## University Calculus: Early Transcendentals (3rd Edition)

The area of the region is $18$.
$x=2y^2$, $x=0$ and $y=3$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $y$ here. 2) Find the limits of integration: With respect to $y$, the region runs from the point $y=0$ to $y=3$, so the upper limit is $3$ and the lower one is $0$. 3) Find the area: The region from $y=0$ to $y=3$ is bounded on the right by $x=2y^2$ and on the left by $x=0$. Therefore, the area of the region is $$A=\int^3_02y^2dy$$ $$A=\frac{2}{3}y^3\Big]^3_0$$ $$A=\frac{2}{3}\times3^3=18$$