- #1

karush

Gold Member

MHB

- 3,269

- 5

$\text{Find the volume of the region bounded above by the surface }\\$

$\text{$z=4-y^2$

and below by the rectangle

$R: 0 \ge x \ge 1 \quad 0\ge y \ge 2$.}$

$\begin{align*}\displaystyle

I&=\iint \limits_{R}4-y^2 \, dA&(1)\\

&=\int_{0}^{2}\int_{0}^{1}4-y^2 \, dxdy

=\int_{0}^{2}4-y^2 \, dy&(2)\\

&=\left[4y- \frac{y^3}{3} \right] \biggr|_0^2

=8-\frac{8}{3}=\frac{16}{3}&(3)

\end{align*}$

$\textit{ok I followed an example to do this

but didn't understand in (2) dropping the $_0^1$ limits}$