Surface Area Problem

  • Thread starter JaysFan31
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JaysFan31

Find the surface area of the cone z=2sqrt(x^2+y^2) and above a region in the xy-plane with area 5.

If anyone could help me with this problem, I would really appreciate it.
Thanks.

Mike
 

quasar987

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Do you mean to find the area of the cone of equation z=2sqrt(x^2+y^2) whose base has area 5 ?
 

JaysFan31

No. Not sure what it means. Just one of my Calculus III Multiple Integration homework problems word for word and I have no idea.

Mike
 

HallsofIvy

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quasar987's point is that the cone [itex]z= 2\sqrt{x^2+ y^2}[/itex] has its vertex in the xy-plane, not a base. Perhaps you mean, as he suggested, the slant area of that cone up to the point the base would have an area of 5. (Obviously, the radius of the base of the cone is [itex]\frac{\sqrt{z}}{2}[/itex]. What value of z gives area 5?)
 

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