# Thread: Surface Area Calc 3

1. ## Surface Area Calc 3

My problem says find the surface area of the part of the surface
z=y^2-x^2 that is inside the cylinder x^2+y^2=2

So im thinking of polar coordinates
saying f(x,y)=y^2-x^2
and fx=-2x, fy=2y
and the area=
double integral of (sqrt(1+4x^2+4y^2))

but my problem is how to go about finding the limits
Im not sure exactly, can i get direction so i can learn

2. Originally Posted by begin
My problem says find the surface area of the part of the surface
z=y^2-x^2 that is inside the cylinder x^2+y^2=2

So im thinking of polar coordinates
saying f(x,y)=y^2-x^2
and fx=-2x, fy=2y
and the area=
double integral of (sqrt(1+4x^2+4y^2))

but my problem is how to go about finding the limits
Im not sure exactly, can i get direction so i can learn
The part of the surfact that is it inside the cylindar is above the disk

$\displaystyle x^2+y^2 \le 2$ and yes polar coordinates will be the best system.

$\displaystyle \displaysytle \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}f(r,\theta)rdrd\ theta$

3. If your integrating from 0-2pie first,
would the limits of integration be the other way?

4. Originally Posted by begin
If your integrating from 0-2pie first,
would the limits of integration be the other way?
You are correct. I have fixed the above post.

5. okk all done, thanks for the help...