Question

Let T be the triangular region with vertices (7, 0, 0), (0, 7, 0), and (0, 0, 7) oriented with an upward-pointing normal vector. Assume distances are in meters. A fluid flows with constant velocity field v = 2k meters per second or m/s. Calculate the flow rate through T.
Calculate the flow rate through the projection of T onto the xy-plane [the triangle with vertices (5, 0, 0), (0, 5, 0), and (0, 0, 0)].

Answer 1

Parameterize $T$ by

$\vec s(u,v)=(1-v)\bigg((1-u)(7\,\vec\imath)+u(7\,\vec\jmath\bigg)+v(7\,\vec k)=7(1-u)(1-v)\,\vec\imath+7u(1-v)\,\vec\jmath+7v\,\vec k$

with both $u,v$ ranging from 0 to 1.

The fluid flows with velocity $\vec v=2\,\vec k$, so it's flowing upward and we should take the normal vector to $T$ to be pointing away from the origin:

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(49-49v)(\vec\imath+\vec\jmath+\vec k)$

Then the flow rate of $\vec v$ across $T$ is

$\displaystyle\iint_T\vec v\cdot\mathrm d\vec S=\int_0^1\int_0^1(49-49v)(2\,\vec k)\cdot(\vec\imath+\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle98\int_0^1\int_0^1(1-v)\mathrm du\,\mathrm dv=\boxed{49}$

(in m/s)

The projection of $T$ on the $x,y$ plane should have vertices (7, 0, 0), (0, 7, 0), and (0, 0, 0). (unless you're talking about another $T$ altogether). Parameterize $T$ by

$\vec s(u,v)=(1-v)\bigg((1-u)(7\vec\imath)+u(7\,\vec\jmath)\bigg)=7(1-u)(1-v)\,\vec\imath+7u(1-v)\,\vec\jmath$

again with both $u,v$ between 0 and 1.

The normal vector is then

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(49-49v)\,\vec k$

and so the flow rate is

$\displaystyle\iint_T\vec v\cdot\mathrm d\vec S=\int_0^1\int_0^1(49-49v)(2\,\vec k)\cdot(\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle98\int_0^1\int_0^1(1-v)\,\mathrm du\,\mathrm dv=\boxed{49}$