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wrongusername
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Homework Statement
Find the area of the surface: The part of the plane 3x + 2y + z = 10 that lies in the first octant
Homework Equations
[tex]A=\iint_{D}|\vec{r}_{u}\times\vec{r}_{v}|dA[/tex]
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?