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Math Help - Vector calculus q

  1. #1
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    Vector calculus q

    Evaluate

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

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



    So I get I = \oint \bold F \cdot d\bold A but can't work out what to replace d\bold A by, for example (x\bold e_1 + y^2\bold e_2 + \bold e_3) dA. I think the answer is (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 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

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

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



    So I get I = \oint \bold F \cdot d\bold A but can't work out what to replace d\bold A by, for example (x\bold e_1 + y^2\bold e_2 + \bold e_3) dA. I think the answer is (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 R^2A which seems reasonable.


    Many thanks for any comments. ^_^
    Since this is a cylinder, use cylindrical coordinates: x= r cos(\theta), 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: \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:
    \vec{u}= \frac{\partial \vec{r}}{\partial \theta}= -Rsin(\theta)\vec{i}+ Rcos(\theta)\vec{j}
    \vec{v}= \frac{\partial \vec{r}}{\partial z}= \vec{k}

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