- #1

karush

Gold Member

MHB

- 3,269

- 5

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

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

$\tiny{15.4.20}$

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

$\tiny{15.4.20}$

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