babeeb0oL

2021-03-05

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.

averes8

Step 1

So we need, Set up only the appropriate triple integrals in cylindrical and spherical coordinates needed to find the volume of the solid.
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.

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