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Thread: Another Surface Integral!

  1. #1
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    Another Surface Integral!

    Hey, I tried converting co-ordinates and all but I am having difficulties.

    Evaluate

    $\displaystyle \int \int_S F \cdot dS$

    where $\displaystyle F= xi+yj +z^4 k$ and $\displaystyle S$ is the part of the cone $\displaystyle z=\sqrt{x^2+y^2}$ beneath the plane $\displaystyle z=1$ with downward orientation.


    Thanks for any help.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Nguyen View Post
    Hey, I tried converting co-ordinates and all but I am having difficulties.

    Evaluate

    $\displaystyle \int \int_S F \cdot dS$

    where $\displaystyle F= xi+yj +z^4 k$ and $\displaystyle S$ is the part of the cone $\displaystyle z=\sqrt{x^2+y^2}$ beneath the plane $\displaystyle z=1$ with downward orientation.


    Thanks for any help.
    $\displaystyle d\mathbf{S}=-\left( \frac{\partial z}{\partial x} \right)\mathbf{i} - \left( \frac{\partial z}{\partial y} \right)\mathbf{j}+\mathbf{k} = -\left( \frac{x}{\sqrt{x^2+y^2}} \right)\mathbf{i} - \left( \frac{y}{\sqrt{x^2+y^2}} \right)\mathbf{j}+\mathbf{k} $

    $\displaystyle \mathbf{F} \cdot d\mathbf{S}= -\left( \frac{x^2}{\sqrt{x^2+y^2}} \right) - \left( \frac{y^2}{\sqrt{x^2+y^2}} \right)+(x^2+y^2)^2 $

    In cylindrical coordinates this gives


    $\displaystyle \mathbf{F} \cdot d\mathbf{S}= (-r+r^4)rdrd\theta $

    Note we need to take the negative of the above because in the derivative I assumed and upward normal the k component is positive.

    Can you finish from here?
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  3. #3
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    Or, working with cylindrical coordinates from the start, we can write the position vector of any point on the cone as $\displaystyle \vec{p}(r, \theta)= rcos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ z\vec{k}= rcos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}$.

    The derivatives give vectors in the tangent plane:
    $\displaystyle \vec{p}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$
    $\displaystyle \vec{p}_\theta= -r sin(\theta)\vec{i}+ cos(\theta)\vec{j}$

    The cross product of those two vectors gives the normal vector
    $\displaystyle rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}- r\vec{k}$
    where the order of multiplication has been chose to make the $\displaystyle \vec{k}$ component negative ("with downward orientation").

    The "vector differential of surface area" is $\displaystyle (rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}- r\vec{k})drd\theta$

    Of course, $\displaystyle \vec{F}= x\vec{i}+ y\vec{j}+ z^4\vec{k}= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ r^4\vec{k}$ so that $\displaystyle \vec{F}\cdot d\vec{S}= (r^2cos^(\theta)+ r^2sin^2(\theta)- r^5 )drd\theta= (r^2- r^5)drd\theta$ just as TheEmptySet says (with the sign changed of course).
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