Did I set up my integral correctly?

[qq545282501]

Homework Statement

use triple integral to find volume of the solid in the first octant that is bounded above byx+2y+3z=6 and laterally by the clyinder x^2+y^2=4

Homework Equations

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 can't just set z=x^2+y^2-4 ? because the Z value of an infinite cylinder does not depend on x or y?

 

[Mark44]

Homework Statement

use triple integral to find volume of the solid in the first octant that is bounded above byx+2y+3z=6 and laterally by the clyinder x^2+y^2=4

Homework Equations

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

Looks fine to me.

my concern is with Z's lower bound, I think it should be 0 as well

Yes. You're in the first octant, where $x \ge 0, y \ge 0$, and $z \ge 0$.

, but if using z=x^2+y^2-4 seems like also okay

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.

, but being this way, this cylinder became a different solid, it became an infinite paraboloid. so that means I can't just set z=x^2+y^2-4 ? because the Z value of an infinite cylinder does not depend on x or y?

 

[Ray Vickson]

Homework Statement

use triple integral to find volume of the solid in the first octant that is bounded above byx+2y+3z=6 and laterally by the clyinder x^2+y^2=4

Homework Equations

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 can't just set z=x^2+y^2-4 ? because the Z value of an infinite cylinder does not depend on x or y?

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.

 

[qq545282501]

got it. thank you all.