# VC5.48

Here the given cone $z=2-\sqrt{(x^2-y^2)}\,$ meets the xy plane in a circle C given by $x^2+y^2=4,z=0\,$.Let R be the plane region of the circular disc bounded by C.Let S1 denote the entire closed surface made up of S and R and Let V1 be the volume enclosed by S1.

Then,we have $\iint_{S_1}(\nabla\times F)\cdot n dS=\iint_{S_1}\mathrm{curl}F\cdot n dS\,$

=$\mathrm{div}\mathrm{curl}F dV\,$ by divergence theorm.

=$0\,$ [As $\mathrm{div}\mathrm{curl}F=0\,$]

Therefore,$\iint_S(\nabla\times F)\cdot n dS+\iint_R(\nabla\times F)\cdot n dS=0\,$

<$\iint_S(\nabla\times F)\cdot n dS=-\iint_R \mathrm{curl}F\cdot n dS\,$ --(1)

Here,$\mathrm{curl}F=\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (x-z) & (x^3+yz) & -3xy^2\end{vmatrix}\,$

=$(-6xy-y)i+(-1+3y^2)j+(3x^2-0)k\,$ --(2)

Since on R,n=-k,therefore,we have

$\mathrm{curl}F\cdot n=\mathrm{curl}F\cdot(-k)=-3x^2\,$,using (2)

Therefore,from (1),$\iint_S(\nabla\times f)\cdot n dS\,$

=$\iint_R 3x^2 dS=3\int_{r=0}^{2}\int_{\theta=0}^{2\pi}(r\cos\theta)^2\times r d\theta dr\,$,changing to polars

=$3\int_{r=0}^{2}\int_{\theta=0}^{2\pi} r^3[\frac{1+\cos 2\theta}{2}]dr d\theta\,$

=$3\int_0^2 r^3[\frac{\theta}{2}+\frac{\sin 2\theta}{4}]_{0}^{2\pi} dr\,$

=$3\pi\int_0^2 r^3 dr=3\pi[\frac{r^4}{4}]_0^2=12\pi\,$

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