# Compute the downwards flux of F

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi Univman,

The links to your work on the problem seem not to work. Anyway, you have been given the vector field,

$\mathbf{F}(x,y,z)=\mathbf{i} \cos\left( \frac{y}{z}\right)-\mathbf{j} \frac{x}{z} \sin \left(\frac{y}{z}\right)+\mathbf{k} \frac{xy}{z^2} \sin \left( \frac{y}{z} \right)$

The downward flux of $$\mathbf{F}$$ over $$S$$ is given by,

$\iint_{S}\mathbf{F}.(-\mathbf{k})\,dS=-\iint_{S}\mathbf{F}.\mathbf{k}\,dS=-\iint_{S}\frac{xy}{z^2}\sin\left(\frac{y}{z}\right)\,dS$

On the surface $$S$$; $$z=1$$. Therefore we have,

$\iint_{S}\mathbf{F}.(-\mathbf{k})\,dS=-\iint_{S}xy\sin y\,dS=-\int_{x=0}^{1}\int_{y=0}^{\pi}xy\sin y\,dy\,dx$

Hope you can continue.

Kind Regards,
Sudharaka.