Find the area bounded by the X axis and the parabola $y = -x^2 + 4$
Given: $y = -x^2 + 4$ $\;\;\; \cdots \; (1)$
$\implies$ $x^2 = - \left(y - 4\right)$ $\;\;\; \cdots \; (2)$
Equation $(2)$ represents a parabola.
It cuts the X axis at the points $\left(\pm 2 , 0\right)$.
Its vertex is $\left(0,4\right)$.
Equation of X axis is $y = 0$.
$\therefore$ $\;$ Required area (shaded portion) $= A = 2 \left|I\right|$ where
$I = \displaystyle \int \limits_{0}^{4} x \; dy$ $\;\;\; \cdots \; (3)$
From equation $(1)$, $x = \sqrt{4 - y}$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ can be written as
$\begin{aligned}
I & = \int \limits_{0}^{4} \sqrt{4 - y} \; dy \\\\
& = \dfrac{-2}{3} \left[\left(4 - y\right)^{3/2}\right]_{0}^{4} \\\\
& = \dfrac{-2}{3} \left[0 - \left(4\right)^{3/2}\right] \\\\
& = \dfrac{2}{3} \times 8 = \dfrac{16}{3}
\end{aligned}$
$\therefore$ $\;$ Required area $= 2 I = \dfrac{32}{3}$ square units