# Help Finding Surface area with Double Integrals. (work shown)

#### KSCphysics

find the surface area of $$f(x,y)=\sqrt{x^2+y^2}$$ above the region
$$R=\{(x,y):0\leq f(x,y) \leq 1\}$$

well.. heres what the answer should be.. $$\sqrt{2} \ \theta$$

1. formula for Surface area:
$$\int_{R} \int \sqrt{1+f_{x}(x,y)^2+f_{y}(x,y)^2} \ dA$$

2. next i need to find the region:
so if $$r^2=x^2+y^2$$
then... $$r=\sqrt{x^2+y^2}$$
which means that our region is from
$$0 \leq r \leq 1$$
and we know $$0 \leq \theta \leq \frac{\Pi}{2}$$

3. $$f_{x}(x,y)=\frac{x}{\sqrt{x^2+y^2}}$$ $$f_{y}(x,y)=\frac{y}{\sqrt{x^2+y^2}}$$

4. $$\int_{0}^\frac{\Pi}{2} \int_{0}^1 \sqrt{1+(\frac{x}{\sqrt{x^2+y^2}})^2+(\frac{y}{\sqrt{x^2+y^2}})^2} \ dr \ d\theta$$

so $$\int_{0}^\frac{\Pi}{2} \int_{0}^1 \sqrt{2} \ dr d\theta$$

somehow..... something doesnt work..... HELP!

Wait, you're going to get an answer in terms of a dummy variable theta?

#### KSCphysics

no i should get a Number.

oops the answer is sqrt(2) * Pi not sqrt(2) * theta

Okay. You have two errors.

First off, the area you're integrating over is slightly off. You're only integrating over the quarter circle in the first quadrant. 0<f(x,y)<1 in an entire circular region of radius one.

Second off, you forgot to add the change of variables correction when you changed from Cartesian to Polar coordinates. So there should be one more thing in the integral along with the sqrt(2).

#### KSCphysics

is that one more thing the r in the r dr dtheta?

You bet.

#### KSCphysics

your pretty smart... are you a professor? or teacher?

#### KSCphysics

hrm.. tried it.. now i get (sqrt(2)*Pi)/4

I'm a lowly student...

The 1/4 seems to indicate that you might still be integrating only the first quadrant.

#### KSCphysics

so i should integrate 0<Theta<2*Pi???

Well, yes... That is a full circle.

Remember that 0<f(x,y)<1 stipulates that the region to be integrated over has radius 1. And since we've got x^2 and y^2, it doesn't matter whether x or y is positive or negative. They both end up positive anyway.

Integrating to pi/2 only yields a quarter of the circle. The quarter in the 1st quadrant, in particular.

Anyway, since the whole thing is isotropic, you can literally just take the integral where you used 0 to pi/2 and multiply it by 4 to get the integral for 0 to 2pi. So the problem's done!