# Area of a surface using integration. Confusion with aspect of formal definition.

My textbook explains that, when finding the area of a surface using integration, we approximate each surface element by $$\left| \Delta u \dfrac{\partial \overrightarrow{r}}{\partial u} \times \Delta v \dfrac{\partial \overrightarrow{r}}{\partial v}\right|.$$

$\partial \overrightarrow{r}$ is defined to be the tangent vector to $C_1: v = v_0$ and $C_2: u = u_0$ at $P_0$.

I don't understand why we are multiplying $\dfrac{\partial \overrightarrow{r}}{\partial u}$ and $\dfrac{\partial \overrightarrow{r}}{\partial v}$ by $\Delta u$ and $\Delta v$. We know that $\partial \overrightarrow{r}$ is tangent to $C_1$ and $C_2$, and we know that $\Delta v$ and $\Delta u$ are scalars. Multiplying a vector (in this case, $\partial \overrightarrow{r}$) by a scalar scales that vector by a factor of the scalar/constant. But what purpose does this serve with $\Delta u$ and $\Delta v$? I don't understand how this is supposed to change the tangent vectors $\dfrac{\partial \overrightarrow{r}}{\partial u}$ and $\dfrac{\partial \overrightarrow{r}}{\partial v}$? How is this multiplication/scaling changing these tangent vectors from what they originally were (from $\dfrac{\partial \overrightarrow{r}}{\partial u}$ and $\dfrac{\partial \overrightarrow{r}}{\partial v}$)? Why is it necessary to define it in this way?

I would greatly appreciate it if people could please take the time to explain this.

• $\partial \overrightarrow{r}$ is just change. it doesnt give change with respect to what. Take one dimensioanl analogy of measuring length, one would take dS/dx, as the rate of change of length of curve and then integrate it with respect to dx. – jnyan Apr 17 '17 at 4:57

Suppose the surface you're interested in integrating is called $\bf\Sigma$.

You're working with "some vector $\vec r$", but you need to ask yourself - what is $\vec r$? From your working and notation, you already know that it depends on $u, v$. More formally, $\vec r$ is a function mapping from some regular coordinate space (e.g. cartesian space) into $\bf\Sigma$. Given any two points $u_0, v_0$ in your "regular surface", $\vec r(u_0, v_0)$ is a point in $\bf\Sigma$.

Using this we can figure out what the surface element will be. We want the surface element to be given by $\Delta\bf{\Sigma}$ = $\Delta a\times \Delta b$ where $\Delta a$ and $\Delta b$ are the sides of the surface element. You have no idea how $a$ and $b$ behave, but you do know exactly what $u$ and $v$ are. (Generally they will be cartesian or polar.) Hence you want to define $a$ and $b$ in terms of $u$ and $v$.

Suppose you set $u_0$ and $v_0$, and consider some small change $\Delta u$. Then $\vec r(u_0, v_0)$ and $\vec r(u_0+\Delta u, v_0)$ will be two points on your surface $\bf\Sigma$ "which are close to each other". We can denote the distance between the two points as $\Delta a$. Hence $\Delta a = \vec r(u_0+\Delta u, v_0) - \vec r(u_0, v_0)$. By the definition of a derivative, you also know that $\dfrac{\partial \vec r}{\partial u} = \lim_{\Delta u \rightarrow 0}\dfrac{\vec r(u_0+\Delta u, v_0) - \vec r(u_0, v_0)}{\Delta u}$.

By combining these, you get $\dfrac{\partial \vec r}{\partial u} = \lim_{\Delta u \rightarrow 0}\dfrac{\Delta a}{\Delta u}$. Hence $\Delta a = \Delta u \dfrac{\partial \vec r}{\partial u}$ "for small $\Delta u$".

In exactly the same way, you can show that $\Delta b = \Delta v \dfrac{\partial \vec r}{\partial v}$.

Hence $\Delta\bf{\Sigma}$ = $\Delta a\times \Delta b = \Delta u \dfrac{\partial \vec r}{\partial u}\times\Delta v \dfrac{\partial \vec r}{\partial v}$ is the surface element.

• Good response! I especially like the way you tied everything together by using the same notation and concepts that were in my question. I think you made a single typo: "Hence you want to define $a$ and $b$ in terms of $u$ and $b$." You mean $u$ and $v$? Thank you for the assistance. – The Pointer Apr 17 '17 at 9:46
• yes that's definitely a typo! I'll fix it now. I actually started off by trying to explain it using my own intuition but then I was like "wait I should try to explain it in accordance with your diagrams" – Harambe Apr 17 '17 at 9:51

The intuition is that you are locally approximating the surface $S: = \vec{r}(u, v)$ by its tangent plane. When we define the surface element, we want its magnitude to express the factor by which small areas in the domain of $\vec{r}$ are scaled in the range of $\vec{r}$. To do this we locally approximate $\vec{r}$ using its linear approximation. This approximation accurately captures how $\vec{r}$ changes small areas.

Let $p = \vec{r}(u_0, v_0)$ be a point on the surface. We can compute the velocity vectors of $S$ at the point $p$, which are given by $\vec{r_u} = \frac{\partial \vec{r}}{\partial u}$ and $\vec{r_v} = \frac{\partial \vec{r}}{\partial v}$.

If $\Delta u$, $\Delta v$ are small, $\vec{r}(u_0 + \Delta u, v_0 + \Delta v)$ is approximately $p + \Delta u \vec{r_u} + \Delta v \vec{r_v}$. A rectangle in the domain with vertices $(u_0, v_0), (u_0 + \Delta u, v_0)$, $(u_0, v_0 + \Delta v)$, $(u_0 + \Delta u, v_0 + \Delta v)$ approximately gets mapped to a parallelogram with vertices $p$, $p + \Delta u \vec{r_u}$, $p+ \Delta v \vec{r_v}$, $p + \Delta u \vec{r_u} + \Delta v \vec{r_v}$. The rectangle has area $|\Delta u \Delta v|$, and the parallelogram has area $|\vec{r_u} \times \vec{r_v}||\Delta u \Delta v|$. That explains why the formula looks the way it does.