Determining the range for surface intergal.

Question

I have to find $$ \iint \limits_S (F.n) ds$$ where,

$$ F = 2i + 5j + 3k $$ and S is the portion of the cone $z = \sqrt{x^2 + y^2}$ and is inside a cylinder $x^2 + y^2 = 1$

let $$x = u , y = v, z = \sqrt{u^2 + v^2}$$ and, Position vector r is,

$$ r = ui + vj + \sqrt{u^2 + v^2}$$

also, $$N = \frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}$$

$$\implies N = \frac{-u}{\sqrt{u^2 + v^2}}i - \frac{v}{\sqrt{u^2 + v^2}}j + k $$

so , $$F \cdot N = \frac{-2u}{\sqrt{u^2 + v^2}} - 5 \frac{v}{\sqrt{u^2 + v^2}} + 3$$

So,

$$ \iint \limits_S (F.n) ds = \iint \limits_R (F.N) du dv$$.

Can Someone help me with the concept of R, how to determine R or simply the range of integration.

Answer 1

The notation is important here.

If ${\bf F}$ is a continuous vector field defined over a surface oriented $S$ with a unit normal vector ${\bf n}$, then the surface integral of ${\bf F}$ on $S$ is given by $$\iint_{S}{\bf F}\cdot {\rm d}{\bf S}=\iint_{S}{\bf F}\cdot {\bf n}{\rm d}S=\iint_{S}{\bf F}\cdot \frac{{\bf r}_{u}\times {\bf r}_{v}}{|{\bf r}_{u}\times {\bf r}_{v}|}{\rm d}S=\iint_{R}\left[{\bf F}({\bf r}(u,v))\cdot \frac{{\bf r}_{u}\times {\bf r}_{v}}{|{\bf r}_{u}\times {\bf r}_{v}|} \right] |{\bf r}_{u}\times {\bf r}_{v}|{\rm d}A.$$ where $S$ is defined by a vectorial function ${\bf r}(u,v)$ and $R$ is the domain of the parameter. Hence, $$\bbox[5px,#ffd]{\iint_{S}{\bf F}\cdot {\rm d}{\bf S}=\iint_{R}{\bf F}\cdot ({\bf r}_{u}\times {\bf r}_{v}){\rm d}A}$$

Notice that $S$ is the portion of the cone $z=\sqrt{x^{2}+y^{2}}$ and is inside a cylinder $x^2+y^2=1$. Then in cylindrical coordinates the cone is $z=r$, so using parameters $r,\theta$ we have the parametrization $${\bf r}(r,\theta)=(r\cos \theta,r\sin \theta,r),\quad 0\leqslant r\leqslant 1,\quad 0\leqslant \theta \leqslant 2\pi$$

Then $${\bf r}_{r}\times {\bf r}_{\theta}=(-r\cos \theta,-r\sin \theta,r).$$ Since the $z$-coordinate is positive, then ${\bf r}_{r}\times {\bf r}_{\theta}$ which has upward orientation.

Then $${\bf F}\cdot ({\bf r}_{r}\times {\bf r}_{\theta})={\bf F}({\bf r}(r,\theta))\cdot ({\bf r}_{r}\times {\bf r}_{\theta})=(2,5,3)\cdot (-r\cos \theta,-r\sin \theta,r)=-2r\cos \theta-5r\sin \theta+3r.$$ Hence, $$\iint_{S}{\bf F}\cdot d{\bf S}=\int_{0}^{2\pi}\int_{0}^{1}(-2r\cos \theta-5r\sin \theta+3r){\rm d}r{\rm d}\theta=3\pi$$

Therefore the answer is $3\pi$, this represents the flow of ${\bf F}$ through $S$.