Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

math_proof Group Title

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

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. tkhunny Group Title
    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 years ago
  2. math_proof Group Title
    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

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

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

    • 2 years ago
  4. malevolence19 Group Title
    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

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

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

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

    thanks

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

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

    • 2 years ago
  8. math_proof Group Title
    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

    • 2 years ago
  9. malevolence19 Group Title
    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.

    • 2 years ago
  10. malevolence19 Group Title
    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.

    • 2 years ago
    • Attachments:

See more questions >>>

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.