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## Main Question or Discussion Point

Thomas calculus Chapter 15, section "additional and advanced exercises" exercise 8 says:

"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.....

"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.....