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

Marks: 6M

Year: May 2015

$$\int_C{\overline{F}.\ \overline{dr}}=\ \int{\int_C{\overline{N(}\mathrm{\nabla }X}}\overline{F})\ ds$$

In the xy plane $r=xi+jy+0k$

$$\therefore \ \partial r=\ \partial x\ i+\ \partial y\ j$$

$\overline{F}.\ \overline{dr}$ = xy dx + x $y^2$ dy , we get

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

Now $\overline{N}=k$ and ds = dx.dy

$$\int{\int_C{\overline{N(}\mathrm{\nabla }X}}\overline{F})\ ds=\ \int{\int_S{\left(y^2-x\right)}}\ dx\ dy$$ Now for square what we will do is we just calculate the surface area in the first quadrant and multiply it by 4 so we will get for the whole square. Thus for the portion in the first quadrant. Now this portion is bounded by x=0 , y=0 and the slant line equation can be found by two point form. i.e. $$\therefore \ \frac{x-x_1}{y-y_1} = \frac{x_1-x_2}{y_1-y_2}$$ $$\therefore \ \frac{x-1}{y-0}=\ \frac{1-0}{0-1}$$ $$\therefore x-1=\ -y$$ $$\therefore x+y=1$$ $$\int{\int_S{\left(y^2-x\right)}}\ dx\ dy$$ Now the horizontal strip will slide from 0 to 1 in horizontal direction and this is the outer integration limits where as for inner integration limits upper limit is the slant line equation which can be written as y = 1-x and lower limit is y=0 $$\therefore \int{\int_S{\left(y^2-x\right)}}\ dx\ dy = \int^1_{x=0}{\int^{y=1-x}_{y=0}{\left(y^2-x\right)dx\ dy}}$$ $$ = \int^1_{x=0}{{[\frac{y^3}{3}-yx]}^{1-x}_0\ dx}$$ $$ = \int^1_{x=0}{\left[\frac{{\left(1-x\right)}^3}{3}-\left(1-x\right)\right]dx}$$ $$ = \int^1_0{\left[\frac{1-3x-3x^2-x^3}{3}+\ x^2-x\ \right]}\ dx$$ $$ = \frac{1}{3}\int^1_0{\left(1-6x+6x^2-\ x^3\right)dx}$$ $$ = \frac{1}{3}\ {[x-\frac{6x^2}{2}+\frac{{6x}^3}{3}-\frac{x^4}{4}]}^1_0$$ $$ = \frac{1}{3}\left[1-3+2-\frac{1}{4}\right]=\ -\frac{1}{12}$$

For complete square multiplying it by 4 we get , -1/3

Therefore, as area cannot be negative thus -1/3 is equivalent to 1/3.