I have to go with the 14, sorry.
What was your argument when you calculated the mass? It wasn't just , was it?
Can you use a double integral? It's about 1000 easier to understand this way.
Looking for somone to validate my answer or maybe tell me what Im doing wrong, I get an answer of x=0,y=98/5......my worksheet claims the answer is x=0,y=14
find the center of mass of the region bounded by y=49-x^2 and the x-axis with density of 7x^2
x=0 because its centered on the orgin
area is over the interval of -7 and 7
I get a mass of (1372p)/3; (p=density)
From my understanding and how Im doing the problem the density cancels out, which is were Im confused on how it will affect my answer, thats were Im probably messing up but my textbook doesnt shine much light on problems with non-uniform density. Can someone please help me out?
-- Not sure what you have in mind --you are treating the density as if it were constant--you are correct about the x coordinate
but be careful. In general you need symmetry over the region and the density function over this region to make this assumption. Here 7x^2 is symmetric over the region so x = 0 is the x coordinate of the center of mass.
for Calculation of the y coordinate see attachment
By the way the density is integrated over the region for the mass and y times the density is integrated
over the region for the moment.
The density does not cancel