- #1
qq545282501
- 31
- 1
Homework Statement
use triple integral to find volume of the solid in the first octant that is bounded above by[tex] x+2y+3z=6[/tex] and laterally by the clyinder [tex]x^2+y^2=4[/tex]
Homework Equations
The Attempt at a Solution
From the given plane I got:
[tex]Z=2-\frac{1} {3}x-\frac{2} {3}y[/tex]
from the given Cylinder i got:
[tex]y=\sqrt{4-x^2}[/tex]
since its only interested in the first octant, lower bounds should all be zero.
so I get [tex]\int_{x=0}^2\int_{y=0}^\sqrt{4-x^2} \int_{z=0}^{2-\frac{1} {3}x-\frac{2} {3}y} \,dz \, dy \, dx[/tex]
my concern is with Z's lower bound, I think it should be 0 as well, but if using [tex]z=x^2+y^2-4[/tex] seems like also okay, but being this way, this cylinder became a different solid, it became an infinite paraboloid. so that means I can't just set [tex]z=x^2+y^2-4[/tex] ? because the Z value of an infinite cylinder does not depend on x or y?