# Consider the solid that is bounded below by the cone z = sqrt{3x^{2}+3y^{2}} and above by the sphere x^{2} +y^{2} + z^{2} = 16..Set up only the appropriate triple integrals in cylindrical and spherical coordinates needed to find the volume of the solid.

Consider the solid that is bounded below by the cone $z=\sqrt{3{x}^{2}+3{y}^{2}}$
and above by the sphere ${x}^{2}+{y}^{2}+{z}^{2}=16.$.Set up only the appropriate triple integrals in cylindrical and spherical coordinates needed to find the volume of the solid.
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averes8

Step 1
To set the triple integral in cylindrical coordinates
By using relation,
$x=r\mathrm{cos}\theta$
$y=r\mathrm{sin}\theta$
$z=z$
Thus,
The cone $z=\sqrt{3{x}^{2}+3{y}^{2}}$ in cylindrical coordinate becomes,
$z=\sqrt{3{r}^{2}}=\sqrt{3r}$
And the sphere become ${r}^{2}+{z}^{2}=16$
To find the limit of r,
Consider,
$⇒\sqrt{3{r}^{2}}=\sqrt{16-{r}^{2}}$
$⇒3{r}^{2}=16-{r}^{2}$

$⇒{r}^{2}=4$
$⇒r=2$
Step 2
Thus, we can describe the region as
$E=\left\{\left(r,\theta ,z\right)|0\le \theta \le 2\pi |0\le r\le 2\sqrt{3r}\le z\le \sqrt{16-{r}^{2}}\right\}$
Hence, the triple integral for the volume by cylindrical coordinates is
$V={\int }_{0}^{2\pi }{\int }_{0}^{2}{\int }_{\sqrt{3r}}^{\sqrt{16-{r}^{2}}}rdzdrd\theta$
Step 3
Now, to set the triple integral in spherical coordinates
Since,
${p}^{2}={x}^{2}+{y}^{2}+{z}^{2}$
$\mathrm{tan}\theta =\frac{y}{x}$
$\phi =\mathrm{arccos}\left(\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\right)$
The sphere
From the cone
$z=\sqrt{3{x}^{2}+3{y}^{2}}=\sqrt{3r}$
$⇒p\mathrm{cos}\left(\phi \right)=\sqrt{3}p\mathrm{sin}\left(\phi \right)$
$⇒p\mathrm{tan}\left(\phi \right)=\frac{1}{\sqrt{3}}$
$\phi =\frac{\pi }{6}$
Step 4
Thus, we can describe the region as
$E=\left\{\left(p,\phi ,\theta \right)|0\le p\le 4,0\le \phi \le \frac{\pi }{6},0\le \theta \le 2\pi \right\}$
Hence, the triple integral for the volume of the solid by spherical coordinate is

Step 5
Now, evaluating the integral of cylindrical coordinate we get.
$⇒V={\int }_{0}^{2\pi }{\int }_{0}^{2}{\int }_{\sqrt{3r}}^{\sqrt{16-{r}^{2}}}rdzdrd\theta$
$⇒V=17.9582$
And evaluating the integral of spherical coordinate
$⇒V={\int }_{0}^{2\pi }{\int }_{0}^{\frac{\pi }{6}}{\int }_{0}^{\sqrt{4}}{r}^{2}\mathrm{sin}\left(\phi \right)dpd\phi d\theta$
$⇒V=17.9582$
Thus, by both coordinate systems, we get the same volume.
Therefore, both triple integrals are appropriate.