- 31

- 0

[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!