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

Marks: 8M

Year: May 2015

Given:

$$\int_C{\left(xy+y^2\right)dx+\ x^2dy}------\left(1\right)$$ $$\int{P\ dx+Qdy------\left(2\right)}$$ Comparing equation (1) and equation (2) we get

P = $xy+y^2$ and Q = $x^2$

$\mathrm{\therefore }$ $\frac{\partial P}{\partial y}=x+2y\ $ and $\frac{\partial Q}{\partial x}=2x$

Given : y = x and y = $x^2$

Solving both equations we get,

x = $x^2$

$0 = x^2-x$

$\mathrm{\therefore }$ x= or x=1

$\mathrm{\therefore }$ Point of intersection is (0,0) and (1,1)

$\mathrm{\therefore }$ By Green theorem, $$\int{P\ dx+Q\ dy=\ \int{\int_R{(}}}\frac{\partial Q}{\partial x}-\ \frac{\partial P}{\partial y})\ dx\ dy$$ 1) Outer limit x : The vertical strip will slide from x=0 to x=1

2)Inner limit y

a) Upper limit is equation of line : y=x

b) Lower limit is equation of parabola : y =$x^2$ $$\int{\int_R{\left(\frac{\partial Q}{\partial x}-\ \frac{\partial P}{\partial y}\right)dx\ dy=\ \int^1_0{\int^x_{x^2}{\left(2x-x-2y\right)dx\ dy}}}}$$ $$=\int^1_0{\int^x_{x^2}{\left(x-2y\right)dx\ dy}}$$ $$=\int^1_0{{\left[xy-\frac{2y^2}{2}\right]}^x_{x^2}\ dx}$$ $$=\int^1_0{[}x^2-x^2]-[x^3-x^4]\ dx$$ $$={[\frac{-x^4}{4}+\frac{x^5}{5}]}^1_0$$ $$=[ 1/5 - ¼]$$

$$= -1/20$$ $$\mathrm{\therefore }\int_{\boldsymbol{C}}{\left(\boldsymbol{xy}\boldsymbol{+}{\boldsymbol{y}}^{\boldsymbol{2}}\right)\boldsymbol{dx}\boldsymbol{+\ }{\boldsymbol{x}}^{\boldsymbol{2}}\boldsymbol{dy}}\boldsymbol{=\ -}\boldsymbol{1}\boldsymbol{/}\boldsymbol{20}$$