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Thread: Integral under a closed curve

  1. #1
    Super Member Showcase_22's Avatar
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    Integral under a closed curve

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

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

    Question: Evaluate $\displaystyle \int_C \underline{v} \cdot d \underline{r}$ where $\displaystyle C$ 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:

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

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

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

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

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

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

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

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

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

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

    So the integral for the x coordinate is:

    $\displaystyle \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; Apr 24th 2009 at 02:56 AM.
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  2. #2
    MHF Contributor

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    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 $\displaystyle \vec{r}\cdot\vec{dr}$ rather than $\displaystyle \vec{r}|\vec{dr}|$ which is what you have.

    Also, you need to integrate along the given curve from t= 0 to $\displaystyle t= \pi$, then integrate along the straight line from (0, -1, 0) back to (0, 1, 0).
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  3. #3
    Super Member Showcase_22's Avatar
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    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 $\displaystyle \vec{r}\cdot\vec{dr}$ rather than $\displaystyle \vec{r}|\vec{dr}|$ which is what you have.
    Okay, so I need something like:

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

    In which case I would have:

    $\displaystyle \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 $\displaystyle t= \pi$, 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 )
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