# Surface Area 1

The part of the plane 3x+2y+z=6 that lies in the first octant.

$$A(s)=\int_0^2\int_0^3\sqrt{14}dydy$$

Are my limits not correct? B/c my answer is just off by a little.

me: 6\sqr(t14), answer: 3\sqrt(14)

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Science Advisor
Homework Helper
No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).

No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).
dope! Gotcha, thanks.