Respuesta :

LammettHash
In Cartesian coordinates, the region is given by [tex]-3\le x\le3[/tex], [tex]-\sqrt{9-x^2}\le y\le\sqrt{9-x^2}[/tex], and [tex]-\sqrt{16-x^2-y^2}\le z\le\sqrt{16-x^2-y^2}[/tex]. Converting to cylindrical coordinates, using

[tex]\begin{cases}\mathbf x(r,\theta,\zeta)=r\cos\theta\\\mathbf y(r,\theta,\zeta)=r\sin\theta\\\mathbf z(r,\theta,\zeta)=\zeta\end{cases}[/tex]

we get a Jacobian determinant of [tex]r[/tex], and the region is given in cylindrical coordinates by [tex]0\le\theta\le2\pi[/tex], [tex]0\le r\le3[/tex], and [tex]-\sqrt{16-r^2}\le z\le\sqrt{16-r^2}[/tex].

The volume is then

[tex]\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=-\sqrt{16-r^2}}^{z=\sqrt{16-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\dfrac{4(64-7\sqrt7)\pi}3[/tex]
Cricetus

The volume of the solid will be "[tex]\frac{4\pi(64-7\sqrt{7} )}{3} \ units^3[/tex]".

Given:

Cylinder,

  • [tex]x^2+y^2=9[/tex]

and

Sphere,

  • [tex]x^2+y^2+z^2 = 16[/tex]

Now,

The desired volume will be:

→ [tex]V = 2\int\limits^{2x}_0 \int\limits^{3}_0 \int\limits^{16-r^2}_0 \ r dz \ dr \ d\Theta[/tex]

Integrating from the inside out yields,

→ [tex]\int\limits^{\sqrt{16-r^2} }_0 r \ dz = [rz]^{\sqrt{16-r^2} }_0[/tex]

                    [tex]= r\sqrt{16-r^2}[/tex]

→ [tex]\int\limits^3_0 r\sqrt{16-r^2} \ dr = [-\frac{1}{3}(16-r^2)^{\frac{3}{2} } ]^3_0[/tex]

                           [tex]= \frac{64-7\sqrt{7}}{3}[/tex]

Again by integrating, we get

→ [tex]\int\limits^{2x}_0 {\frac{64-7\sqrt{7} }{3} } \ d\Theta = [(\frac{64-7\sqrt{7} }{3} )\Theta]^{2x}_0[/tex]

→                      [tex]= \frac{2 \pi(64-7\sqrt{7} )}{3}[/tex]

hence,

→ [tex]V = 2(\frac{2\pi(64-7\sqrt{7} )}{3} )[/tex]

      [tex]= \frac{4\pi(64-7\sqrt{7} )}{3} \ units^3[/tex]

Learn more about cylindrical coordinates here:

https://brainly.com/question/14004645

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