MHB Level Surfaces & Intersection of a Graph: Exploring $f(x,y,z) = x^2+y^2$

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The level surfaces of the function f(x, y, z) = x² + y² are defined as {(x, y, z) | x² + y² = c}. For c = 0, the level set is the z-axis, while for c < 0, it is the empty set. For c > 0, the level set forms a cylinder with radius √c. The intersection of the graph with a vertical plane, such as the xz-plane or yz-plane, results in parabolas defined by z = x² or z = y², respectively. Clarification on the choice of vertical plane for intersection is suggested, along with the potential inclusion of radius details when describing the cylinder.
mathmari
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Hey! :o

Draw or describe the level surface and an intersection of the graph for the function $$f: \mathbb{R}^3 \rightarrow \mathbb{R}, (x, y, z) \rightarrow x^2+y^2$$

I have done the following:

The level surfaces are defined by $$\{(x, y, z) \mid x^2+y^2=c\}$$

- For $c=0$ we have that $x^2+y^2=0$. So for $c=0$, the level set consists of the $z-$axis.
- For $c<0$, the level set is the empty set.

For $c>0$, the level set is the cylinder $x^2+y^2=c$.

Is this correct?? (Wondering)

Could I improve something?? (Wondering)

How can we describe an intersection?? (Wondering)
 
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Hi! (Blush)

mathmari said:
Is this correct?? (Wondering)

Yep. (Nod)

Could I improve something?? (Wondering)

Nope. (Shake)

How can we describe an intersection?? (Wondering)

I'm not sure what is intended here. Can you clarify? Or give an example? (Wondering)

As I see it, the cylinder surface is an intersection of the function with the level $c$.
Or perhaps an intersection with a plane is intended, in which case an ellipse will come out (possibly degenerated). (Thinking)
 
I like Serena said:
Nope. (Shake)

When I describe the level set at the case when $c>0$ is it enough to say that it is a cylinder or do I have to say also something else for example to mention the radius?? (Wondering)
I like Serena said:
I'm not sure what is intended here. Can you clarify? Or give an example? (Wondering)

As I see it, the cylinder surface is an intersection of the function with the level $c$.
Or perhaps an intersection with a plane is intended, in which case an ellipse will come out (possibly degenerated). (Thinking)

In my book there is the following definition:

The intersection of the graph of $f$ is the intersection of the graph with a vertical plane.

For example, if we have $f(x, y)=x^2+y^2$ we have the following:

If $P_1$ is the plane $xz$ in $\mathbb{R}^3$ that is defined by $y=0$, then the intersection of $f$ is the set $$P_1 \cap \text{ graph } f=\{(x, y, z) \mid y=0, z=x^2\}$$
that is a parabola in the plane $xz$.
Similarily, if $P_2$ is the plane $yz$, that is defined by $x=0$, then the intersection $$P_2 \cap \text{ graph } f=\{(x, y, z) \mid x=0, z=y^2\}$$ is a parabola in the plane $yz$.
So, do we take which vertical plane we want?? (Wondering)
 
It would be nice to mention the radius.

Isn't your example the same as your problem? (Wondering)
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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