Results 1 to 6 of 6

Thread: Volume with Parametric Equations

  1. Feb 23rd 2008, 09:07 AM #1
    Quasar
    Quasar is offline
    Newbie
    Joined
    Dec 2006
    Posts
    8

    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.
    Follow Math Help Forum on Facebook and Google+
  2. Feb 23rd 2008, 01:53 PM #2
    Quasar
    Quasar is offline
    Newbie
    Joined
    Dec 2006
    Posts
    8
    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
    Follow Math Help Forum on Facebook and Google+
  3. Feb 23rd 2008, 03:15 PM #3
    galactus
    galactus is offline
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+
  4. Feb 23rd 2008, 03:25 PM #4
    Quasar
    Quasar is offline
    Newbie
    Joined
    Dec 2006
    Posts
    8
    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.
    Follow Math Help Forum on Facebook and Google+
  5. Oct 19th 2009, 08:26 AM #5
    CubiX
    CubiX is offline
    Newbie
    Joined
    Oct 2009
    Posts
    2
    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
    Follow Math Help Forum on Facebook and Google+
  6. Oct 21st 2009, 03:35 AM #6
    CubiX
    CubiX is offline
    Newbie
    Joined
    Oct 2009
    Posts
    2
    Never mind. It turns out that I got the correct answer after all.

    Thx
    Follow Math Help Forum on Facebook and Google+
« trig substitution | Taylor series and radius of convergence »

Similar Math Help Forum Discussions

  1. Parametric Curves and Volume of the area rotated about y-axis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 8th 2011, 05:36 AM
  2. volume of parametric equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 22nd 2010, 03:59 AM
  3. Rotated parametric curve : calculate the volume
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jul 8th 2009, 01:12 PM
  4. How to convert cartesian equations into vector or parametric equations?
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Dec 2nd 2008, 10:54 AM
  5. Help with finding Cartesian Equations from Parametric Equations using Chain Rule
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Sep 1st 2007, 06:35 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum