# What are the following shapes in 3-space?

I have these 3 equations. I tried to use wolfram alpha to graph in 3d, but did't succeed.

1) $x^2 + 2y^2 - 6x + 4y + 7 = 0$
2) $z^2 - 4z - 6x = 2$
3) $z = -y + 2$

I think that:
1) is a cylinder
2) is a cylinder
3) is just a sheet/plane

How close am I?

EDIT:

I also have: $x^2 + y^2 - 2x - 6y + 6z = 0$

I put the equation in the form $(x-1)^2 + (y-3)^2 + 6z = -10$... and I think this is also a cylinder... ?

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You might consider accepting some of the answers posted to your other questions if you found them satisfying. –  talmid Jun 10 '12 at 21:56

Try to put these equations in more familiar forms to know for sure what shape they represent. For example, 3) could be written as $y + z = 2$. In general, some algebraic manipulation takes these polynomial equations into the canonical expressions that help us gain some intuition and make a mental image of the shapes. Hint for 1) and 2): complete the squares.

1) is an elliptic cylinder, and 2) is a parabolic cylinder and 3) is a plane.

Added for 1): From $x^2 + 2y^2 - 6x + 4y + 7 = 0$ we see that we have quadratic terms in $x$ and $y$. I would try to isolate $x$ and $y$ like this: $(x^2 - 6x) + 2(y^2 + 2y) + 7 = 0$. Now, we complete the squares: $(x^2 - 6x + 9) - 9 + 2(y^2 + 2y + 1) - 2 + 7 = 0$ (we add and substract the necessary quantities). Then $(x-3)^2 + 2(y+1)^2 = 4$. Dividing through by $4$ we have $\frac{(x-3)^2}{4} + \frac{(y+1)^2}{2} = 1$, which is the equation of an ellipse in the $xy$-plane. Since $z$ is absent in the equation, for all $z$ the shape will be the same, and we have an elliptic cylinder with axis parallel to the $z$ axis, i.e., in a "vertical" position.

Added: $(x-1)^2 + (y-3)^2 + 6z = -10$ isn't a cylinder. Note that since all variables are involved in the equation, you could rearrange to get: $z = -\frac{1}{6}(x-1)^2 - \frac{1}{6}(y-3)^2 -\frac{10}{6}$, which is the graph of an "upside down" paraboloid.

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Alright I was able to draw and visualize 2 and 3, but 1 I am struggling.. any suggestions? –  Nick Jun 10 '12 at 21:32
Complete the squares. You have a 'free variable' in each equation so you basically only have to figure out how the projections look in the respective planes. –  M.B. Jun 10 '12 at 21:34
@OldJohn A parabolic cylinder, not a usual cylinder. –  talmid Jun 10 '12 at 21:43
I also have another shape, I will edit the original question too –  Nick Jun 10 '12 at 21:48