# SURFACE INTEGRALS appying to z=f(x,y) x=f(y,z) and y=f(x,z)

1. Aug 20, 2008

### PhysicsHelp12

You know when a definition is given in terms of z=f(x,y) like the surface integral

and its assmed to apply to y=f(x,z)and x=f(y,z) too ...

Why is this?

I know theyre just variables ...but since x y and z mean something specifically wrt the

coordinate system

Would it be trivial to say By Symmetry, x=f(y,z) and y=f(x,z) applies too ?

which way is better to think about it?

2. Aug 20, 2008

### NoMoreExams

Why would it matter? Labeling the axis x, y and z is your own choice, you can label them in any manner you want. Yes there is a convention but the labels we give are our own choice and the reason we choose a particular convention is so others would have an idea when they try to do the problem.

3. Aug 20, 2008

### Defennder

Note that it may not always be possible to express an equation z=f(x,y) as x=f(y,z) or y=f(x,z). That has something to do with the implicit function thereom.

4. Aug 21, 2008

### HallsofIvy

In general a surface, being a two dimensional geometric object, can be written in terms of 2 parameters: $\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}$
or simply $x= f(u,v)$, $y= g(u,v)$, $z= h(u,v)$.

IF the surface is given as z= h(x,y), then you can use x and y themselves as "u" and "v": x= f(x,y)= x, y= g(x,y)= y, z= h(x,y).

IF it is possible to solve for y, say y= g(x, z), then you can use x and z as "u" and "v": x= f(x,z)= x, y= g(x,z), z= h(x,z)= z.

IF it is possible to solve for x, say x= f(y,z), then you can use y and z as "u" and "v":
x= f(x,z), y= g(y,z)= y, z= h(y,z)= z.

But it is quite possible that NONE of those can be used. For example, with the surface of the unit sphere, none of x, y, or z can be written as a function of the other two. What you can do is use spherical coordinates with $\rho$ set to 1:
$x= cos(\theta)sin(\phi)$, $y= sin(\theta)sin(\phi)$, $z= cos(\phi)$.

A good way to find the "differential of surface area" when given x= f(u,v), y= g(u,v), z= h(u,v) or, equivalently, $\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}$ is to take the derivatives: $\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}$ and $\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}$ and then take their cross product. The differential of surface area is $dS= ||\vec{r}_u\times\vec{r}_v|| dudv$.

Even simpler is the "vector differential of surface area" (calculating the flux of a vector field through a surface, say): $d\vec{S}= \vec{r}_u\times\vec{r}_v dudv$

For example, if you are given z= f(x,y) then $\vec{r}= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}$ so $\vec{r}_x= \vec{i}+ f_x\vec{j}$ and $\vec{r}_y= \vec{j}+ f_y\vec{k}$ so the cross product is $f_x\vec{i}+ f_y\vec{j}+ \vec{k}$ (or its negative, depending on orientation) and the length of that is
$\sqrt{f_x^2+ f_y^2+ 1}$ so the differential of surface area is just the "usual" formula $\sqrt{f_x^2+ f_y^2+ 1}dx dy$.

Last edited by a moderator: Aug 21, 2008