Multivariable Calculus 3D co-ord. system help

[Luongo]

1. Write inequalities to describe the region: The solid cylinder that lies on or below the plane z=8 and on or above the disk in the xy-plane with a center the origin and radius 2.


I don't understand because I'm using stewart's calculus i have no idea what the equation for a cylinder is? they only show you the eqn of spheres (x-h)^2+(y-k)^2+(z-l)^2 = r^2

 

[VeeEight]

In three dimensions, the general equation for a cylinder with cross section ellipse is x²/a² + y²/b² = 1.
If the cross sections are circles then you can say x² + y² = r²

 

[Luongo]

In three dimensions, the general equation for a cylinder with cross section ellipse is x²/a² + y²/b² = 1.
If the cross sections are circles then you can say x² + y² = r²

I understand that, however i don't understand how i can have the radius of 2 in this inequality if the cylinder fills the z axis from z=0 to z=8 how do i show this? is it 0<x^2+y^2<8?
where do i show the radius is 2 in this inequality? because the height of the cylinder is 8. the radius is 2. i can't show both? please help it would be greatly appreciated as there is a test on this material soon

 

[gabbagabbahey]

The cylinders axis is coincident with the z-axis, so the fact that it lies between z=0 and z=8 tell you its height is 8 units.

You need two inequalities to describe the region enclosed by this cylinder

Start with a simpler problem...what inequality would represent the area enclosed by the circle $x^2+y^2=4$?

 

[Luongo]

The cylinders axis is coincident with the z-axis, so the fact that it lies between z=0 and z=8 tell you its height is 8 units.

You need two inequalities to describe the region enclosed by this cylinder

Start with a simpler problem...what inequality would represent the area enclosed by the circle $x^2+y^2=4$?

-2 < x^2+y^2 < 2 on the x-axis but i don't understand the z-axis can you just tell me how to incorperate the z=0 to z=8 in this? it doesn't fit in.

 

[gabbagabbahey]

No, the smallest possible value of $x^2+y^2$ is zero (remember, $x^2+y^2$ represents the square of distance of the point(x,y) from the z-axis)...the furthest a point inside the circle $x^2+y^2=4$ can be from the z-axis is if it lies on the perimeter of the circle (2 units away from the z-axis), and the closest it can be to the z-axis is if it actually lies on the z-axis...so $0\leq x^2+y^2\leq 4$...make sense?

Points inside the cylinder will satisfy the same inequailty right?

They will also be somewhere between z=0 and z=8 right?

So...$0\leq x^2+y^2\leq 4$ and $0\leq z\leq 8$ describes the cylinder...make sense?

 

[Luongo]

yes, that makes sense because 0<x^2+y^2<4 describes the circle portion of the cylinder, but what is the point of saying it's greater than 0 if 0 is the lowest and would just confuse more could you also say x^2+y^2=4, 0<z<8?. because if it was 1<x^y+y^2<4 it would be a cylinder with a hole kind of like a toilet paper roll? but for cylinders do you always have to state the inequality for z independently because otherwise you would simply get a circular plane in R^3? thank you so much for this i really do appreciate it i was so confused but you cleared it up i think!