find the surface area of [tex]f(x,y)=\sqrt{x^2+y^2}[/tex] above the region(adsbygoogle = window.adsbygoogle || []).push({});

[tex]R=\{(x,y):0\leq f(x,y) \leq 1\}[/tex]

well.. heres what the answer should be.. [tex]\sqrt{2} \ \theta[/tex]

1. formula for Surface area:

[tex]\int_{R} \int \sqrt{1+f_{x}(x,y)^2+f_{y}(x,y)^2} \ dA[/tex]

2. next i need to find the region:

so if [tex]r^2=x^2+y^2[/tex]

then... [tex]r=\sqrt{x^2+y^2}[/tex]

which means that our region is from

[tex]0 \leq r \leq 1 [/tex]

and we know [tex]0 \leq \theta \leq \frac{\Pi}{2} [/tex]

3. [tex]f_{x}(x,y)=\frac{x}{\sqrt{x^2+y^2}} [/tex] [tex]f_{y}(x,y)=\frac{y}{\sqrt{x^2+y^2}} [/tex]

4. [tex]\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[/tex]

so [tex]\int_{0}^\frac{\Pi}{2} \int_{0}^1 \sqrt{2} \ dr d\theta[/tex]

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**