The region bounded by the parabola $x^{2} = 6y$ and the line $y = 6$ is found with vertical strips. Between which limits of $x$ is the integral taken?

Explanation

1. 01

   The line meets the parabola where $x^{2} = 6 \times 6 = 36$, so $x = \pm 6$.

2. 02

   Between them the line lies above the parabola: $\int_{-6}^{6} \left(6 - \frac{x^{2}}{6}\right)dx$.