals: One for the two horizontal lines that go from points (1,\frac{5}{2}) and (1,1) to the y-axis; and one for the parametric curve.

    I have been trying to use the shell method with the shell radius being y, and y=\frac{3}{2}t^2+1 but I don't feel comfortable with anything I have set up.

    I would greatly appreciate any help you can offer.
    Thank you in advance.
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  2. #2
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    Isn't that formula for surface area? I am trying to find the volume.

    Now I am thinking \pi\int_a^b (y\frac{dx}{dt}-x\frac{dy}{dt})dt

    Does that sound right?
    I have been all over the internet and through all the books that I have and nobody seems to talk about volume when using parametric equations. I've been working on this problem for so long now, I'm starting to feel like this little guy
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  3. #3
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    Come to think of it, I beleive the formula for a revolution about the x-axis in parametric is given by

    {\pi}\int_{a}^{b}[y(t)]^{2}\cdot{(x'(t))}]dt

    In your case, {\pi}\int_{-\sqrt{3}}^{\sqrt{3}}\frac{3(t^{2}-1)(3t^{2}+2)^{2}}{4}dt=\int_{-\sqrt{3}}^{\sqrt{3}}[\frac{27t^{6}}{4}+\frac{9t^{4}}{4}-6t^{2}-3]=\frac{1476{\pi}\sqrt{3}}{35}\approx{229.47}

    I hope this helps. Let me know.
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  4. #4
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    Well, if that is correct, then at leaset that means I was getting warmer!

    Can you help me think through the process of getting the integral set up? I have a fair understanding of more basic volume problems, but the parametric equations confuse me. I feel like I almost understand.......and then I fall on my face.
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  5. #5
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    I'm working on a similar problem. I have a parametric equation like so:
    x(t) = 2t^3 - \frac{3t}{2} + 2
    y(t) = 3t^2 + 2.

    I first need to find the value of the parameter t, where the curve intersects itself. I have found this to be values -\frac{\sqrt{3}}{2} and \frac{\sqrt{3}}{2}.

    Now I need to find the volume of the revolution of the curve with the above lower and upper limit, both over the x-axis and y-axis.

    This is supposed to be the formula for the x revolution:
    V_x=\pi\int_{t1}^{t2}y(t)^2x'(t)dt.

    However I'm not getting the correct result. Am I using the formula incorrectly?
    \pi\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}(3t^2 + 2)^2(2t^3 - \frac{3t}{2} + 2)'dt
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  6. #6
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    Never mind. It turns out that I got the correct answer after all.

    Thx
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