3 years ago

12, 15, 18 pleaseeee.

Mathematics

2 answers:

vagabundo [1.1K]3 years ago

8 0

Man that's so easy I think you can do this

Sonbull [250]3 years ago

6 0

Lol i think skip counting sorry if I'm wrong, 21,24,27,30

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

$\begin{cases}\mathbf x(r,\theta,\zeta)=r\cos\theta\\\mathbf y(r,\theta,\zeta)=r\sin\theta\\\mathbf z(r,\theta,\zeta)=\zeta\end{cases}$

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

The volume is then

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