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Let d be the region bounded below by the plane zequals​0, above by the sphere x squared plus y squared plus z squared equals 144​, and on the sides by the cylinder x squared plus y squared equals 16. set up the triple integral in cylindrical coordinates that give the volume of
d.

Respuesta :

LammettHash
The cylinder is given straight away by [tex]x^2+y^2=r^2=16\implies r=4[/tex]. To get the cylinder, we complete one revolution, so that [tex]0\le\theta\le2\pi[/tex]. The upper limit in [tex]z[/tex] is a spherical cap determined by

[tex]x^2+y^2+z^2=144\iff z^2=144-r^2\implies z=\sqrt{144-r^2}[/tex]

So the volume is given by

[tex]\displaystyle\iiint_{\mathcal D}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=0}^{z=\sqrt{144-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

and has a value of [tex]\dfrac{128(27-16\sqrt2)\pi}3[/tex] (not that we care)

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