Double Surface Integrals in Polar Coordinates

[allllllll]

Homework Statement

Find the surface area of the cone z=3x^2+y^2and above a region in the xy-plane with area 4.

Homework Equations

double integral sqrt( (dz/dx)^2 + (dz/dy)^2 +1)

The Attempt at a Solution

I was able to simplify the equation, I just don't know what to do with the area part of the question:

dbl int sqrt ( (3x/sqrt(x^2+y^2))^2 + (3y/sqrt(x^2+y^2))^2 + 1)
dbl int sqrt (9 (x^2+y^2)/(x^2+y^2) + 1 )
dbl int sqrt (10)

change into polar coordinates

integral [0-1] dtheta integral [0-2pi] sqrt(10) r dr
= 1*2pi sqrt(10)

what am I doing wrong?

 

[Dick]

That's not a cone It's a paraboloid. Did you actually mean to say z^2=3*(x^2+y^2)? That's a cone. I'm just guessing that from your work. But if you do that you don't get sqrt(10) for the integrand of your double integral. And you can't integrate over the xy domain using polar coordinates. You don't even know what the xy domain IS! All you know is it has area 4. Is that enough?

 

[allllllll]

yeah sorry should have said z=3sqrt(x^2+y^2)..

 

[allllllll]

answer is sqrt(10)*4.. don't know why

 

[Dick]

You have sqrt(10) for the integrand of the surface integral. It's a constant. If you integrate a constant over a domain of area 4, what do you get? I can repeat this if it's too simple. I really don't mind. I can cut and paste.

 

[HallsofIvy]

yeah sorry should have said z=3sqrt(x^2+y^2)..

So $\sqrt{x^2+ y^2}= z/3$

and so (3x/sqrt(x^2+y^2))^2 + (3y/sqrt(x^2+y^2))^2 + 1)= (9x+ 9y+ z)/z.

 

[Dick]

So $\sqrt{x^2+ y^2}= z/3$

and so (3x/sqrt(x^2+y^2))^2 + (3y/sqrt(x^2+y^2))^2 + 1)= (9x+ 9y+ z)/z.

Isn't that 9x^2/(x^2+y^2)+9y^2/(x^2+y^2)+1=9*(x^2+y^2)/(x^2+y^2)+1=9+1=10?

 

[HallsofIvy]

Yes, It is. We clearly have solved this problem! I hope allllllll has too!

 

[Dick]

alllllllllllllllll had the sqrt(10) from the beginning. Just couldn't figure out how to handle the domain. Yes, let's hope it's got.