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

[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?

[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.

[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.

[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.