(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the area of the surface: The part of the plane 3x + 2y + z = 10 that lies in the first octant

2. Relevant equations

[tex]A=\iint_{D}|\vec{r}_{u}\times\vec{r}_{v}|dA[/tex]

3. The attempt at a solution

I parametrized the plane:

[tex]\vec{r}\left(u,v\right)=<u,v,6-3u-2v>[/tex]

and found the partial derivatives of u and v

[tex]\vec{r}_{u}=<1,0,-3>[/tex]

[tex]\vec{r}_{v}=<0,1,-2>[/tex]

then got this for their cross product:

[tex]\vec{r}_{u}\times\vec{r}_{v}=<3,2,1>[/tex]

[tex]|\vec{r}_{u}\times\vec{r}_{v}|=\sqrt{14}[/tex]

I think u goes from 0 to 2 because that's the highest and lowest values for u before the plane exits the first octant. This is also why I think the integral of v goes from 0 to 3. I obtain this answer:

[tex]\int_{0}^{2}\int_{0}^{3}\sqrt{14}dvdu=6\sqrt{14}[/tex]

when the actual answer is half of that. What have I done wrong?

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

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!

# Homework Help: Finding the area of a surface

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