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mknut389
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Homework Statement
Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 16, above the xy plane, and outside the cone z = 3 [tex]\sqrt{{x^2+y^2}}[/tex].
Homework Equations
spherical system:
x=[tex]\rho[/tex]cos[tex]\theta[/tex]sin[tex]\phi[/tex]
y=[tex]\rho[/tex]sin[tex]\theta[/tex]sin[tex]\phi[/tex]
z=[tex]\rho[/tex]cos[tex]\phi[/tex]
cylindrical system
[tex]x^2+y^2=r^2[/tex]
z=z
The Attempt at a Solution
I have tried this using both the spherical and cylindrical systems and arrived at the same answer, cylindrical is easiest here, so Ill use it to demonstrate what I have done
[tex]x^2+y^2+z^2=16[/tex]
[tex]r^2+z^2=16[/tex]
z=[tex]\sqrt{{16-r^2}}[/tex]
z = 3 [tex]\sqrt{x^2+y^2}[/tex]
z=3[tex]\sqrt{r^2}[/tex]
z=3r
when z=0
0=[tex]\sqrt{16+r^2}[/tex]
r=4
0=3r
r=0
Therefore the bounds are
z: [tex]\left[3r,\sqrt{16-r^2}\right][/tex]
r:[tex]\left[0,4\right][/tex]
[tex]\theta[/tex]: [tex]\left[0,2\pi\right][/tex]
Which gives me
[tex]\int_{0}^{2\pi}\int_{0}^{4}\int_{3r}^{\sqrt{16-r^2}}r dzdrd\theta[/tex]
After going through the steps i get
[tex]-(256\pi)/3[/tex]
First off it is negative, so that tells me I am completely off base here, but when doing the same problem with spherical coordinates, I get [tex](256\pi)/3[/tex]. Since this answer is wrong, I am at a loss of what I should be doing. Please help.
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