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Find the flux through the surface limited by the plane $y+z=2$, the cylinder $x^2+y^2=1$ and the plane $z=0$, for the vector field $\vec{F}(x,y,z)=(4x,-3y,2z)$.
I'm trying to find the flux through this surface, it's a truncated cylinder, so I tried to find the flux trought three diferent parts of the surface, the top $S_1$, the cylinder $S_2$, and the bottom $S_3$, and for each one there is a diferent normal vector ¿true? then applied the formula $flux=\displaystyle{\int\int_S \vec{F}\cdot \vec{n}d\sigma}$.

The normal vector for $S_1: \vec{n_1}=(0,1,1), \quad$ for $S_2: \vec{n_2}=(2x,2y,0),\quad $ and for $S_3: \vec{n_3}=(0,0,-1) $

Flux of $S_1$: $\displaystyle{\sqrt{2}\int_0^{2\pi}\int_0^1(-5r^2sin(\theta)+4r)drd\theta=4\sqrt{2}\pi }$

Flux of $S_2$: $\displaystyle{\int_0^{2\pi}\int_0^{2-sin(\theta)}(8cos^2(\theta)-6sin^2(\theta))dzd\theta =4\pi }$

And flux of $S_3$ is zero because $z=0$. So the total flux is the sum of each flux, $(4+4\sqrt{2})\pi$.

Is all this correct ? I'm asking about this exercise because I want verify the divergence theorem

$\displaystyle{\int\int_S=\vec{F}\cdot\vec{n}d\sigma=\int\int\int_D \nabla\cdot\vec{F}dV}$.

I calculated the right side and I got: $\displaystyle{\int\int\int_D \nabla\cdot\vec{F}dV=3\int_0^{2\pi}\int_0^1\int_0^{2-rsin(\theta)}rdzdrd\theta=6\pi}$

It's not the same for the left side, Please could you explain me what is wrong?

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    $\begingroup$ Your normalizations for the normal vectors are incorrect. You found $n dS$ (sort of, the normalizations are still all over the place) but then you never calculated the Jacobian for $dS$ $\endgroup$ Mar 12, 2021 at 2:39

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a) Flux through $S_1$ (top disk in plane $y+z=2$):

Mistake: you multiplied by $\sqrt2$. Dot product with normal vector $(0, 1, 1)$ takes care of the factor for surface area element. If you are multiplying by $\sqrt2$, then you also need to make sure you have dot product with unit normal vector which is $\hat{n}=\frac{1}{\sqrt2}(0, 1, 1)$.

Parametrization of the surface: $(r \cos t, r \sin t, 2 - r \sin t), 0 \leq r \leq 1, 0 \leq t \leq 2\pi$ and $\vec{n} = (0, 1, 1)$.

$I_{S_1} = \displaystyle \int_0^{2\pi} \int_0^1 (4 r \cos t, -3r \sin t, 4 - 2 r \sin t) \cdot (0, 1, 1) \ r \ dr \ dt$

b) Flux through $S_2$ (cylindrical surface $x^2+y^2 = 1$):

Mistake: you hade a mistake in your normal vector and that is doubling the flux.

Parametrization of the surface: $s(z, t)=(\cos t, \sin t, z), 0 \leq t \leq 2\pi, 0 \leq z \leq 2 - \sin t$ and $\vec{n} = (x, y, 0) = (\cos t, \sin t, 0)$.

$I_{S_2} = \displaystyle \int_0^{2\pi} \int_0^{2-\sin t} (4 \cos t, -3 \sin t, 2z) \cdot (\cos t, \sin t, 0) \ dz \ dt$

[For outward normal vector to the cylinder it is easy to see that from axis of the cylinder $(0, 0, z)$ to a point on the cylinder orthogonal to the surface $(x, y, z)$, the vector is $(x, y, 0)$. Otherwise you can take partial derivatives of $s(z, t)$ and do the cross product $s'_t \times s'_z$ and that would give you the same.]

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    $\begingroup$ Thank you so much for your answer. I want to explain why I multiplied by $\sqrt{2}$ in the flux of $S_1$, I used the next theorem: $\displaystyle{\int\int_S} g(x,y,z)d\sigma=\int\int_D g(x,y,f(x,y))\sqrt{(f_x)^2+(f_y)^2+1}dA $ where $S$ is a surface of the form $z=f(x,y), D$ is the proyection of $S $ over the plane $xy$ and $f_x, f_y$ are the partial derivates of $f(x,y)$ Is this theorem used incorrectly ? $\endgroup$
    – ifthenelse
    Mar 12, 2021 at 16:44
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    $\begingroup$ That theorem is correct for surface area element but we use that when $g(x,y,z)$ is a scalar function. Here we have a vector field we do dot product with $\vec n = (0, 1, 1)$ instead of multiplying by $|\vec n| = \sqrt2$ $\endgroup$
    – Math Lover
    Mar 12, 2021 at 16:49
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    $\begingroup$ As I wrote in my answer if you are multiplying by $\sqrt 2$, you can but then we need to make sure we are doing dot product with unit normal vector (unit magnitude) otherwise we are effectively multiplying by $\sqrt2$ and again multiplying by $(0, 1, 1)$ which has magnitude of $\sqrt2$. $\endgroup$
    – Math Lover
    Mar 12, 2021 at 16:52
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    $\begingroup$ I got it, Thanks. $\endgroup$
    – ifthenelse
    Mar 12, 2021 at 16:52

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