# If $\vec A=6z\hat i+(2x+y)\hat j-x\hat k$ evaluate $\iint_S \vec A\cdot \hat n\,dS$

Suppose $$\vec A=6z\hat i+(2x+y)\hat j-x\hat k$$. Evaluate $$\iint_S \vec A\cdot\hat n\,dS$$ over the entire surface $$S$$ of the region bounded by the cylinder: $$x^2+z^2=9,x=0,y=0,z=0$$ and $$y=8$$. Here $$\hat n$$ is unit normal vector to surface $$S$$.

I got this question in Schaum's outline of Vector Analysis. Answer is given $$18\pi$$. But I don't understand on which plane I would make projection of the surface.

• You should take projection in $XZ$ plane Jan 12 at 7:39

If you can use divergence theorem, that is more straightforward.

Find divergence of the vector field using $$\ \nabla \cdot \vec{A}$$ and integrate over the cylinder volume in the first octant. Applying divergence theorem will give you flux through the entire closed surface.

In this case, you realize that divergence is simply $$1$$ so you are left with finding volume of the cylinder in the first octant. It may not even require an integral.

But if you have to do double integral then

Parametrize your surface as $$\ \vec{r}(y, \theta) = (3 \cos \theta, y, 3 \sin \theta)$$

$$\vec{r'}_{\theta} \times \vec{r'}_y = (3 \cos \theta, 0, 3 \sin \theta)$$

So your integral to find flux through cylindrical surface is

$$\displaystyle \int_0^{\pi/2} \int_0^8 \vec{A}{(r(y,\theta))} \cdot (3 \cos \theta, 0, 3 \sin \theta) \ dy \ d\theta$$

Then you also need to consider planes $$x = 0, z = 0, y = 0, y = 8$$ with normal vectors as $$(-1, 0, 0), (0, 0, -1), (0, -1, 0), (0, 1, 0) \$$ respectively as the question seeks flux through the entire surface.