- #1
Telemachus
- 835
- 30
Hi, I have to compute de volume for the region:
[tex]\{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}[/tex]
I've tried to do by two different parametrizations, in spherics and cylindrical coordinates
For cylindrical coordinates I've made:
[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}[/tex]
For spherics I've considered [tex]4\sin \phi=\sqrt[ ]{15}\Rightarrow{\phi=Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}=u[/tex]
[tex]3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }[/tex]
I've used wolfram to make the calculus, and it gives that for cylindrical coord. I'm having twice the volume that I have for spherical coord. I don't know where I'm committing the mistake. Bye, thanks for helping :)
Note: I'm multiplying by 3 because I was going to use for Gauss theorem and is the divergence for the vector field given by the problem
[tex]\{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}[/tex]
I've tried to do by two different parametrizations, in spherics and cylindrical coordinates
For cylindrical coordinates I've made:
[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}[/tex]
For spherics I've considered [tex]4\sin \phi=\sqrt[ ]{15}\Rightarrow{\phi=Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}=u[/tex]
[tex]3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }[/tex]
I've used wolfram to make the calculus, and it gives that for cylindrical coord. I'm having twice the volume that I have for spherical coord. I don't know where I'm committing the mistake. Bye, thanks for helping :)
Note: I'm multiplying by 3 because I was going to use for Gauss theorem and is the divergence for the vector field given by the problem