compute $\iint_Y F.N\ dS $ with Gauss

Question

The question is: $$ \iint_Y F.N\ dS \quad Y=(x-z)^2+(y-z)^2=1+z^2,\ \ 0\leq z\leq 1 \quad N=\text{pointing outward} $$ $$ F=(y,x,1+x^2z)$$ Here is how I have tried to solve it: I tried to solve it with Gauss theorem like this, $\gamma=\text{upper lid}$, $\sigma=\text{lower lid} $ $$ \iint_{Y+\sigma+\gamma}F.N\ dS=\iiint_K \text{div}f\ dA \iff \iint_Y F.N\ dS=\iiint_k-\iint_\sigma-\iint_\gamma$$ $$ \iint_\sigma=-\pi, \iint_\gamma=2+\pi$$ But I have trouble to integrate this triple integral $$ \iiint_kx^2 dxdydz$$ I don't know if I have been calculating right so far, and how should I proceed? Any suggestion would be great, thanks

Answer 1

Surface is $(x-z)^2+(y-z)^2 = 1 + z^2$

$\vec F = (y,x,1+x^2z)$

To apply divergence theorem, we close the surface with discs at $z = 0$ and at $z = 1$.

a) Flux through disc at $z = 1$,

Plugging in $z = 1$ in the equation of hyperboloid, we get the equation of the disc as

$(x-1)^2+(y-1)^2 = 2$

In polar coordinates, $x = 1 + \rho \cos\theta, y = 1+\rho \sin\theta, z = 1 \ (0 \leq \rho \leq \sqrt2, 0 \leq \theta \leq 2\pi)$

Outward normal vector is $(0, 0, 1)$.

So the integral to find flux becomes

$\displaystyle \int_0^{2\pi} \int_0^{\sqrt2} [1+(1+ \rho \cos\theta)^2] \ \rho \ d\rho \ d\theta = 5 \pi$

b) Flux through the disc at $z = 0$ is simply the area of the disc with negative sign as $\vec F \cdot (0,0,-1) = -1$. So it is $-\pi$ as you correctly calculated.

c) Now to find flux through the closed surface, apply divergence theorem.

$\nabla \cdot \vec F = x^2$ as you mentioned.

Now the region for triple integral is a hyperboloid which is difficult to calculate without a Jacobian. So as the other answer suggests,

use change of variable $x = u + w, y = v + w, z = w$. Its Jacobian is $1$.

So the transformed region is $u^2 + v^2 - w^2 = 1$. In cylindrical coordinates,

$u = \rho \cos \theta, v = \rho \sin \theta, w$

$0 \leq \rho \leq \sqrt{1+w^2}, 0 \leq w \leq 1, 0 \leq \theta \leq 2\pi $

Integrand $x^2 = (u+w)^2 = (\rho \cos \theta + w)^2$

So integral to find flux,

$\displaystyle \int_0^{2\pi} \int_{0}^{1} \int_0^{\sqrt{1+w^2}} (\rho \cos \theta + w)^2 \ \rho \ d\rho \ dw \ d\theta = \pi$.

d) So flux through hyperboloid surface $ = \pi - 5 \pi - (-\pi) = - 3 \pi$

Answer 2

You can use transformation $f(x,y,z)=(x+z,y+z,z)$. The jacobian of this transform is 1. You get:$$ \iiint_{k'}(x+z)^2dxdydz $$ where $x^2+y^2-z^2=1$.

Using cylindric coordinates, you can transform it to: $$ \int_{0}^{1}\int_0^{2\pi}\int_0^{\sqrt{1+z^2}}(\rho\cos\phi+z)^2\rho d\rho d\phi dz $$ Does this help you?