Evaluate the flux integral $\displaystyle \int \int_S {\bf F \cdot n} \ dS$ Where ${\bf F}(x,y,z) = z^2 {\bf k}$ where S is the part of the cone $z^2 = x^2 + y^2$ that lies between the planes $z = 1$ and $z = 2$.

My course teaches the specific case formula $\displaystyle \int \int_S {\bf F \cdot n} \ dS = \int \int_R (-F_1 f_x - F_2 f_y + F_3) \ dx \ dy$

I get $\displaystyle \int \int_S {\bf F \cdot n} \ dS = -\int \int_R -z^2 \ dx \ dy = \int \int_R x^2+y^2 \ dx \ dy = \int \int_R r^3 \ dr \ d\theta$, with bounds $0\leq r\leq{2}$ and $0\leq\theta\leq{2}\pi$

the negative sign in front of the integral with $z^2$ is because the unit normal is pointing downwards (or so I think...)

Where have I made the mistake?

  • $\begingroup$ the answer is 15pi/2 but I get 8pi... $\endgroup$ – Kevin Mar 27 '15 at 9:56

Since the region of integration is only from $z=1$ to $z=2$, so your bounds should be from $r=1$, $r=2$. You will get $\dfrac{15}{2}\pi$.

  • $\begingroup$ Ah, I see. I messed up the projection onto the xy plane by visualizing the cone as being solid on the inside... when it's just the surface. Thank you very much! $\endgroup$ – Kevin Mar 27 '15 at 10:08
  • $\begingroup$ Yeah, it is a 'hollow' cone. (without a top and bottom even!) $\endgroup$ – cgo Mar 27 '15 at 10:09

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