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Thread: Vector calculus q

  1. #1
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    Vector calculus q

    Evaluate

    $\displaystyle I = \int \nabla \cdot \bold F dV$

    for $\displaystyle \bold F = \sqrt{x^2+y^2}(x\bold e_1 + y\bold e_2)$ over the volume $\displaystyle x^2 + y^2 \leq R^2, 0 \leq z \leq L$



    So I get $\displaystyle I = \oint \bold F \cdot d\bold A$ but can't work out what to replace $\displaystyle d\bold A$ by, for example $\displaystyle (x\bold e_1 + y^2\bold e_2 + \bold e_3) dA$. I think the answer is $\displaystyle (x\bold e_1 + y\bold e_2) dA$ but I don't have a clue how to find that. If it is true which makes it parallel to F leading to the answer $\displaystyle R^2A$ which seems reasonable.


    Many thanks for any comments. ^_^
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  2. #2
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    Quote Originally Posted by Thomas154321 View Post
    Evaluate

    $\displaystyle I = \int \nabla \cdot \bold F dV$

    for $\displaystyle \bold F = \sqrt{x^2+y^2}(x\bold e_1 + y\bold e_2)$ over the volume $\displaystyle x^2 + y^2 \leq R^2, 0 \leq z \leq L$



    So I get $\displaystyle I = \oint \bold F \cdot d\bold A$ but can't work out what to replace $\displaystyle d\bold A$ by, for example $\displaystyle (x\bold e_1 + y^2\bold e_2 + \bold e_3) dA$. I think the answer is $\displaystyle (x\bold e_1 + y\bold e_2) dA$ but I don't have a clue how to find that. If it is true which makes it parallel to F leading to the answer $\displaystyle R^2A$ which seems reasonable.


    Many thanks for any comments. ^_^
    Since this is a cylinder, use cylindrical coordinates: $\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$, z= z. On the boundary of the cylinder, r is the constant R. In particular, that means we can write the cylindrical surface as the vector function of two parameters: $\displaystyle \vec{r}= Rcos(\theta)\vec{i}+ R sin(\theta)\vec{j}+ z\vec{k}$

    Now take the derivatives of that with respect to the two variables:
    $\displaystyle \vec{u}= \frac{\partial \vec{r}}{\partial \theta}= -Rsin(\theta)\vec{i}+ Rcos(\theta)\vec{j}$
    $\displaystyle \vec{v}= \frac{\partial \vec{r}}{\partial z}= \vec{k}$

    The cross product of those two vectors is the "fundamental vector product". It, times $\displaystyle d\theta dz$, is dA.
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