Question

Determine the type of solid
0 <= theta <= pi/2; r <= z <= 7
Would this be a quarter of a cone or a quarter of a paraboloid?

Answer 1

Think about what happens along the x-axis. There
\[ r =\sqrt{x^2 + y^2} = \sqrt{x^2 + 0^2} = |x| \]

Hence if \( z = |x| \), is that consistent with a cone or a paraboloid?

Answer 2

I still don't understand...

Answer 3

So would this be looking like a cone? Since parabolas are curvy, which doesn't closely represent |x|? Whereas, a cone on a 2D graph would look like |x|??

Answer 4

You have that
r <= z <= 7
Hence
|x| <= z <= 7

Now at the boundary of the object, r = z, or along the x-axis
z = r = |x|.

If the object were a cone, would we expect the cross section along the x-axis to be ... what? A straight line, a parabola, what?

If the object were a paraboloid, what would you expect of the cross-section? A straight line, parabola or something else?

Given that z = |x|, is that straight line, parabola or something else?

Therefore is your shape consistent with a cone or a paraboloid.

Answer 5

Therefore is your shape consistent with a cone or a paraboloid?

Answer 6

It would have to be more consistent with a cone.

Answer 7

Yes, because z = |x| is a straight line for x > 0 and a different straight line for x < 0

Answer 8

and a cone has straight edges along the x-axis, while a paraboloid does not.

Answer 9

Thanks James! I don't know how you're so boss at calculus, but I really appreciate your help and detailed explanations :)

Answer 10

I'm a fallen angel. Had a few millennia practicing conic sections.

Answer 11

Are you a professor???

Answer 12

In a parallel universe.

Answer 13

I love you man. All I have to say :).