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

Marks: 6M

Year: May 2014

Given:

$$\overline{\boldsymbol{F}}\boldsymbol{=}\boldsymbol{4}\boldsymbol{x}\overline{\boldsymbol{i}}\boldsymbol{-}\boldsymbol{2}{\boldsymbol{y}}^{\boldsymbol{2}}\boldsymbol{\ }\overline{\boldsymbol{j}} + {\boldsymbol{z}}^{\boldsymbol{2}}\overline{\boldsymbol{k}}$$

$$\boldsymbol{\therefore } \boldsymbol{\nabla }.\overline{\boldsymbol{F}}\boldsymbol{=\ }\frac{\boldsymbol{\partial }\boldsymbol{(}{\boldsymbol{F}}_{\boldsymbol{x}}\boldsymbol{)}}{\boldsymbol{\partial }\boldsymbol{x}}\boldsymbol{+\ \ }\frac{\boldsymbol{\partial }\boldsymbol{(}{\boldsymbol{F}}_{\boldsymbol{y}}\boldsymbol{)}}{\boldsymbol{\partial }\boldsymbol{y}}\boldsymbol{+\ }\frac{\boldsymbol{\partial }\boldsymbol{(}{\boldsymbol{F}}_{\boldsymbol{z}}\boldsymbol{)}}{\boldsymbol{\partial }\boldsymbol{z}}\boldsymbol{\ \ \ }$$

$$\boldsymbol{\therefore }\boldsymbol{\mathrm{\nabla }}.\overline{\boldsymbol{F}}\boldsymbol{\mathrm{=\ }}\frac{\boldsymbol{\mathrm{\partial }}\boldsymbol{\mathrm{(}}\boldsymbol{\mathrm{4x}}\boldsymbol{)}}{\boldsymbol{\mathrm{\partial }}\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{+\ \ }}\frac{\boldsymbol{\mathrm{\partial }}\boldsymbol{\mathrm{(}}\boldsymbol{-}\boldsymbol{2}{\boldsymbol{y}}^{\boldsymbol{2}}\boldsymbol{)}}{\boldsymbol{\mathrm{\partial }}\boldsymbol{\mathrm{y}}}\boldsymbol{\mathrm{+\ }}\frac{\boldsymbol{\mathrm{\partial }}\boldsymbol{\mathrm{(}}{\boldsymbol{z}}^{\boldsymbol{2}}\boldsymbol{)}}{\boldsymbol{\mathrm{\partial }}\boldsymbol{\mathrm{z}}}\boldsymbol{\mathrm{\ \ \ }} = 4 - 4y +2z$$

1) The limit of z is directly given as z = 0 to z = 3

Given : ${\boldsymbol{y}}^{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{4}\boldsymbol{\ }$which is parabola and x = 1 is a line.

The parabola and the line intersect at x=1 in equation of parabola $$\boldsymbol{\therefore } {\boldsymbol{y}}^{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{4}$$ $$\boldsymbol{\therefore }\boldsymbol{y}\boldsymbol{=\ \pm }\boldsymbol{2}$$

The point of intersection is (1,-2) and (1,2)

2) Outer limit y, As the horizontal strip will slide from y -2 to y = -2

3) Inner limit x

a) Upper limit is equation of line $\boldsymbol{\therefore }\boldsymbol{x}\boldsymbol{=}\boldsymbol{1}$

b) Lower limit is equation of parabola $\boldsymbol{\therefore }\boldsymbol{\ }\boldsymbol{x}\boldsymbol{=}\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}$

$$\boldsymbol{\therefore }\boldsymbol{I}\boldsymbol{=\ }\int{\int{\int_{\boldsymbol{V}}{\boldsymbol{\mathrm{\ }}\boldsymbol{\mathrm{\nabla }}\boldsymbol{\mathrm{\ .\ }}}}}\overline{\boldsymbol{F}}\boldsymbol{\ }\boldsymbol{dv}$$ $$ = \int^2_{y=-2}{\int^1_{x=\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}}{\int^3_{z=0}{\left(4-4y+2z\right)dz\ dy\ dx}}}$$

Integrating w.r.t z

$$= \int^2_{y=-2}{\int^1_{x=\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}}{}}{\left[4z-4yz+2*\frac{z^2}{2}\right]}^3_0dx dy$$

$$= \int^2_{y=-2}{\int^1_{x=\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}}{}}{\left[12-12y+9\right]}dx dy$$

$$ = \int^2_{y=-2}{\int^1_{x=\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}}{\left[21-12y\right]dxdy}}$$ $$ = \int^2_{-2}{{[21x-12xy]}^1_{\frac{{\boldsymbol{y}}^{\boldsymbol{2}}}{\boldsymbol{4}}}} dy$$ $$= \int^2_{-2}{[21-12y-\ \frac{21y^2}{4}}+\frac{12y^3}{4}]\ dy$$ $$={\left[\mathrm{21y\ }--\mathrm{\ }\frac{12y^2}{2}-\frac{21y^3}{4*3}+\frac{{12y}^4}{4*4}\right]}^2_{-2}$$ $$=[42-6*4 -- 7*2 +12] - [21*2 -- 6*4 +7*2 +12] = 56$$

$\boldsymbol{\therefore }$ By Gauss divergence theorem

$$\int{\int_{\boldsymbol{S}}{\boldsymbol{\ }\overline{\boldsymbol{F}}}}\boldsymbol{\ }\boldsymbol{ds}\boldsymbol{\ } = \int{\int{\int_{\boldsymbol{V}}{\boldsymbol{\mathrm{\ }}\boldsymbol{\mathrm{\nabla }}\boldsymbol{\mathrm{\ .\ }}}}}\overline{\boldsymbol{F}}\boldsymbol{\ }\boldsymbol{dv} = 56$$