# parametrization of surface element in surface integrals

I don't understand this

How $dS = \sqrt{ \left ( \partial g \over \partial x\right )^2 + \left ( \partial g \over \partial y\right )^2 + 1 } \; dA \; \;$ ?? Is $dA = dx\times dy$??

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The surface in question is given by $z=g(x,y)$.

The vector in the surface that comes as a small change, $\mathrm{d}x$, to $x$ is $$\left(1,0,\frac{\partial g}{\partial x}\right)\mathrm{d}x\tag{1}$$ The vector in the surface that comes as a small change, $\mathrm{d}y$, to $y$ is $$\left(0,1,\frac{\partial g}{\partial y}\right)\mathrm{d}y\tag{2}$$ If we take the cross product of $(1)$ and $(2)$, we get $$\left(1,0,\frac{\partial g}{\partial x}\right)\times\left(0,1,\frac{\partial g}{\partial y}\right)\,\mathrm{d}x\,\mathrm{d}y = \left(-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1\right)\,\mathrm{d}x\,\mathrm{d}y\tag{3}$$ The area represented is the absolute value of $(3)$: $$\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}\quad\mathrm{d}x\,\mathrm{d}y\tag{4}$$

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@experimentX: a Jacobian is used for a change of variables. This is not a change of variables, so normally, a Jacobian would not be used. –  robjohn Jun 19 '12 at 22:36
@experimentX: however, if one were very adventuresome, one could look at a surface area as the volume of a very thin layer of uniform thickness, divided by that thickness. One would get the change of variables $$(x,y,z)\mapsto(x,y,g(x,y))+\frac{\left(-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1\right)}{\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}}z$$ where the coefficient of $z$ is the unit surface normal. The Jacobian of this change of variables would give you the same element of surface area. –  robjohn Jun 19 '12 at 22:37

The surface integral is given by: $$\int_{S} f \,dS = \iint_{D} f(\mathbf{\sigma}(u, v)) \left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\| \,du\,dv$$

Where $\mathbf{\sigma}(u, v)$ is a parametrization of the surface $S$.

Now, if $S$ is given by the function $z = g(x, y)$, we have: $$\sigma(u, v) = \sigma(x, y) = (x, y, g(x, y))$$

Therefore: $$\left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\| = \left\|\left(1, 0, {\partial g \over \partial x}\right)\times \left(0, 1, {\partial g \over \partial y}\right)\right\|$$

Calculate the cross product on the RHS to find that:

$$\left(1, 0, {\partial g \over \partial x}\right)\times \left(0, 1, {\partial g \over \partial y}\right) = \left(-{\partial g \over \partial x}, -{\partial g \over \partial y}, 1\right)$$

And its norm is equal to: $$\sqrt{\left({\partial g \over \partial x}\right)^2+\left({\partial g \over \partial y}\right)^2+1}$$

Therefore, the surface integral for $S$ given by $z = f(x, y)$ is: $$\iint_{D} f(x, y, g(x, y)) \sqrt{\left({\partial g \over \partial x}\right)^2+\left({\partial g \over \partial y}\right)^2+1} \,dx\,dy$$

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thank you both for answers :) –  Santosh Linkha Jun 19 '12 at 22:09
@experimentX - Happy to help! –  Ayman Hourieh Jun 19 '12 at 22:13

For a surface integral (or line integral, or path integral, or change of variables...), the computation breaks into four steps. Notation: let $f(x,y,z)$ denote the integrand and let $S$ be the surface parametrized by $(x(u,v),y(u,v),z(u,v))$ so that $u$ and $v$ are the parameters.

1. Convert integrand to $u$'s and $v$'s by plugging in the parametrizations for $(x,y,z)$,
2. Convert differential by $dS = ||T_u \times T_v||dudv$, where $$T_u = \left( \frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u} \right),$$ $$T_v = \left( \frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v} \right),$$
3. Convert integral bounds so that they are in terms of $u$ and $v$
4. Put steps 1-3 together and compute!

As applies to your situation, your surface is defined by the portion of the graph of $z = g(x,y)$ which lies over the region $D$ in the $xy$-plane. Thus, we can parametrize the surface using the $x$ and $y$ as the parameters. Pedantically, we have $x=u$, $y=v$, and $z=g(u,v)$.

Going through the steps now: First, we have the integrand $f(x,y,z)$. Plugging in $x,y$ and $z$ gives us $f(u,v,g(u,v))$. But that's just $f(x,y,g(x,y))$.

Step 2: Calculate $T_u$ and $T_v$: $$T_u = \left( 1,0,\frac{\partial g}{\partial u} \right)$$ $$T_v = \left( 0,1,\frac{\partial g}{\partial v} \right)$$ $$T_u \times T_v = \left( -\frac{\partial g}{\partial u}, -\frac{\partial g}{\partial v}, 1 \right)$$ $$||T_u \times T_v|| = \sqrt{ \left( \frac{\partial g}{\partial u} \right)^2+\left( \frac{\partial g}{\partial v} \right)^2+1}$$ $$dS = ||T_u \times T_v||dudv = \sqrt{ \left( \frac{\partial g}{\partial u} \right)^2+\left( \frac{\partial g}{\partial v} \right)^2+1}dudv$$ Now, $x=u$ and $y=v$, so we may as well write $$dS = \sqrt{ \left( \frac{\partial g}{\partial x} \right)^2+\left( \frac{\partial g}{\partial y} \right)^2+1}dxdy$$ And yes, $dA=dxdy$.

Step 3: We need ranges on $u$ and $v$, which for this problem means ranges on $x$ and $y$. we know that $(x,y)$ must lie in $D$.

That's how all the pieces come together. It's a standard surface integral with the wrinkle that we get to use $x$ and $y$ as the parameters.

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thank you too :) –  Santosh Linkha Jun 19 '12 at 22:24

It is simple

$$dS = \sqrt{(dxdy)^2+(dydz)^2+(dxdz)^2} = \sqrt{1+\left(\dfrac{dz}{dx}\right)^2+\left(\dfrac{dz}{dy}\right)^2}\hspace{1mm}‌​dxdy$$

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