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 one year ago
Evaluate the surface integral:
\[\iint _{ S }^{ }{ z } dS\]
Where S is the surface \(x=y+3z^2\) for \(0\le y\le 1\) and \(0\le z\le 1\)
 one year ago
Evaluate the surface integral: \[\iint _{ S }^{ }{ z } dS\] Where S is the surface \(x=y+3z^2\) for \(0\le y\le 1\) and \(0\le z\le 1\)

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PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0I'm stuck. I think I have to use the formula \[\iint _{ S }^{ }{ f(x,y,z) } dS=\iint _{ D }^{ }{ f(x,y,g(x,y)) \sqrt { { (\frac { \delta z }{ \delta x } ) }^{ 2 }+{ (\frac { \delta z }{ \delta y } ) }^{ 2 }+1 } } dA\] But I keep getting lost. Can anyone help me out?

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0Not sure if I should be able to project it onto the yzplane and change the formula for x=g(y,z), etc. Or if I'm supposed to rearrange \(x=y+3z^2\) for z, and continue with the normal way.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2\[\int_{ S }f(x,y,z) dS=\iint_{ D } f(g(y,z),y,z) \sqrt {\left (\frac { \partial x }{ \partial y} \right)^{ 2 }+\left(\frac { \partial x }{ \partial z } \right)^{ 2}+1}~dA\]

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0That's what I thought, but it didn't seem right. Oh well, I'll just go with it. Thanks @Zarkon

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2you can always switch the roles of x and z and then use your original formula (same thing)
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