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Flux of $\boldsymbol{\vec{F}=-\hat{i}+2\hat{j}+3\hat{k}}$ across the surface $S : \boldsymbol{z=0,0\leq x\leq 2,0\leq y\leq 3}$ in direction of $\boldsymbol{\hat{k}}$ is equal to?

So we have to calculate $\int F.\hat{n} dS$,but i dont know how to convert $dS$ to $dxdy$. What is the general method? Any help appreciated.

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    I think only the last term survives. – 2017-02-11
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    @Nameless yes, but how to convet dS into dxdy – 2017-02-11
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    I mean isn't it just $\int \int F \cdot kdS = \int \int 3 dx dy$? You are integrating over the rectangle, so the surface area is $3(2)(3) = 18$ – 2017-02-11
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    In your space R^3 you define a frame of reference, here (i,j,k) on which you choose a coordinate system (cartesian, polar, spheric ....), here cartesian coordinates. In these coordinates you have to express the infinitesimal element of your surface S, dS = dxdy. http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf – 2017-02-11

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General method:

  1. Parametrize the surface as $$ \vec x(s,t) = (x_1(s,t),x_2(s,t),x_3(s,t)).$$
  2. The surface element is given by $$ \left| \frac{\partial \vec x}{\partial s} \times \frac{\partial \vec x}{\partial t}\right|dsdt$$ and you integrate over the parameters $(s,t).$ The $\times$ is a cross product and $|\cdot|$ is the length of the vector.

Your question:

Your surface is just a rectangle in the $x-y$ plane. $dS = dxdy.$

Connection between them:

Let $\vec x(s,t) = (s,t,0)$ for $0

Plugging in the above formula you'll find the surface element is $dsdt$. Then just call $s$ and $t$ by their more natural names $x$ and $y.$

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    what if the surface was a sphere $x^2+y^2+z^2=a^2$, in the first quadrant, the we would have three parameters $r, \theta , \phi$,, what will be the formula now? – 2017-02-11
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    Well you need to parametrize the sphere. The most natural parametrization just uses spherical coordinates: $\vec x(\theta,\phi) = (a\sin(\theta)\cos(\phi),a\sin(\theta)\sin(\phi),a\cos(\theta))$ where $0<\theta<\pi$ and $0<\phi< 2\pi$ ($\phi$ is the azimuth here). Then you can apply the formula above. After doing the tedious cross product you should find $dS = a^2\sin(\theta)d\theta d\phi.$ (which you should be able to also infer from drawing pictures and doing a little geometry) – 2017-02-11