- #1
Damascus Road
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Greetings,
I'm trying to find the surface area of the part of the sphere [tex]x^{2}+y^{2} + z^{2}=1[/tex] above the cone [tex]z=\sqrt{x^{2}+y^{2}}[/tex].
I know, that a surface area of a surface r(u,v) = x(u,v) + y(u,v) + z(u,v) can be given by,
A(S) = [tex]\int\int | r_{u} \times r_{v} | dA [/tex]
A function, z=f(x,y) can utilize
A(S) = [tex]\int \int \sqrt{1 + (\frac{d z}{d x})^{2}+(\frac{dz}{dy})^{2}}[/tex] dA
It seems part of my surface requires the first equation and the second part requires the second equation. (One is an equation z=f(xy) and the other is x y z = 1, how do I approach this?
Thanks in advance!
I'm trying to find the surface area of the part of the sphere [tex]x^{2}+y^{2} + z^{2}=1[/tex] above the cone [tex]z=\sqrt{x^{2}+y^{2}}[/tex].
I know, that a surface area of a surface r(u,v) = x(u,v) + y(u,v) + z(u,v) can be given by,
A(S) = [tex]\int\int | r_{u} \times r_{v} | dA [/tex]
A function, z=f(x,y) can utilize
A(S) = [tex]\int \int \sqrt{1 + (\frac{d z}{d x})^{2}+(\frac{dz}{dy})^{2}}[/tex] dA
It seems part of my surface requires the first equation and the second part requires the second equation. (One is an equation z=f(xy) and the other is x y z = 1, how do I approach this?
Thanks in advance!
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