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 900x2+y2+z2=900, and on the sides by the cylinder x squared plus y squared equals 100x2+y2=100. set up the triple integrals in cylindrical coordinates that give the volume of d using the following orders of integration.

a. dz dr dthetaθ
b. dr dz dthetaθ
c. dthetaθ dz dr

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

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

So the volume is given by

\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

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

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Step-by-step explanation: