Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
math_proof
Group Title
Center of mass for the region bounded by the paraboloid z=4x^2y^2 and z=0 with p(x,y,z)=5z
 one year ago
 one year ago
math_proof Group Title
Center of mass for the region bounded by the paraboloid z=4x^2y^2 and z=0 with p(x,y,z)=5z
 one year ago
 one year ago

This Question is Closed

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Well, youhave three integral sin your future. Can you find the first? Usually "M"?
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
first we have to find the mass. but i have trouble setting up the integrals
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
would it be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4r}(6z) rdzdrd \Theta\]
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.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: \[0 \le z \le 4r^2\] \[0 \le \phi \le 2 \pi \] Now we only need r. So look at z=4r^2 if z=0 r=2 and at the max height r will be zero. So we have what you have :D
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
except the top bound should be 4r^2 not 4r
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
malevolence19 is there easier way to find \[\frac{ }{ x }\]
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
instead of going through all of the integrals etc, since the difference is that you use only x
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago
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
 Engagement 19 Mad Hatter
 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.