Why the following integral means the area of surface $f(x,y)=z$? $$\iint \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\:dx\:dy$$

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    $\begingroup$ It's not clear to me what you're asking. What is your definition of surface area? $\endgroup$ – Git Gud Mar 24 '13 at 15:24
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    $\begingroup$ Do you understand why $\int \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$ is the arc length of a curve $y=f(x)$? It's the same idea but extended to an extra dimension. $\endgroup$ – Sp3000 Mar 24 '13 at 15:28
  • $\begingroup$ I can derive are length of a cuvr $y=f(x)$ by using pythagorean theorem but I don't know how to apply it to surface. $\endgroup$ – Guillermo Mar 24 '13 at 15:37
  • $\begingroup$ Actually, I don't know definition of surface area... $\endgroup$ – Guillermo Mar 24 '13 at 15:38

$\frac{\partial f}{\partial x}$ - Think of this as saying "If x increases by $\partial x$, how much does f increase by?". Now above every little square patch $dx dy$, you are computing the area of a small plane. The plane can be represented by the four corners $$(x, y, f(x,y))$$ $$ (x + \partial x, y, f(x,y) + \frac{\partial f}{\partial x} \partial x)$$ $$(x, y + \partial y, f(x,y) + \frac{\partial f}{\partial y} \partial y)$$ $$ (x + \partial x, y + \partial y, f(x,y) + \frac{\partial f}{\partial x}\partial x + \frac{\partial f}{\partial y} \partial y)$$

If you compute the area of this parallelogram, you get $$\sqrt{\partial x^2 \partial y^2 + \left(\frac{\partial f}{\partial x}\right)^2 (\partial x)^2 (\partial y)^2 + \left(\frac{\partial f}{\partial y}\right)^2(\partial y)^2 (\partial x)^2}$$

Taking out the $(\partial x)^2(\partial y)^2$ gets you the result.


There is a fact that $||\vec{v}\times\vec{w}||$ is the area of the parallelogram spanned by $\vec{v}$ and $\vec{w}$. So, a good approximation of the area around a point $x$ on a surface is $||\vec{T_1}\times \vec{T_2}||$, where $\vec{T_1}$ and $\vec{T_2}$ are linearly independent tangent vectors at $x$. Your surface is given by $z=f(x,y)$. Think of this as all the points $(x,y,f(x,y))$. Two tangent vectors will then be $(1,0,f_x(x,y))$ and $(0,1,f_y(x,y))$. These are just the coordinate-wise partial derivatives of $(x,y,f(x,y))$. Therefore your good approximation to the area around any point on the surface is given by the magnitude of the cross product, which works out to be $$ \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2} $$ Now just integrate over your surface.


There's a pretty good summary here, but it uses some elementary differential geometry concepts. The truth is that surface area is a notoriously tricky concept to define precisely. In any case, the square root factor results from the definition of surface area in terms of a cross-product of tangent vectors in directions defined by a parametrization $(u,v)$.

For such a general parametrization, the surface area is defined in terms of the first fundamental form

$$\iint_S du\,dv \sqrt{E\,G-F^2}$$


$$E=\left(\frac{\partial x}{\partial u} \right)^2+\left(\frac{\partial y}{\partial u} \right)^2+\left(\frac{\partial z}{\partial u} \right)^2$$

$$F=\left(\frac{\partial x}{\partial u} \right)\left(\frac{\partial x}{\partial v} \right) + \left(\frac{\partial y}{\partial u} \right)\left(\frac{\partial y}{\partial v} \right) + \left(\frac{\partial z}{\partial u} \right)\left(\frac{\partial z}{\partial v} \right) $$

$$G=\left(\frac{\partial x}{\partial v} \right)^2+\left(\frac{\partial y}{\partial v} \right)^2+\left(\frac{\partial z}{\partial v} \right)^2$$

When $z=f(x,y)$, $x=u$, $y=v$. Then

$$E\,G-F^2 = \left[1+\left(\frac{\partial z}{\partial x} \right)^2\right] \left[1+\left(\frac{\partial z}{\partial y} \right)^2\right] - \left(\frac{\partial z}{\partial x} \right)^2\left(\frac{\partial z}{\partial y} \right)^2$$

The result follows.


The following is redundant as an answer, but since I had already typed it and it's too long for a comment, I decided to post it anyway.

Let $\vec {r}\colon \operatorname{dom}(f)\longrightarrow \Bbb R^3$, be defined by $\vec {r}(x,y)=(x,y,f(x,y))$, for all $(x,y)\in \operatorname{dom}(f)$.

Let $1_\varphi\colon \Bbb R^3\longrightarrow \Bbb R$ be constantly $1$.

If $\vec{r},f$ and $\operatorname{dom}(f)$ respect certain conditions then the area is given by

$$\displaystyle \iint \limits_{\operatorname{dom}(f)} (1_\varphi \circ \vec{r})(x,y)\left\Vert\frac{\partial \vec{r}}{\partial x}(x,y) \times \frac{\partial \vec{r}}{\partial y}(x,y) \right\Vert\mathrm dx\mathrm dy$$

Since for all $(x,y)\in \operatorname{dom}(f)$ it is true that $(1_\varphi \circ \vec{r})(x,y)=1\\$, $$\displaystyle \frac{\partial \vec{r}}{\partial x}(x,y)=\left(1,0,\frac{\partial f}{\partial x}(x,y)\right),$$ $$\displaystyle \frac{\partial \vec{r}}{\partial y}(x,y)=\left(0,1,\frac{\partial f}{\partial y}(x,y)\right),$$ $$\displaystyle \frac{\partial \vec{r}}{\partial x}(x,y) \times \frac{\partial \vec{r}}{\partial y}(x,y)=\left(\displaystyle -\frac{\partial f}{\partial x}(x,y),-\frac{\partial f}{\partial x}(x,y),1\right)$$ and $$\displaystyle \left\Vert\left(\displaystyle -\frac{\partial f}{\partial x}(x,y),-\frac{\partial f}{\partial x}(x,y),1\right)\right\Vert=\sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial y}(x,y)\right)^2,}$$ it follows that

$$\displaystyle \iint \limits_{\operatorname{dom}(f)} (1_\varphi \circ \vec{r})(x,y)\left\Vert\frac{\partial \vec{r}}{\partial x}(x,y) \times \frac{\partial \vec{r}}{\partial y}(x,y) \right\Vert\mathrm dx\mathrm dy=\\=\iint \limits_{\operatorname{dom}(f)}\sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial y}(x,y)\right)^2}\mathrm dx\mathrm dy.$$


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