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

Marks: 8M

Year: May 2014

Given :

$y^2=x$ and $y=4$

Solving both we get,

$x^2=4$

$\therefore$ x = $\pm$ 4

$\therefore$ Curve and the line intersect at (-2,4) and (2,4)

Part I:

Let I = $\int{Pdx+Qdy} \\ = \int^c_c{{(x}^2-y)\ dx+(2y^2+x)dy}$

Along curve AOD

$\therefore $ y = $x^2$

$\therefore dy=2x\ dx$

And $x\ $varies from -2 to 2

$\therefore I_{AOB}=\ \int^2_{-2}{(}x^2-\ x^2)\ dx+(2x^4+x)\ 2xdx \\ \therefore I_{AOB}=\ \int^2_{-2}{(}\ 4\ x^5+2x^2)\ dx\ \\ = [\frac{4x^6}{6}+\ \frac{2x^3}{3} ]^2_{-2} \\ = [\frac{4*64}{6}+\ \frac{2+8}{3} ]-[\frac{4*64}{6}-\ \frac{2+8}{3} ] = 32/3$

Along line AB

y = 4

dy = 0

And x varies from 2 to -2

$\therefore I_{BA}=\ \int^2_{-2}{(}x^2-\ 4)\ dx+0 \\ = [\frac{x^3}{3}-4x ]^2_{-2} \\ \therefore I_{BA}=\left[\ \frac{-8}{3}+8\right]-\left[\frac{8}{3}-8\right]=32/3 \\ \therefore I_{BA}+\ I_{AOB}=I \\ \therefore I=\frac{32}{3}+\frac{32}{3} = 64/3$

Part 2 :

$\int{Pdx+Qdy} = \mathop{\int\int}\nolimits_R{(\frac{\partial Q}{\partial x}-\ \frac{\partial P}{\partial y}}\ )\ dxdy$

Now evaluating R.HS. of this equation ,

Here P = $x^2-y$ and Q = 2$y^2+x$

$\therefore \frac{\partial P}{\partial y}=\ -1\ and\ \frac{\partial Q}{\partial x}=1$

Now here the vertical strip will slide from -2 to 2 in x axis

1) Outer limit x : x = -2 to x=2

2) Inner limit y:

a) Upper limit is equation of the line : y=4

b) Lower limit is equation of the parabola : y = $x^2$

I = $\mathop{\int\int}\nolimits_R{(\frac{\partial Q}{\partial x}-\ \frac{\partial P}{\partial y}}\ )\ dxdy=\ \int^2_{-2}{\int^4_{x^2}{\left(1+1\right)dx\ dy}}$

Integrating w.r.t y

$\therefore $ I = $\int^2_{-2}{2[y]^4_{x^{2\ \ }}dx} \\ = 2 \int^2_{-2}[4-x^{2}]^4_{x^{2}}dx \\ = 2 [4x-\frac{x^{3}}{3}]^2_{-2} \\ = 2[8-8/3] - [ -8 +8/3] = 2(32/3)$

= 64/3

= L.H.S

Hence verified.