# Volume of the fluid in the upward direction

Let $$\mathbf{V}$$ the velocity vector of a fluid particle at the point $$(x,y,z)$$ in a steady-state fluid flow. $$\mathbf{V}=x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+3z\hat{\mathbf{k}}$$

Let $$S$$ be the graph of the function $$f:D\to \mathbb{R},\ (x,y)\mapsto xy(2y+x-2)$$ where $$D$$ is the triangle with the vertices $$(0,0), (2,0)$$ and $$(0,1)$$. Let $$E$$ be the solid between $$S$$ and the $$xy$$-plane. Find the net volume that passes through $$S$$ in the upward direction in $$1$$ second directly as a surface integral.

My attempt:

Parametrization of $$S$$ can be given as $$\mathbf{r}(u,v)=[u,v,uv(2v+u-2)],\ (u,v)\in D.$$ I am getting $$\mathbf{r}_u=[1,0, 2v^2+2uv-2v]\text{ and } \mathbf{r}_v=[0,1,4uv+u^2-2u]$$ and therefore $$\mathbf{r}_u\times \mathbf{r}_v=[-(2v^2+2uv-2v), -(4uv+u^2-2u),1] .$$ Hence the volume is $$\iint_D\mathbf{V}(u,v)\cdot (\mathbf{r}_u\times\mathbf{r}_v)~du~dv=-\dfrac{1}{3}.$$

I don't know why am I getting a minus sign in the final answer. What is the significance of the upward direction.

• Negative flux indicates the net flow is in the opposite direction of the orientation or $\mathbf{r}_u\times\mathbf{r}_v$. Note that $z$ is negative throughout most of the surface, so that the vector field has a negative vertical component. Of course you cannot consider just a single component but should consider the relationship of the vector field to the surface normal. Throughout the interior of the surface the flow is from above to below the surface. – Michael E2 Mar 31 at 21:19