Question

Find the area of the region bounded by the curve x=y(2-y) at y-axis.

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:x = y(2 - y)$

Now, we first find the point of intersection with y -axis.

We know, on y - axis, x = 0.

So, on substituting the value in given curve, we get

$\rm :\longmapsto\:0 = y(2 - y)$

$\bf\implies \:y = 0 \: \: or \: \: y = 2$

Hence,

The point of intersection of curve x = y(2 - y) with y - axis are (0, 0) and (0, 2)

Now,

Required area with respect to y - axis is

$\rm \: = \: \displaystyle\int_0^2 \rm x \: dy$

$\rm \: = \: \displaystyle\int_0^2 \rm y(2 - y) \: dy$

$\rm \: = \: \displaystyle\int_0^2 \rm (2y - {y}^{2} ) \: dy$

We know,

$\underbrace{ \boxed{ \bf \: \int \: {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} \: + \: c}}$

$\rm \: = \: \bigg(2 \: \times \dfrac{ {y}^{1 + 1} }{1 + 1} - \dfrac{ {y}^{2 + 1} }{2 + 1} \bigg)_0^2 \rm$

$\rm \: = \: \bigg(2 \: \times \dfrac{ {y}^{2} }{2} - \dfrac{ {y}^{3} }{3} \bigg)_0^2 \rm$

$\rm \: = \: \bigg({y}^{2} - \dfrac{ {y}^{3} }{3} \bigg)_0^2 \rm$

$\rm \: = \: \bigg( {2}^{2} - \dfrac{ {2}^{3} }{3} \bigg) - \bigg(0 - 0 \bigg)$

$\rm \: = \: \bigg( 4 - \dfrac{8}{3} \bigg)$

$\rm \: = \: \dfrac{12 - 8}{3}$

$\rm \: = \: \dfrac{4}{3} \: sq. \: units$

Hence,

$\: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: Required \: area \: is \: \frac{4}{3} \: sq. \: units}}$