Question

Find the area of the region bounded by the curves $x=4y-y^2$ and the y-axis.

Answer 1

$x=4y-y^2$
i.e., $y^2-4y=-x$
$(y-2{)}^2=-(x-4)$
Which represents a left parabola with vertex at A(4, 2).
The parabola meets Y-axis i.e., x = 0
$4y-y^2=0$
i.e., at $y=0,4$.

The area enclosed between the curve and the Y-axis.
${\int }_0^{4}xdy={\int }_0^4(4y-y^2)dy$
$={[4.\frac{y^2}{2}-\frac{y^3}{3}]}_0^4$
$=[32-\frac{64}{3}]-[0,0]=\left|\frac{32}{3}\right|sq.unit=\frac{32}{3}sq.unit$