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Math Help - stoke's thrm

  1. #1
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    stoke's thrm

    Let  \vec{F} = y\vec{i}  + z\vec{j}  + x\vec{k} and C be the boundary of  z = 5 - (x^2 + y^2) for z ≥ 3, oriented upward. Find  \int\limits_c\vec{F}   d\vec{r} .
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  2. #2
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    Quote Originally Posted by TheRekz View Post
    Let  \vec{F} = y\vec{i} + z\vec{j} + x\vec{k} and C be the boundary of  z = 5 - (x^2 + y^2) for z ≥ 3, oriented upward. Find  \int\limits_c\vec{F}  d\vec{r} .
    So to use stokes theorem we need to find the curl of the vector

    \begin{vmatrix}<br />
\vec i & \vec j & \vec k \\<br />
\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\<br />
y & z & x<br />
\end{vmatrix} = (0-1)\vec i-(1-0)\vec j+(0-1)\vec k=-\vec i -\vec j -\vec k

    Now we need the normal vector to the surfe

    F(x,y,z) = x^2+y^2+z-5

    \nabla F = 2x \vec i + 2y \vec j + \vec k

    |\nabla F| = \sqrt{4x^2 + 4y^2 + 1}

    dS=\sqrt{1+(z_x)^2+(z_y)^2}= \sqrt{1+4x^2+4y^2}dA

    \oint \vec F \cdot d\vec r = \iint_S \nabla \times \vec F\cdot d \vec S= \iint_S \nabla \times \vec F\cdot \frac{\nabla \vec F}{|\nabla F|}\sqrt{1+(z_x)^2+(z_y)^2}dA

    -\iint_S(2x+2y+1)dA

    We need to integrate over the circle x^2+y^2=(\sqrt{2})^2

    Just change to polar coordinates to finish
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  3. #3
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    here's what I got:

     - \int_0^{2\pi}\int_0^{\sqrt(2)} (2\sqrt(2)cos\theta + 2\sqrt(2)sin\theta + 1) r dr d\theta

    is this correct?
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  4. #4
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    Quote Originally Posted by TheRekz View Post
    here's what I got:

     - \int_0^{2\pi}\int_0^{\sqrt(2)} (2\sqrt(2)cos\theta + 2\sqrt(2)sin\theta + 1) r dr d\theta

    is this correct?
    Note quite you jumped the gun on the \sqrt{2}

     - \int_0^{2\pi}\int_0^{\sqrt(2)} (2rcos\theta + 2r\sin\theta + 1) r dr d\theta

    This will do it
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  5. #5
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    I got it! thanks
    Last edited by TheRekz; May 6th 2009 at 09:19 PM.
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  6. #6
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    Quote Originally Posted by TheRekz View Post
    I got -2\pi\sqrt(2) but it marks the answer as wrong

     - \int_0^{2\pi} \frac{4\sqrt(2)cos(\theta)}{3} + \frac{4\sqrt(2)sin(\theta)}{3} + \sqrt(2) d\theta
     - \frac{4\sqrt(2)sin(\theta)}{3} + \frac{4\sqrt(2)cos(\theta)}{3} + \sqrt(2)\theta evaluated from 0 to 2pi
    and I got the above answer

    - \int_0^{2\pi}\int_0^{\sqrt(2)} (2r^2cos\theta + 2r^2\sin\theta + r) dr d\theta=-\int_{0}^{2\pi} \frac{2}{3}r^3\cos(\theta)+\frac{2}{3}r^3\sin(\the  ta)+\frac{1}{2}r^2 \bigg|_{0}^{\sqrt{2}}d\theta

    =-\int_{0}^{2\pi} \frac{4\sqrt{2}}{3}\cos(\theta)+\frac{4\sqrt{2}}{3  }\sin(\theta)+1 d\theta =

    -\left(\frac{4\sqrt{2}}{3}\sin(\theta)-\frac{4\sqrt{2}}{3}\cos(\theta)+\theta \bigg|_{0}^{2\pi}\right)=-2\pi
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  7. #7
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    what is the normal vector to this surface:

    the surface z = 9 - x^2 for 0 ≤ x ≤ 3 and -6 ≤ y ≤ 6, oriented upward
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  8. #8
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    Quote Originally Posted by TheRekz View Post
    what is the normal vector to this surface:

    the surface z = 9 - x^2 for 0 ≤ x ≤ 3 and -6 ≤ y ≤ 6, oriented upward
    write the function in the form

    F(x,y,z)=0

    z=9-x^2 \iff x^2+z-9=0

    This gives

    F(x,y,z)=x^2+z-9

    \nabla F = \frac{\partial F}{\partial x} \vec i+ \frac{\partial F}{\partial y} \vec j+ \frac{\partial F}{\partial z} \vec k =2x\vec i +\vec k
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    so if  \vec{F} = xy \vec{i}  + yz\vec{j}  + xz\vec{k} and what is dS here as what you did with the question above?
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  10. #10
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    Quote Originally Posted by TheRekz View Post
    so if  \vec{F} = xy \vec{i} + yz\vec{j} + xz\vec{k} and what is dS here as what you did with the question above?
    d\vec S=\vec n \cdot dS

    Where n is a normal unit vector from the surface

    \vec n =\frac{\nabla F}{|\nabla F|}

    This will always give you a normal unti vector.

    Note that IF z can be isolated on your surface then ie

    z=g(x,y)

    then |\nabla F|=dS=\sqrt{1+(z_x)^2+(z_y)^2}

    and then

    \iint_S F \cdot d\vec S = \iint F \cdot \vec n dA
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  11. #11
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    so dS above is equal to 2xi + k dA ??
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  12. #12
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    ok after doing all that I got:

     \iint_S (-2xy - x) d\vec{A}

    I need the integration surface now... having a hard time to set it
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