1

A fluid is poured down through the cone $z = \sqrt{x^2+y^2}$ for $x^2+y^2 \leq 1$, where the vector field describing the fluid flow is $f(x,y,z) = (0,0,-1)$. Determine the flux of the fluid through the cone.

My attempt was to parametrize the cone as $r(x,y)$ as $(x,y,\sqrt{x^2+y^2})$ for $x^2+y^2 \leq 1$. Then $T_x$ would be be $(1,0,\frac{x}{\sqrt{x^2+y^2}}$), and $T_y$ would be $(0,1,\frac{y}{\sqrt{x^2+y^2}}$), and therefore $T_x \times T_y = (-\frac{x}{\sqrt{x^2+y^2}}, -\frac{y}{\sqrt{x^2+y^2}}, 1)$.

So $$\int\int_S f\cdot dR = \int\int_{x^2+y^2\leq1} (0,0,-1) \cdot (T_x \times T_y)dxdy $$

$$\int\int_{x^2+y^2\leq1}-1dxdy$$

Then converting into polar coordinates, $$\int_0^{2\pi}\int_0^1-rdrd\theta = -\pi$$

However, my professor got an answer of $0$ using Stokes' theorem, using f as the curl of $g = (y,0,0)$ and parametrizing the boundary as $r(t) = (\cos(t), \sin(t), 1)$.

I am confused as to why my method didn't work? Thank you!

Rafa Budría
  • 7,364
Jcap
  • 11
  • 1

1 Answers1

1

Your method is correct. Using Stoke's theorem gives the same result. With the parametrization you gave the tangent to the curve is $$ (-\sin(t), \cos(t), 0) $$ so the integrand becomes $$ (\sin(t), 0, 0)\cdot (-\sin(t), \cos(t), 0) = -\sin^2(t)$$ And the integral $$ \int_0^{2\pi} -\sin^2(t) dt = -\pi$$

themaker
  • 2,451