Calculating Surface Area of a Cone with a Vertex in the XY-Plane

[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]

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]

quasar987's point is that the cone z= 2\sqrt{x^2+ y^2} 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 \frac{\sqrt{z}}{2}. What value of z gives area 5?)