15.4.20 volumn via triple integrals

[karush]

$\textsf{The region in the first octant bounded by the coordinate planes and the surface }$
View attachment 8112
$$z=4-x^2-y$$
$\textit{From the given equation we get}$
\begin{align*}\displaystyle
&0 \le z \le 4-x^2-y\\
&0 \le y \le 4-x^2\\
&0 \le x \le z
\end{align*}
$\textit{Thus the triple Integral results:}$
\begin{align*}\displaystyle
V&=\iiint\limits_{E} \, dzdyd
\quad =\int_{0}^{2}
\int_{0}^{4-x^2}
\int_{0}^{4-x^2-y}
\, dzdydx
\end{align*}
So just seeing if this is going the right direction
$\tiny{15.4.20}$

 

[tkhunny]

$\textsf{The region in the first octant bounded by the coordinate planes and the surface }$

$$z=4-x^2-y$$
$\textit{From the given equation we get}$
\begin{align*}\displaystyle
&0 \le z \le 4-x^2-y\\
&0 \le y \le 4-x^2\\
&0 \le x \le z
\end{align*}
$\textit{Thus the triple Integral results:}$
\begin{align*}\displaystyle
V&=\iiint\limits_{E} \, dzdyd
\quad =\int_{0}^{2}
\int_{0}^{4-x^2}
\int_{0}^{4-x^2-y}
\, dzdydx
\end{align*}
So just seeing if this is going the right direction
$\tiny{15.4.20}$

It's a good thing you didn't need that equation on the upper left.

Okay, now do it all over dydzdx and dydxdz.