- #1

karush

Gold Member

MHB

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$\textsf{Find the volume of the given solid region bounded by the cone}$

$$\displaystyle z=\sqrt{x^2+y^2}$$

$\textsf{and bounded above by the sphere}$

$$\displaystyle x^2+y^2+z^2=128$$

$\textsf{ using triple integrals}$

\begin{align*}\displaystyle

V&=\iiint\limits_{R}p(x,y,z) \, dV

\end{align*}

ok I don't think I can get the Integral set up the way the equations are written

I assume we break it up into $x=, y=$ and $z=$

\begin{align*}\displaystyle

z_{cone}&=\sqrt{x^2+y^2}\\

z^2&=x^2+y^2\\

x^2&=z^2-y^2\\

\therefore x_{cone}&=\sqrt{z^2-y^2}\\

y^2&=z^2-x^2\\

\therefore y_{cone}&=\sqrt{z^2-x^2}

\end{align*}.

$$\displaystyle z=\sqrt{x^2+y^2}$$

$\textsf{and bounded above by the sphere}$

$$\displaystyle x^2+y^2+z^2=128$$

$\textsf{ using triple integrals}$

\begin{align*}\displaystyle

V&=\iiint\limits_{R}p(x,y,z) \, dV

\end{align*}

ok I don't think I can get the Integral set up the way the equations are written

I assume we break it up into $x=, y=$ and $z=$

\begin{align*}\displaystyle

z_{cone}&=\sqrt{x^2+y^2}\\

z^2&=x^2+y^2\\

x^2&=z^2-y^2\\

\therefore x_{cone}&=\sqrt{z^2-y^2}\\

y^2&=z^2-x^2\\

\therefore y_{cone}&=\sqrt{z^2-x^2}

\end{align*}.

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