Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : DEC 2015

Consider the first equation i.e. $axy=4$. i.e. $ax=(4/y)$

Let $x=0$ then $y=∞ $

Here $(0, ∞)$

Let $x=∞$ then $y=0;$

When $x=1 ,y=4/9$

When $y=1 ,x=4/9$

Plotting through diagram we get

Now consider 2nd equation

i.e. $2x+y = 2 ………………………….(2)$

let $x=0 ,y=2 $

let $y=0 ,x=1 $

$\therefore $Equation (2) is a straight line passing from $(0,2)$ and $(1,0)$

$\therefore$ this line is intersecting curves at solving equation (2)

$$xy = (4/9) ,x=(4/9y)$$

Substituting value of x in equation (2)

$$\therefore \dfrac {2\times 4}{9y}+y=2$$

$ 8+9y^2=18y\\ 9y^2-18y+8=0\\ \therefore y=\dfrac 43 \space \&\space \dfrac 23 \\ \therefore \text{ Corresponding x is }\\ x=\dfrac 4{9\times \frac 43}=\dfrac 13\\ x=\dfrac 23$

Hence the points of intersections of the hyperbola and the line are

$(2/3 , 2/3)$ and $(1/3 , 4/3)$

1) Now considering the horizontal strip which will slide along x axis i.e. from $1/3$ to $2/3$ which is

∴ outer limit

2.a) Upper limit is equation of line $2x+y=2$ in terms of y we get $y=2-2x$

2.b) Lower limit is equation of curves $9xy=4$ in terms of y we get $y=(4/9x)$

$$\therefore A=\int\limits_{1/3}^{2/3}\int\limits_{4/9x}^{2-2x}dy\space dx$$

$\text{ Integrating w.r.t. y we get}\\ A= \int\limits_{1/3}^{2/3}[y]^{2-2x}_{4/9x}dx \\ \int\limits_{1/3}^{2/3} 2-2x-\dfrac 4{9x}dx \\ A=\Bigg[2x-\dfrac {2x^2}2-\dfrac 49\log x\Bigg]_{1/3}^{2/3}\\ =\dfrac 43-\dfrac 49-\dfrac 49\log \dfrac 23-\dfrac 23+\dfrac 19+\dfrac 49\log \dfrac 13\\ =\dfrac 13-\dfrac 49[\log 2-\log 3-\log 1+\log 3]\\ =\dfrac 13-\dfrac 49\log 2$