- #1
kudoushinichi88
- 129
- 2
I'm working on this problem,
Let S be the cone described by
[tex]z=\sqrt{x^2+y^2}[/tex]
where [itex]0\leq z \leq 4[/itex]
If
[tex]\textbf{F}(x,y,z)=y\textbf{i}-x\textbf{j}+z^2\textbf{k}[/tex]
find the surface integral
[tex]\int\int_S \textbf{F} \bullet d\textbf{S}[/tex]
where the orientation of S is given by the inner normal vector.So I first parameterized S into
[tex]\textbf{r}(u,v)=u\textbf{i}+v\textbf{j}+\sqrt{u^2+v^2}\textbf{k}[/tex]
Then I find the cross product
[tex]\textbf{r}_u \times \textbf{r}_v = -\frac{u}{\sqrt{u^2+v^2}}\textbf{i}-\frac{v}{\sqrt{u^2+v^2}}\textbf{j}+\textbf{k}[/tex]
So when I plug this into the surface integral formula, do I take the negative of this cross product? Because the orientation of S is given by the inner normal vector.
Let S be the cone described by
[tex]z=\sqrt{x^2+y^2}[/tex]
where [itex]0\leq z \leq 4[/itex]
If
[tex]\textbf{F}(x,y,z)=y\textbf{i}-x\textbf{j}+z^2\textbf{k}[/tex]
find the surface integral
[tex]\int\int_S \textbf{F} \bullet d\textbf{S}[/tex]
where the orientation of S is given by the inner normal vector.So I first parameterized S into
[tex]\textbf{r}(u,v)=u\textbf{i}+v\textbf{j}+\sqrt{u^2+v^2}\textbf{k}[/tex]
Then I find the cross product
[tex]\textbf{r}_u \times \textbf{r}_v = -\frac{u}{\sqrt{u^2+v^2}}\textbf{i}-\frac{v}{\sqrt{u^2+v^2}}\textbf{j}+\textbf{k}[/tex]
So when I plug this into the surface integral formula, do I take the negative of this cross product? Because the orientation of S is given by the inner normal vector.