Results 1 to 3 of 3

Math Help - Integral under a closed curve

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Integral under a closed curve

    Let H be the curve parameterised by \underline{r}(t):=(sin ~t, cos~t, sin~2t), \ 0 \leq t \leq \pi, and let L be the horizontal line from (0,-1,0) to (0,1,0).

    Let \underline{v} be the vector field \underline{v}(x,y,z):=(2z \ , \ y(x^2+y^2) \ ,\ x).

    Question: Evaluate \int_C \underline{v} \cdot d \underline{r} where C is the closed curve formed by joining H and L (You may use either orientation of C).
    I can't picture geometrically what this question is asking me to do.

    Here's my attempt at a solution:

    \int_C \underline{v} \cdot d \underline{r}=\int_a^b\underline{v}(\underline{r}(  t))|\underline{r}'(t)|~dt where a=(0,-1,0) and b=(0,1,0).

    =\int_a^b (2sin(2t), \ sin(2t), \ sin(t)) |r'(t)| \ dt

    r'(t)=(cos(t), \ -sin(t), \ 2sin(t)cos(t)) \Rightarrow \ |r'(t)|=\sqrt{1+4sin^2(t)cos^2(t)} =\sqrt{1+sin^2(2t)}

    So I have to work out \int_a^b (2sin(2t), \ sin(2t), \ sin(t)) \sqrt{1+sin^2(2t)} \ dt

    = \int_a^b (2sin(2t) \sqrt{1+sin^2(2t)}, sin(2t) \sqrt{1+sin^2(2t)}, sin(t) \sqrt{1+sin^2(2t)} \ dt

    From here is it okay to integrate each of the x,y,z values with respect to t?

    Here's what happens when I do (this is for the x coordinate. I thought it would be easier doing each one separately):

    Let u=1+sin^2(2t) \Rightarrow \ \frac{du}{dt}=4sin(2t)cos(2t)

    This gives: \int_a^b \frac{1}{2}\frac{\sqrt{u}}{cos(2t)}~du

    But we also know that u=1+(1-cos^2(2t))=2-cos^2(2t) \Rightarrow \ cos(2t)=\sqrt{2-u}

    So the integral for the x coordinate is:

    \frac{1}{2}\int_a^b \sqrt{ \frac{u}{2-u}}~du ....which I can't think of a way of integrating
    Last edited by Showcase_22; April 24th 2009 at 02:56 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,579
    Thanks
    1418
    Your final integral is clearly wrong: you have an integral of a vector when it should be the integral of a scalar function. You want \vec{r}\cdot\vec{dr} rather than \vec{r}|\vec{dr}| which is what you have.

    Also, you need to integrate along the given curve from t= 0 to t= \pi, then integrate along the straight line from (0, -1, 0) back to (0, 1, 0).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Quote Originally Posted by HallsofIvy View Post
    Your final integral is clearly wrong: you have an integral of a vector when it should be the integral of a scalar function. You want \vec{r}\cdot\vec{dr} rather than \vec{r}|\vec{dr}| which is what you have.
    Okay, so I need something like:

    \int_C \underline{v} \cdot d \underline{r}=\int_a^b \underline{v} (\underline{r}(t)) \cdot r'(t)~dt?

    In which case I would have:

    \int_a^b (2sin(2t), \ sin(2t), \ sin(t)) \cdot (cos(t), \ -sin(t), \ 4sin(2t)cos(2t))~dt

    Also, you need to integrate along the given curve from t= 0 to t= \pi, then integrate along the straight line from (0, -1, 0) back to (0, 1, 0).
    So I should have a double integral? ( which makes me question what i've put above this )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Closed Curve Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 12th 2012, 09:27 AM
  2. Replies: 1
    Last Post: April 8th 2010, 04:58 PM
  3. integral, closed path
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: January 29th 2010, 01:58 PM
  4. closed curve, trace
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 27th 2010, 07:19 PM
  5. closed curve, trace
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 27th 2010, 06:59 PM

Search Tags


/mathhelpforum @mathhelpforum