0
1.1kviews
Evaluate $\int\int\limits_{s} (\nabla \times \overline{F}ds$ where $\overline{F} = (2x - y + z)i + (x + y - z^2)j + (3x - 2y + 4z)k$

And s is the surface of the cylinder $x^2 + y^2 = 4$ bounded by the plane z = 9 and open at the other end.

Marks: 8M

Year: Dec 2015

1 Answer
0
2views

$$\overline{\boldsymbol{F}}\boldsymbol{=}\left(\boldsymbol{2}\boldsymbol{x}\boldsymbol{-}\boldsymbol{y}\boldsymbol{+}\boldsymbol{z}\right)\overline{\boldsymbol{i}}\boldsymbol{+}\left(\boldsymbol{x}\boldsymbol{+}\boldsymbol{y}\boldsymbol{-}\boldsymbol{\ }{\boldsymbol{z}}^{\boldsymbol{2}}\right)\overline{\boldsymbol{j}}\boldsymbol{+(}\boldsymbol{3}\boldsymbol{x}\boldsymbol{-}\boldsymbol{2}\boldsymbol{y}\boldsymbol{+}\boldsymbol{4}\boldsymbol{z}\boldsymbol{)}\overline{\boldsymbol{k}}$$ $$\mathrm{\therefore } \mathrm{\nabla }\ X\ \overline{F} = \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \ 2x-y+z & x+y-\ z^2 & 3x-2y+4z \end{array} \right|$$ $$ = \overline{i}\ [\frac{\partial }{\partial y}(3x-2y+4z)-\ \frac{\partial }{\partial z}(2x-y+z)]-\overline{j}(\ \frac{\partial }{\partial x}(3x-2y+4z)-\ \frac{\partial }{\partial z}(2x-y+z)]$$ $$+\ \overline{k}\left[\frac{\partial }{\partial x}\left(x+y-\ z^2\right)-\frac{\partial }{\partial y}\left(2x-y+z\right)\right]=0$$ $$ = \overline{i}\ \left(-2-2z\ \right)--\overline{j}\left(3-1\right)+\overline{k}\left(1+1\right)$$ $$\ \ \ \ \ \ \ \ \ \ \ \ =\ -2\left(1+z\right)\mathrm{\ }\overline{i}-2\overline{j}+2\overline{k}$$

In the plane z = 9 , Unit Normal $\overline{N}=\ \overline{K}$ and $ds=dx\ dy$ $$\mathrm{\therefore } \overline{N}\ \left(\mathrm{\nabla }\ X\ \overline{F}\right)=\ \overline{K}(-2\left(1+z\right)\mathrm{\ }\overline{i}-2\overline{j}+2\overline{k}) = 0-0 +2 = 2$$

enter image description here

Now the boundary of the curved surface s and the top surface of the cylinder $S_1$ are same

So we find the surface integration of top surface, $$\therefore \int{\int_S{\overline{N}(\mathrm{\nabla }\ X\ \overline{F}}})\ ds=\ \int{\int_{S_1}{2dx\ dy}}$$ $$ = 2 \int{\int_{S_1}{dx\ dy}}$$ $$ = 2 * Area \ of \ circle \ S_1$$ $$= 2 * \Pi * 2^2$$ $$ = 8 * \Pi$$

$\boldsymbol{\therefore }\int{\int_{\boldsymbol{S}}{\overline{N}\boldsymbol{(}\boldsymbol{\mathrm{\nabla }}\boldsymbol{\ }\boldsymbol{X}\boldsymbol{\ }\overline{\boldsymbol{F}}}}\boldsymbol{)\ }\boldsymbol{ds}\boldsymbol{=}\boldsymbol{8}$ $\boldsymbol{\Pi}$

Please log in to add an answer.