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

[math]\int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz drd\theta[/math].

 

[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".
[math]\int_r^{\sqrt{128- r^2}} r dz= \left[rz\right]_r^{\sqrt{128- r^2}}=r\sqrt{128- r^2}- r^2[/math]

So the next integral is [math]\int_0^8 r\sqrt{128- r^2}- r^2 dr[/math].

Of course, [math]\int_0^{2\pi} d\theta= 2\pi[/math] and just multiplies the other integral.

(And, by the way, the integral of "[math]rdr[/math]" is NOT "[math]r^2[/math]".)