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Surface Area (help me to prove something:)
I was studying a bit about multiple integrals and found this theorem:
If we have function z=f(x,y) which is defined over the region R, surface S over the region is
[tex]S=\iint_R\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\,dA[/tex]
I wanted to prove this, because I doesn't seem to me to be trivial and I had to use nasty gradients and nontrivial things. I wonder if there is some easier proof out there?
You know, if we want to count arc length of curve given by function y=f(x), integral looks similar
[tex]L=\int_R\sqrt{1+\left(\frac{df}{dx}\right)^2}\,dx[/tex]
but in this case it is obvious...
I was studying a bit about multiple integrals and found this theorem:
If we have function z=f(x,y) which is defined over the region R, surface S over the region is
[tex]S=\iint_R\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\,dA[/tex]
I wanted to prove this, because I doesn't seem to me to be trivial and I had to use nasty gradients and nontrivial things. I wonder if there is some easier proof out there?
You know, if we want to count arc length of curve given by function y=f(x), integral looks similar
[tex]L=\int_R\sqrt{1+\left(\frac{df}{dx}\right)^2}\,dx[/tex]
but in this case it is obvious...
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