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

  1. Aug 20, 2008 #1
    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. jcsd
  3. Aug 20, 2008 #2
    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.
     
  4. Aug 20, 2008 #3

    Defennder

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    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.
     
  5. Aug 21, 2008 #4

    HallsofIvy

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

    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 [itex]\rho[/itex] set to 1:
    [itex]x= cos(\theta)sin(\phi)[/itex], [itex]y= sin(\theta)sin(\phi)[/itex], [itex]z= cos(\phi)[/itex].

    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, [itex]\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex] is to take the derivatives: [itex]\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}[/itex] and [itex]\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}[/itex] and then take their cross product. The differential of surface area is [itex]dS= ||\vec{r}_u\times\vec{r}_v|| dudv[/itex].

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

    For example, if you are given z= f(x,y) then [itex]\vec{r}= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}[/itex] so [itex]\vec{r}_x= \vec{i}+ f_x\vec{j}[/itex] and [itex]\vec{r}_y= \vec{j}+ f_y\vec{k}[/itex] so the cross product is [itex]f_x\vec{i}+ f_y\vec{j}+ \vec{k}[/itex] (or its negative, depending on orientation) and the length of that is
    [itex]\sqrt{f_x^2+ f_y^2+ 1}[/itex] so the differential of surface area is just the "usual" formula [itex]\sqrt{f_x^2+ f_y^2+ 1}dx dy[/itex].
     
    Last edited: Aug 21, 2008
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