# Did I set up my integral correctly?

1. Nov 12, 2015

### qq545282501

1. The problem statement, all variables and given/known data
use triple integral to find volume of the solid in the first octant that is bounded above by$$x+2y+3z=6$$ and laterally by the clyinder $$x^2+y^2=4$$

2. Relevant equations

3. The attempt at a solution
From the given plane I got:
$$Z=2-\frac{1} {3}x-\frac{2} {3}y$$
from the given Cylinder i got:
$$y=\sqrt{4-x^2}$$

since its only interested in the first octant, lower bounds should all be zero.

so I get $$\int_{x=0}^2\int_{y=0}^\sqrt{4-x^2} \int_{z=0}^{2-\frac{1} {3}x-\frac{2} {3}y} \,dz \, dy \, dx$$

my concern is with Z's lower bound, I think it should be 0 as well, but if using $$z=x^2+y^2-4$$ seems like also okay, but being this way, this cylinder became a different solid, it became an infinite paraboloid. so that means I cant just set $$z=x^2+y^2-4$$ ? because the Z value of an infinite cylinder does not depend on x or y?

2. Nov 12, 2015

### Staff: Mentor

Looks fine to me.
Yes. You're in the first octant, where $x \ge 0, y \ge 0$, and $z \ge 0$.
No, it's not. The equation of the cylinder is $x^2 + y^2 = 4$. z is completely arbitrary, so you can't just stick it in the equation.

3. Nov 12, 2015

### Ray Vickson

The "equation" $z = x^2+y^2-4$ has no connection to the problem, because the lowest point on the plane and the curve $x^2+y^2=4$ has positive $z$. Therefore, the lowest value of $z$ is $z = 0$ and the top face of the cylinder does not come down that far. In short, you setup is correct.

BTW: thanks for using a typed version, which in this case is do-able because no diagrams are needed.

4. Nov 12, 2015

### qq545282501

got it. thank you all.