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

Marks : 6

Year : 2014

$$xy= a^2 \text{ is an hyperbola }$$

$y=x$ now to find intersection of line $x=y$ and hyperbola $xy=a^2$

Put $x=y$ in Equation of hyperbola $xy= a^2$

Put $x=y$ in equation of hyperbola

$$ \therefore x^2 = a^2 \therefore x=a$$

When $x=a , y=a $

$\therefore$ point of intersection is $(a,a)$

The region is divided into two parts as upper limit changes.

Region 1

1) ouiter limit x

$X=0$ to $x=0$

2) Inner limit y

a) Upper limit is equation of line

$$y=x$$

b) lower limit $y=0$

Region 2

1) Outer limit x

$X=a$ to $x=2a $

2) Inner limit y

a) Upper limit is equation of hyperbola i.e $xy=a^2$ i.e. $y=a^2/x$

b) Lower limit $y=0$

$$\therefore I=\int\limits_0^a\int\limits_0^x x^2dydx+\int\limits_a^{2a}\int\limits_0^{\frac {a^2}x}x^2dydx$$

$\therefore I=\int\limits_0^a\int\limits_0^x x^2[y]dx+\int\limits_0^{2a}x^2[y]_0^{\frac {a^2}x}\\ =\int\limits_0^a x^{2(x)}dx+ \int\limits_a^{2a}x^2\times \dfrac {a^2}xdx\\ =\int\limits_0^ax^3dx+ a^2\int\limits_a^{2a}x\space dx\\ =\Bigg[\dfrac {x^4}4\Bigg]^a_0 +a^2\Bigg[\dfrac {x^2}2\Bigg]_a^{2a}\\ =\dfrac {a^4}4+\dfrac {4a^4}2 -\dfrac {a^4}2\\ =\dfrac {a^4+8a^4-2a^4}4\\ =\dfrac {7a^4}4$