Find $$\oint_S (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS$$ where $S$ is the surface bounded by $x^2+y^2=z^2$ and the plane $z=1$.
By divergence theorem the integration is $$\begin{align}\oint_S (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS=\underset{V}{\iiint}\operatorname{div(x \hat i+y\hat j+z^2 \hat k)}\,dV=\underset{V}{\iiint}2+2z\,dV\\=\underset{R}{\iint}\,\int_0^1 (2+2z)\,dz\,dA=3\underset{R}{\iint}\,dA=3\pi\end{align}$$
where $R$ is the area inside $x^2+y^2=1$.
But If I try to do this by first finding the integral on the top surface (in the plane $z=1$) and then on the bottom surface (surface of the cone) then I'm getting different answer.
For top surface
$\hat n=\hat k$, so $$\iint_{S_1} (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS=\iint_{S_1}z^2\,dA$$ since here $z=1$, we get $$=\iint_{S_1}\,dA=\pi$$
For bottom surface
Here normal is towards negative $z-$axis. Since $z^2=x^2+y^2$ so $\frac{\partial z}{\partial x}=\frac x z,\frac{\partial z}{\partial y}=\frac y z$ so we'd have $$\iint_{S_2} (x \hat i+y\hat j+z^2 \hat k)\cdot\left(\frac{\partial z}{\partial x}\hat i+\frac{\partial z}{\partial y}\hat j-\hat k\right) \,dS$$$$=\iint_{S_2}(x \hat i+y\hat j+z^2 \hat k)\cdot\left(\frac{x}{z}\hat i+\frac{y}{z}\hat j-\hat k\right)\,dA$$$$=\iint_{S_2}\frac{x^2+y^2}{z}-z^2\,dA$$$$=\iint_{S_2}\sqrt{x^2+y^2}-x^2-y^2\,dA\\=\int_0^{2\pi}\int_0^1(r-r^2)r\,dr\,d\theta=\frac {\pi}6$$
I think I did something wrong in bottom surface. Please help.Thank you.
Best Answer
When you are applying divergence theorem, your integral should be
$\displaystyle \int_0^{2\pi} \int_0^1 \int_r^1 (2+2z) \ r \ dz \ dr \ d\theta = \frac{7 \pi}{6}$
As you are integrating over the region which is volume of the cone, the $z$ goes from $r$ to $1$ where $z = 1$ is the top surface of the cone. The radius goes from $0$ to $1$.
If you have $0 \leq z \leq 1$ and $0 \leq r \leq 1$, independent of each other, you will get flux through the closed cylinder of radius $1$ and height $1$.