Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2014

$$\overline{\boldsymbol{F}}\boldsymbol{=}\boldsymbol{4}\boldsymbol{x}\overline{\boldsymbol{i}}\boldsymbol{+}\boldsymbol{3}\boldsymbol{y}\overline{\boldsymbol{j}}\boldsymbol{-}\boldsymbol{2}\boldsymbol{z}\overline{\boldsymbol{k}}$$ $$\mathrm{\nabla }\mathrm{.}\overline{F}=\boldsymbol{\ }\frac{\partial (4x)}{\partial x}\ +\ \frac{\partial (3y)}{\partial y}\ dy+\ \frac{\partial (-2z)}{\partial z}\ dz$$ $$= 4+ 3-2 = 5$$ ![enter image description here][1] $$\int{\int_{\boldsymbol{S}}{\overline{\boldsymbol{F}}.\overline{\boldsymbol{N}}\boldsymbol{\ }\boldsymbol{ds}\boldsymbol{\ }}}$\textbf{ = }$\int{\int{\int{\mathrm{\nabla }\mathrm{.}\overline{F}}}}dv$$ $$ = \int^2_{x=0}{\int^{2-x}_{y=0}{\int^{4-2x-2y}_{z=0}{\ 5\ dz\ dy\ dx}}}$$ $$ = \int^2_{y=-2}{\int^1_{x=\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}}{5{\left[z\right]}^{4-2x-2y}_0\ dy\ dx}}$$ $$ = 5 \int^2_0{\int^{2-x}_0{\left(4-2x-2y\right)dy\ dx}}$$ $$ = 5 \int^2_0{{[4y-2xy-\frac{2y^2}{2}]}^{2-x\ }_0\ dx}$$ $$ = 5 \int^2_0{\left[4\left(2-x\right)-2x\left(2-x\right)-{\left(2-x\right)}^2\right]dx}$$ $$= 5\int^2_0{[x^2}-4x+4]dx$$ $$ = 5[{\frac{x^3}{3}-4\frac{x^2}{3}\ +4x]}^2_0$$ $$= 40/3$$ $$\therefore \int{\int_{\boldsymbol{S}}{\overline{\boldsymbol{F}}.\overline{\boldsymbol{N}}\boldsymbol{\ }\boldsymbol{ds}\boldsymbol{\ }}}\boldsymbol{=}\boldsymbol{40}\boldsymbol{/}\boldsymbol{3}$$