The area enclosed by the parabola $x^2 = 6y$ and the line $x-6y+2=0$ is:

$\frac{3}{4}$

$x^2 = 6y$   ...(1)

$x - 6y+2=0$

$⇒6y=x+2$   ...(2)

Placing value of 6y from (2) in (1)

$x^2=x+2$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$x(x-2)+1(x-2)=0$

$x=-1,2$

So both intersect at $x=-1$ and $x=2$

area = area under line - area under parabola

$⇒\int\limits_{-1}^2\frac{x}{6}+\frac{1}{3}-\frac{x^2}{6}dx$  $[∵x^2=6y⇒y=\frac{x^2}{6},\,6y=x+2=y=\frac{1}{6}x+\frac{1}{3}]$

$=\left[\frac{x^2}{12}+\frac{x}{3}-\frac{x^3}{18}\right]_{-1}^2$

$=\frac{4}{12}+\frac{2}{3}-\frac{8}{18}-\frac{1}{12}+\frac{1}{3}-\frac{1}{18}$

$=\frac{3}{12}+\frac{3}{3}-\frac{9}{18}=\frac{1}{4}-1-\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ sq. units