- #1

- 39

- 0

"Sphere and cylinder" Find the volume of material cut from the

solid sphere [tex] r^2+z^2 \leq 9[/tex] by the cylinder [tex]r = 3 sin \theta [/tex] "

the answer's book gives de solution

[tex]\int _0^{\pi }\int _0^{3\sin \theta }\int _0^{\sqrt{9-r^2}}rdzdrd\theta =9\pi [/tex]

I think that's wrong, since Z should be [tex] -\sqrt{9-r^2}<z<\sqrt{9-r^2}[/tex], or we would be considering half the cilinder. However, if we subtitute those limits of integrations, we would get as an answer [tex]18\pi [/tex], which can not be correct, since the volume of the sphere is [tex]36\pi [/tex], and (by looking at the graph of the region) the volume of the cylinder should be LESS than half the volume of the sphere.

help please! i'm really confused.....