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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.

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  • $\begingroup$ 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. $\endgroup$ – Michael E2 Mar 31 at 21:19

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