# Vector calculus q

• Feb 22nd 2009, 08:16 AM
Thomas154321
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. ^_^
• Feb 22nd 2009, 10:29 AM
HallsofIvy
Quote:

Originally Posted by Thomas154321
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.