Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

math_proof

  • 2 years ago

Center of mass for the region bounded by the paraboloid z=4-x^2-y^2 and z=0 with p(x,y,z)=5-z

  • This Question is Closed
  1. tkhunny
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well, youhave three integral sin your future. Can you find the first? Usually "M"?

  2. math_proof
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    first we have to find the mass. but i have trouble setting up the integrals

  3. math_proof
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    would it be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4-r}(6-z) rdzdrd \Theta\]

  4. malevolence19
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, first notice that z goes between: \[0 \le z \le 4-(x^2+y^2)\] To me this looks like a good time to switch to cylindrical. So we have: \[0 \le z \le 4-r^2\] \[0 \le \phi \le 2 \pi \] Now we only need r. So look at z=4-r^2 if z=0 r=2 and at the max height r will be zero. So we have what you have :D

  5. malevolence19
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    except the top bound should be 4-r^2 not 4-r

  6. math_proof
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thanks

  7. math_proof
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    malevolence19 is there easier way to find \[\frac{ }{ x }\]

  8. math_proof
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    instead of going through all of the integrals etc, since the difference is that you use only x

  9. malevolence19
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well careful inspection tells you that x and y are both zero. By symmetry and the fact that: \[x=r \cos(\phi);y=r \sin(\phi) \implies \iiint r \sin(\phi) \rho dV; \iiint r \cos(\phi) \rho dV\] But if you realize that integrating sine or cosine over a full period (i.e., 2 pi) are just ZERO. That is: \[\int\limits_0^{2 \pi}\sin(x)dx=\int\limits_0^{2 \pi} \cos(x)dx=0\] So they are zero by symmetry and by the math.

  10. malevolence19
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Since z is just z is cylindrical you see that there is no sine or cosine so the integral is NON zero, as we'd expect.

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.