## anonymous 5 years ago 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?

1. JamesJ

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?

2. anonymous

I still don't understand...

3. anonymous

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|??

4. JamesJ

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.

5. JamesJ

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

6. anonymous

It would have to be more consistent with a cone.

7. JamesJ

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

8. JamesJ

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

9. anonymous

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

10. JamesJ

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

11. anonymous

Are you a professor???

12. JamesJ

In a parallel universe.

13. anonymous

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