# Finding the bounds for double integrals

• Mar 6th 2008, 05:50 PM
dukebdx12
Finding the bounds for double integrals
I do not need help finding the answer. My problem is finding the bounds for the integral. I'm not sure how to find them and am a little confused. Could someone explain how to find these bounds for the 2 problems please.

1) Find the center of mass of the lamina that occupies the region D and has the given density function if D is bounded by the parabola y=64-^2 and the x-asix. p(x,y)=y

2) A lamina occupies the part of the disk x^2+y^2<=100 in the first quadrant. Find its center of mass if the density at any points is proportional to the D of its distance from the origin.

- I found my bounds to be 0,pi/2 and 0,10? When I found M,My, and Mx I did not get the correct answer. I wasn't sure what I was doing on it. I was following an example in the book and they had a similar problem but it was <=1 and solving for M they got Kr^2sin(theta). For My they got Kr^3sin(theta)cos(theta). And for Mx they got kr^3 sin(theta)^2. My question is how did they find the function of each of those for those integrals?
• Mar 6th 2008, 07:49 PM
ThePerfectHacker
Quote:

Originally Posted by dukebdx12
1) Find the center of mass of the lamina that occupies the region D and has the given density function if D is bounded by the parabola y=64-^2 and the x-asix. p(x,y)=y

The upper curve is $y=64-x^2$ and the lower curve is x-axis: $y=0$. Thus, $D = \{(x,y)| -8\leq x\leq 8 \mbox{ and }0\leq y\leq 64 - x^2 \}$.

Quote:

2) A lamina occupies the part of the disk x^2+y^2<=100 in the first quadrant. Find its center of mass if the density at any points is proportional to the D of its distance from the origin.
Note $p(x,y)$ is proportional to the distance, which is $\sqrt{x^2+y^2}$. Which just means $p(x,y) = k\sqrt{x^2+y^2}$, for some constant (non-zero hopefully) of proportionality $k$. Here use polar transform.