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Calculus area element help

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Let f(x,y,z)=x^2yz. Express f*dA in terms of u and v.

    3. The attempt at a solution
    We're given that
    x= u+v
    y= u-v
    z= u^2+v^2.

    The main problem I'm having is that I know I need to take partial derivatives in respect to u and v but I'm not sure how. I did an easier problem that required me to find the area element and I had to take the partial of x in respect to u, then the partial of y in respect to u. In that problem the x, y, z were separated but in this case, x y and z are all together so I'm not sure how to separate them and take the partials to find are element. Thanks :smile:
  2. jcsd
  3. Apr 16, 2007 #2
    chain rule
  4. Apr 16, 2007 #3
    so do i just substitute all of those equations into x^2yz and take the derivatives once in respect to u and then once in respect to v?
  5. Apr 17, 2007 #4


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    dA is given by the "fundamental vector product": First write the surface as a vector function of the two parameters:
    [tex]\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= (u+v)\vec{i}+ (u-v)\vec{j}+ (u^2+ v^2)\vec{k}[/tex]
    Now take the derivatives with respect to each parameter:
    [tex]\vec{r}_u= \vec{i}+ \vec{j}+ 2u\vec{k}[/tex]
    [tex]\vec{r}_v= \vec{i}- \vec{j}+ 2v\vec{k}[/tex]

    The "fundamental vector product" is the cross product of those two vectors. Since they both are tangent to the circus (in the direction of the "coordinate axes for coordinates u and v) their cross product is perpendicular to the surface and its length is the "differential of surface area"
    If I have done the calculation correctly (something I never guarentee!) then
    [tex]dS= 2\sqrt{2u^2+ 2v^2+ 1}dudv[/tex]
    and f(u,v)dA is
    [tex]2(u+v)^2(u-v)(u^2+ v^2)\sqrt{2u^2+ 2v^2+ 1}dudv[/tex]
  6. Apr 17, 2007 #5
    hm.....interesting, I'll definitely be doing more problems for practice. Thanks
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