VC5.48

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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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