Question

find the area bounded both the parabola y^2 = 9x and x^2 = 9y​

Answer 1

Step-by-step explanation:

We have, the upward parabola: x²=9y

Right-handed parabola: y²=9x

Their point of intersection is $(0,0) \: \& \: (9,9)$

Required area

$\int^{9}_{0} \bigg( 3 \sqrt{x} - \frac{ {x}^{2} }{9} \bigg) dx$

$= \bigg[3 \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} } - \frac{ {x}^{3} }{9.3} \bigg] ^{9} _{0}$

$= \bigg[2 {x}^{ \frac{3}{2} } - \frac{ {x}^{3} }{27} \bigg] ^{9} _{0}$

$= 2 . {9}^{ \frac{3}{2} } - \frac{ {9}^{3} }{27}$

$= 54 - \frac{ 9 \times 3 \times 3 \times 9 }{27}$

$= 54 - 27$

$= 27 sq \: units$