Surface integral

  • Thread starter bosox284
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  • #1
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Homework Statement
Evaluate the surface integral, the double integral of zdS if the region is the patch of surface defined by x^2 + y^2 - z^2 = -1 in the first octant with z less than or equal to 4.


The attempt at a solution
I really don't know where to begin. I believe the equation is of a hyperboloid of two sheets. So I want to say I have to parametrize the equation with x = bxcos(u)sinh(v), y = bysin(u)sinh(v), z = bzcosh(v), and then take the partial derivatives of u and v, and take the cross product of the two. Could someone please point me in the right direction? Or instruct me where to go from there?
 

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  • #2
gabbagabbahey
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The attempt at a solution
I really don't know where to begin. I believe the equation is of a hyperboloid of two sheets. So I want to say I have to parametrize the equation with x = bxcos(u)sinh(v), y = bysin(u)sinh(v), z = bzcosh(v), and then take the partial derivatives of u and v, and take the cross product of the two. Could someone please point me in the right direction? Or instruct me where to go from there?
That sounds like a good plan, although I see no need for the [itex]b[/itex]'s...why not show us how far you get with that method?
 
  • #3
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Well the b's would be for if it were like (x^2)/4, where b would be 2. So in this case, the value of b is one and I could have gone without them. The only problem I see myself running into from here, would probably be finding out the values of u and v for when I set up the integral.
 
  • #4
gabbagabbahey
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That shouldn't be too difficult...[itex]x[/itex],[itex]y[/itex], and [itex]z[/itex] all have to be positive in the first octant and [itex]z\leq 4[/itex]....find the range on [itex]v[/itex] first using your limits on [itex]z[/itex] and then see what range of [itex]u[/itex]-values [itex]x[/itex] and [itex]y[/itex] are positive for.
 

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