- #1
Damascus Road
- 120
- 0
Hello,
This is a question in which I'm getting the right answer, I just want to make sure what I'm doing its legitimate, as it seems a bit off to me.
The volume enclosed by
[tex]z=16[x^{2}+y^{2}], z=8-16[x^{2}+y^{2}][/tex]
I set my triple integral up as follows:
[tex]\int^{?r}_{?r}\int^{2\pi}_{0}\int^{8-16r^{2}}_{16r^{2}} r dz d\vartheta dr[/tex]
So, my height is the bottom function to the upper, and obviously completing a full circle.
Now here is what I'm unsure about.
I know I need to either do two integrals for r and add them together 0 to 1 and 0 to -1,
or -1 to 1 so that I'm only left with 2/pi (the answer) but what is the logic behind getting the limits for r?
Thanks in advance
This is a question in which I'm getting the right answer, I just want to make sure what I'm doing its legitimate, as it seems a bit off to me.
The volume enclosed by
[tex]z=16[x^{2}+y^{2}], z=8-16[x^{2}+y^{2}][/tex]
I set my triple integral up as follows:
[tex]\int^{?r}_{?r}\int^{2\pi}_{0}\int^{8-16r^{2}}_{16r^{2}} r dz d\vartheta dr[/tex]
So, my height is the bottom function to the upper, and obviously completing a full circle.
Now here is what I'm unsure about.
I know I need to either do two integrals for r and add them together 0 to 1 and 0 to -1,
or -1 to 1 so that I'm only left with 2/pi (the answer) but what is the logic behind getting the limits for r?
Thanks in advance