Recent content by Michels10

Nevermind, solved it. Needed to substitute v=y^2. Thanks!

Solve y' = y/x + 1/y I get a similar answer to the correct one but I believe I am making a substitution error. Here is my attempt: dy/dx = y/x + 1/y set v = y/x equation now becomes: v + x(dv/dx) = v + 1/(x*v) reduces to: dv/dx = 1/(x^2 * v) Now the equation is...

There is, but the semester is over. The problems we have discussed/were in the book did not involve varying densities(not to mention the concavity at the top of the cone[irregular shapes]). We have done simple double or triple integrals of the equation for density*the density proportionality...

Yes it bothers me. This assignment was due two days ago and clearly I do not understand it. My book examples don't help me at all and I've got nothing to go by. Intuition? As for the "back to the drawing board" -- I don't thinks so. These repeated attempts post due-date are just brewing...

I am indeed using maple. I've already found the volume of the sphere,using triple integrals- int(int(int(vsp(rho, theta, Phi), rho = 0 .. 4), theta = 0 .. 2*Pi), Phi = 0 .. Pi) = 256Pi/3 for the mass of the cone and the center of mass i got: f := ((4-sqrt(7))*1.4)*r^3 mass:=Int(Int(f, r = 0...

Okay, got that, now what should I do for the sphere? Finding the density of the sphere should be relatively easy... Its not just the triple integral of volume * density constant *radius?

The center of mass of the cone is somewhere along the Z axis. This yields a problem when dealing with double integrals doesn't it? I can't substitute a Z into the equation when dealing with polar coordinates. I believe the equation is: 1/mass Int(Int((z*density)rdr)dtheta) edit: I've been...

Okay very cool. I suppose now I have to find out how deep the cone is in the center (along the z axis) due to concavity. Thanks you for your help thus far. I am going to keep going at it! I'll post back soon with an update.

Okay, I'm assuming it has something to do with the slope of the cone, and the increasing radius/size as you move up from the bottom of the cone... Since the radius is 3 and the height is 8, this gives us a slope of 8/3. I'm assuming that I need to use that 8/3 in my equation, I'm not sure...

I think I've about given up on this one. I clearly have no idea what I'm doing. I was trying to rearrange the equations to fit my graph. I saw my professor, he gave me an equation of z^2 = 64/9 * x^2 + y^2 for the cone and said that the height from the center of the sphere to the bottom...

Okay, then how about I change them... sphere- 16 = x2+y2+(z - (8+sqrt(7)))2 cone- x2 + (y-3)2 = (z-8)2

I updated one of my last posts with some integrals... But back to the graph.. The cone has a height of 8. The total height from the bottom of the cone to the center of the sphere is 8+sqrt(7). And the radius is still 4 for the sphere, and still 3 for the cone

i've attached a picture of the cone/sphere graphed.

how do these equations look: cone- z2 = (64/9)(x2+y2) sphere- 4 = x2+y2+z2 Cone: Int(Int(Int(1.4*(64/9)(r2)*r,dr),d(theta)),dz) r=0 to 3 theta = 0 to 2pi z = 0 to 8 Sphere: Int(Int(Int(1.8*p2*sin(phi),d(p),d(theta)),d(phi)) p = 0..2 theta = 0..2pi phi = 0..pi

Thanks for your reply Mark44! Dont I still have two unknowns? density and mass? I guess my problem now is finding the equation for density. Do I need to set the equations equal to and find out where they intersect? I've solved for both volumes: Vcone = 24*pi Vsphere = (32/3)*pi Thanks, Michels10