213.15.4.17 triple integral of bounded by cone and sphere

[karush]

$\textsf{Find the volume of the given solid region bounded by the cone}$
$$\displaystyle z=\sqrt{x^2+y^2}$$
$\textsf{and bounded above by the sphere}$
$$\displaystyle x^2+y^2+z^2=128$$
$\textsf{ using triple integrals}$

\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*}.

 

[HOI]

Because of the symmetry here, I think I would be inclined to use cylindrical coordinates.

In cylindrical coordinates, \(\displaystyle x= r cos(\theta)\), \(\displaystyle y= r sin(\theta)\), and \(\displaystyle z= z\). The cone becomes z= r and the sphere becomes \(\displaystyle r^2+ z^2= 128\). The upper portion of the sphere, above the cone, is given by \(\displaystyle z= \sqrt{128- r^2}\). The cone and sphere meet where \(\displaystyle r^2+ r^2= 2r^2= 128\) so \(\displaystyle r= 8\). To cover the region between the cone and the sphere, \(\displaystyle \theta\) must go from 0 to \(\displaystyle 2\pi\). r must go from 0 to 8, and, for each r, z goes from r, at the cone, to \(\displaystyle \sqrt{128- r^2}\), at the sphere. The "differential of volume" in cylindrical coordinates is \(\displaystyle r drd\theta dz\) so the integral you want is

\(\displaystyle \int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz drd\theta\).

 

[karush]

\begin{align*}\displaystyle
V&=\int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz dr d\theta\\
&=\int_0^{2\pi}\int_0^8\biggr[r^2\biggr]_r^{\sqrt{128- r^2}} dz \, d\theta\\
\end{align*}uhh the $dr$ ?

 

[HOI]

\begin{align*}\displaystyle
V&=\int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz dr d\theta\\
&=\int_0^{2\pi}\int_0^8\biggr[r^2\biggr]_r^{\sqrt{128- r^2}} dz \, d\theta\\
\end{align*}uhh the $dr$ ?

uhh, what did you do with it? The first (inner) integral is with respect to z. When you have done that integral, it is the "dz" that disappears, not "dr".
\(\displaystyle \int_r^{\sqrt{128- r^2}} r dz= \left[rz\right]_r^{\sqrt{128- r^2}}=r\sqrt{128- r^2}- r^2\)

So the next integral is \(\displaystyle \int_0^8 r\sqrt{128- r^2}- r^2 dr\).

Of course, \(\displaystyle \int_0^{2\pi} d\theta= 2\pi\) and just multiplies the other integral.

(And, by the way, the integral of "\(\displaystyle rdr\)" is NOT "\(\displaystyle r^2\)".)