# Confirming my knowledge on surface integrals

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• Isaac0427
In summary: The only thing I’m unsure about is the orientation; as I see it, the surface should only have two possible orientations, so I don’t think all those plus/minuses can be independent.The orientation of the surface is determined by the direction of the gradient relative to the orientation of the surface.

#### Isaac0427

Hi,

I want to make sure my understanding of calculating surface integrals of vector fields is accurate. It was never presented this way in a textbook, but I put this together from pieces of knowledge. To my understanding, surface integrals can be calculated in four different ways (depending on how the surface is given). For this, I'll integrate the vector field ##\vec{A}(\vec{r})## over a surface S.

1. If one has a parametric surface ##\vec{r}(u,v)##, a surface integral would be calculated as $$\pm \iint_S \vec{A}\left(\vec{r}(u,v)\right)\cdot(\vec{r}_u\times \vec{r}_v)dudv$$where the subscripts represent partial derivatives, and the sign is determined by the direction of the cross product relative to the orientation of the surface.
2. If one has a surface in the form ##F(x,y,z)=0##, a surface integral would be calculated as $$\pm \iint_S \frac{\vec{A}\cdot\nabla F}{\left|F_z\right|}dxdy$$where the subscript represents a partial derivative, and the three variables can be interchanged. One must also substitute into get the integrand in terms of only x and y (or whatever the variables of integration are). The sign is determined by the direction of the gradient relative to the orientation of the surface.
3. If one has a surface in the form ##z=f(x,y)## (or alternatively ##y=f(x,z)## or ##x=f(y,z)## with the appropriate adjustments made), the surface integral would be calculated as $$\pm \iint_S \vec{A}\left(x,y,f(x,y) \right)\cdot(-f_x\hat{i}-f_y\hat{j}+\hat{k})dxdy$$where the subscripts represent partial derivatives. The sign is plus if the surface is oriented upwards, and minus if it is oriented downwards.
4. For this last one, one needs a surface that can be expressed relatively easily as ##x=f(y,z)##, ##y=g(x,z)## AND ##z=h(x,y)##. The vector field will be expressed as ##\vec{A}=A_1 \hat{i}+A_2 \hat{j}+A_3 \hat{k}##. The surface integral can be computed as $$\iint_S \left[ \pm A_1\left(f(y,z),y,z \right)dydz \pm A_2\left(x,g(x,z),z \right)dxdz \pm A_3\left(x,y,h(x,y) \right)dxdy\right]$$where the three signs are respectively determined by the orientation of the surface in the x, y, and z directions (plus for oriented in the positive direction, minus for oriented in the negative direction).

I'm not 100% sure about the last one, but I think I have seen something like it before. It also appears as if it would be useful for the divergence theorem. Are all four of these valid ways to compute surface integrals?

Hiero
A couple things to note.
(3) is just (2) with F(x,y,z) = z-f(x,y), so not that interesting in some sense.

For (1) and (2), all you're doing is finding a vector orthogonal to the surface. ##\nabla F## is orthogonal to level surfaces of ##F##, and ##r_u \times r_v## is orthogonal to both ##r_u## and ##r_v##, which are tangent to the plane. The fact that you picked vectors that are scaled correctly requires a bit of thinking (the magnitude is basically a Jacobian determinant for transforming the coordinates)I think (4) looks like it might be right to me, but I haven't seen it before. I feel like you can probably prove it from (3) somehow, I'll think about it.

Hiero
@Isaac0427 Thanks for the food for thought. I had actually never seen (2) before. I had to contemplate over (4) for a while, but I finally realized it can be easily derived from (2) as follows:

According to (2) we have,
$$\pm d\vec S = \frac{\nabla F}{|\partial _z F|}dxdy= \frac{\nabla F}{|\partial _yF|}dxdz = \frac{\nabla F}{|\partial _x F|}dydz$$
But, by your hypothesis, we can define three such functions which implicitly define the surface, namely,
##F_1=x-f(y,z)##
##F_2=y-g(x,z)##
##F_3=z-h(x,y)##
Therefore, (by choosing to divide each by the component which is one) we can say,
$$\pm d\vec S = \nabla F_1 dydz = \nabla F_2 dxdz = \nabla F_3 dxdy$$
Now we compute the integrand by using each of these equivalent forms for one part of the dot product (for dS_i we use the form for which the i’th component is one) and the result follows,
$$\vec A \cdot d\vec S = A_i dS_i =\pm A_xdydz\pm A_ydxdz\pm A_zdxdy$$
(I've suppressed the arguments of A.)
The only thing I’m unsure about is the orientation; as I see it, the surface should only have two possible orientations, so I don’t think all those plus/minuses can be independent.

Anyway those are my thoughts on version (4).

By the way, I would call these integrals “flux integrals,” because in my mind, a “surface integral” would be integrating the vector by the absolute value of the surface area (resulting in a vector). I’m not sure if that terminology is standard though, that’s just how I distinguish the two ideas.
Office_Shredder said:
The fact that you picked vectors that are scaled correctly requires a bit of thinking (the magnitude is basically a Jacobian determinant for transforming the coordinates)
I had actually never seen (2) before, though it’s quite nice. The way I proved it was from (1) (my favorite version) by choosing ##\vec r = xe_x+ye_y+z(x,y)e_z## from which we get,
$$\pm d\vec S = (-\partial_x z e_x-\partial_y z e_y + e_z)dxdy$$
Then we use the fact that on the surface dF=0 to say,
$$dF=\partial_xFdx+\partial_yFdy+\partial_zFdz =0 \implies \partial_x z = -\frac{\partial_x F}{\partial_zF}$$
And likewise for ##\partial_y z##. From these we can observe that ##\pm d\vec S = \frac{\nabla F}{|\partial _z F|}dxdy##

If you can show that the magnitude is correct by using a Jacobian I would be interested to see, as I’m not sure what basis we would be transforming to/from.

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