The area of the region bounded by the curve $y^2 = 4x$, the $y$-axis and the line $y = 3$ is

1. The region is bounded by the $y$-axis, so integrate horizontal strips: $A = \int_0^3 x\,dy$. 2. From $y^2 = 4x$ we get $x = \frac{y^2}{4}$, so $A = \int_0^3 \frac{y^2}{4}\,dy$. 3. $= \frac{1}{4}\left[\frac{y^3}{3}\right]_0^3 = \frac{1}{4}\cdot\frac{27}{3} = \frac{1}{4}\cdot 9 = \frac{9}{4}$. 4. Integrating $x\,dx$ by mistake (vertical strips) would give $\frac{9}{2}$, the wrong setup. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.4_