# Homework Help: Find M, Mx, and My for a thin plate if density=ky

1. Nov 19, 2012

### cloudboy

1. The problem statement, all variables and given/known data
Set up integrals to find M, Mx, and My for the thin plate if d(x)=ky

bounded by y=2-x and y = 4-x^2 or x = sqrt(4-y)

2. Relevant equations

M = $\int$ dm = $\int$d(x)dA
Mx = $\int$ydm
My = $\int$xdm

3. The attempt at a solution
I broke this down into three equations. x1 = 2-y; x2 = -sqrt(4-y), x3= sqrt(4-y)
I used horizontal elements so that d(x) is constant across the entire element.

When calculating M, I got $\int$(ky)(x2-x1)dy = $\int$(0→3)(ky)(sqrt(4-y))-(2-y)dy + $\int$(3→4)(ky)(sqrt(4-y))-(-sqrt(4-y))dy

I got this down to:

k$\int$(0→3)(y2-2y+ysqrt(4-y))dy + 2k$\int$(3→4)(ysqrt(4-y))dy

The problem is basically that I can't figure out how to integrate these to find the total mass. Please help thanks!

2. Nov 19, 2012

### haruspex

I feel it would be much simpler to handle this as a double integral, integrating wrt y first. That will avoid surds in the bounds.

3. Nov 19, 2012

### cloudboy

It very well may, I just haven't learned anything about that yet.

EDIT -- I am looking up double integrals, how would you apply it here?

Last edited: Nov 19, 2012
4. Nov 19, 2012

### haruspex

In x-y co-ordinates, an element of area is dxdy (or dydx - either will do).
So the mass of lamina covering a region A is $\int_{(x,y) \in A}ρ(x,y) dx dy$. If the area is bounded by f(x)<y<g(x) and a<x<b then you can write M = $\int_{x=a}^b\int_{y=f(x)}^{g(x)}\rho(x,y) dy dx$

5. Nov 19, 2012

### cloudboy

Ok I'm in my first calculus class and some of unfortunately I don't even recognize some of your notation. But I do understand somewhat and will try to use this. Thanks very much.

ok i am trying to figure this out. but my density function is ky, where k is some constant. it doesnt have an x value

Last edited: Nov 19, 2012
6. Nov 20, 2012

### HallsofIvy

The plate is bounded by $y= 4- x^2$ and by $y= 2- x$ which intersect when $y= 4- x^2= 2- x$ or $x^2- x- 2= (x- 2)(x+ 1)= 0$

With density ky, the mass is given by
$$M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx$$

7. Nov 20, 2012

### cloudboy

ok I watched a couple videos on youtube and this is what I came up with:

$$M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx$$
⇔ k$\int_{y=2-x}^{4- x^2} ydy$
= k (y2/2) = k [(4-x^2)/2 - (2-x)/2]
= k/2 [4-x^2 - 2+x]
= k/2 [-x^2+x-2]

then

$$M= \int_{x=-1}^2 (k/2 [-x^2+x-2]) dx$$
= (k/2)[(-x^3)/3 + (x^2)/2 +2x]from -1->2
= ... = 9k/4

Last edited: Nov 20, 2012
8. Nov 20, 2012

### haruspex

That's My, right?
No, you've dropped a power of 2. Setting y = 4- x2 in y2/2 gives (4- x2)2/2

9. Nov 21, 2012

### cloudboy

Ok I did it again making sure to get the powers correct and I got M = 44k/5.
I calculated My to be 9k/2.

When I divided My/M to get xbar I got 88/45 which can't be right!

10. Nov 21, 2012

### haruspex

I don't get quite the same. I have M = 31/6, My = 549/310, giving ybar (not xbar!) about 1.8. That's quite close to what you have, and seems perfectly reasonable to me.

11. Nov 21, 2012

### cloudboy

That's strange. I thought xbar = My/M ? Maybe Im getting my limits messed up, I evaluated My from 0->3, because from 3->4 is symmetric.

12. Nov 21, 2012

### haruspex

No. ∫∫ydydx/∫∫dydx must be computing an average value of y.
What is from 0 to 3? x? The x range is determined by where the two curves meet.