The question is: Find $\iint_Y F.N \ dS $ $$ F=(x^4+yz-x^5,5x^4y,z),\quad \text{The surface }Y=x^2+y^2-z^2=1, \ \ 0\leq z\leq 1 \quad N=\text{ the normal points away from z axis}$$

Here is how I thought:

I parameterized the surface with $r(x,y)=(x,y,\sqrt{x^2+y^2-1}) $ and i think the normal is: $(r'_x\times r'_y)= (\frac{x}{\sqrt{x^2+y^2-1}},\frac{y}{\sqrt{x^2+y^2-1}},-1)$ and then I calculated the integral in this way but the integral looks very complicated and awful, have I calculated correctly? and what is the limit of this integration? $$\iint \frac{x^5-x^6+5x^4y^2+x^2+y^2-1+\sqrt{x^2+y^2-1}xy}{\sqrt{x^2+y^2-1}}\ dx\,dy $$ Any suggestion would be great, thanks.


One of the ways would be to use divergence theorem and find the flux through entire closed surface which consists of the hyperboloid between $0 \leq z \leq 1, $ and circular top and bottom surfaces at $z = 0, z = 1$. We can then subtract the flux through circular surfaces at $z = 0$ and $z = 1$ which is an easier calculation (in fact you can see why the flux through the circular bottom at $z = 0 \ $ for the given vector field would be zero. So it is only the flux through the circular top that we may have to find and subtract).

$div\vec{F} = \nabla \cdot \vec{F} = 4x^3-5x^4 + 5x^4 + 1 = 1 + 4x^3$

Also as $x^3$ is an odd function and there is symmetry along $x-$plane ($YZ$), the integral simplifies further.

So, Flux $ = \displaystyle \int_V div\vec{F} \ dV = \int_0^{2\pi} \int_0^1 \int_0^\sqrt{1+z^2} \ r \ dr \ dz \ d\theta$

Now calculate the flux through the circular top and subtract from the above.

Also to comment on your working, the parametrization and your steps for double integral are correct. It is a tedious one! For limits,

From here on, you can see that for $0 \leq z \leq 1, 1 \leq x^2+y^2 \leq 2$. So you can convert it into polar coordinates as $x = r \cos\theta, y = r\sin\theta, z$.

$\int_0^{2\pi} \int_1^\sqrt2 \vec{F}\cdot (r_x \times r_y) \ r \ dr \ d\theta$

Another way to parametrize the hyperboloid would be,

$r(\theta, z) = (\sqrt{1+z^2} \cos \theta, \sqrt{1+z^2} \sin \theta, z)$ and

$(r'_\theta \times r'_z) = (\sqrt{1+z^2} \cos \theta, \sqrt{1+z^2} \sin \theta, z)$

But it does not make the integral any easier.

  • 2
    $\begingroup$ $4x^3$ isan odd function, you can make this simplification $\endgroup$ – Ninad Munshi Jan 17 at 20:57
  • 1
    $\begingroup$ @NinadMunshi thank you for pointing out! I call out $5$ times a day and when I had to do. I missed it :) $\endgroup$ – Math Lover Jan 17 at 21:17
  • $\begingroup$ Thank you, I always wanted to understand "odd function and symmetry" in multivariable function, I understand it regarding one variable function, "if the function $y=f(x)$ is odd and the interval of $x$ is between (-p,p) "symmetric about origin" then the integral is zero, but in our case $x^3$ is odd but I wonder how to know that interval is symmetric and which interval should be symmetric y,z? x,z? or x,y ? and why ? if you are busy can you please suggest an pdf file that I could study more on this subject, Thanks $\endgroup$ – simon Jan 18 at 6:47
  • $\begingroup$ one simple way could be to look at the region. Our region is $x^2 + y^2 - z^2 = 1$ and $(-x)^2 + y^2 - z^2 = x^2 + y^2 - z^2$ so you can see it is symmetric to $x$. So what matters is the limits of $x$. You can see that $0 \leq \theta \leq \pi$ and $\pi \leq \theta \leq 2\pi$ have equal limits of $x$ on both positive and negative sides. $\endgroup$ – Math Lover Jan 18 at 7:28
  • $\begingroup$ Thank you for helping me out $\endgroup$ – simon Jan 18 at 7:58

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