Graphing 3D Equations: Solving x^2 + z^2 ≤ 9

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SUMMARY

The graph of the inequality x² + z² ≤ 9 represents a cylinder extending infinitely along the y-axis, as the absence of a y term indicates that y coordinates can take any value. This conclusion is drawn from the understanding that the set of points (x, y, z) satisfying the inequality forms a circular cross-section in the xz-plane with a radius of 3. Additionally, the discussion clarifies misconceptions regarding the graph's orientation in 3D space, particularly concerning its representation in different octants.

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Miike012
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I am having difficulty graphing equatioins in 3D for instance...

what does the graph in 3-D look like x^2 + z^2 ≤ 9?

Apparently its a cylinder extending along the y-axis. My initial guess was it was a 2-D circle with radius 3 on the xz plane. Anyways how should I have known it extended along the y-axis?
 
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Miike012 said:
I am having difficulty graphing equatioins in 3D for instance...

what does the graph in 3-D look like x^2 + z^2 ≤ 9?

Apparently its a cylinder extending along the y-axis. My initial guess was it was a 2-D circle with radius 3 on the xz plane. Anyways how should I have known it extended along the y-axis?
The statement of the problem probably says something about graphing the set of points (x, y, z) such that x2 + z2 ≤ 9.

Since the inequality doesn't include a y term, the y coordinates are completely arbitrary. That's why the graph extends along the y-axis.
 
Another question... the 3d picture is on the paint doc and I am wondering why the entire surface is only in the first octant because isn't the point (which I hightligted in the paint document) in the 3rd octant?
 

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That graph is screwy.

The graph of the equation z = 2x2 + 2y2 - 4 is a paraboloid that opens up from its vertex at (0, 0, -4), and whose central axis lies along the z-axis. In your picture, the thing appears to be in the first octant.
 
Miike012 said:
Another question... the 3d picture is on the paint doc and I am wondering why the entire surface is only in the first octant because isn't the point (which I hightligted in the paint document) in the 3rd octant?

It isn't in the first octant. It is correct. The planes that look like the coordinate planes aren't ##x=0## and ##y=0##. Look how the axes are labeled.
 
LCKurtz said:
It isn't in the first octant. It is correct. The planes that look like the coordinate planes aren't ##x=0## and ##y=0##. Look how the axes are labeled.
Good point. I saw how the axes are labeled, but I didn't notice that the planes you see aren't the regular coordinate planes.
 

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