Find M, Mx, and My for a thin plate if density=ky

[cloudboy]

Homework Statement

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)

Homework Equations

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

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!

 

[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.

 

[cloudboy]

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.

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?

 

[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 = $\int_{x=a}^b\int_{y=f(x)}^{g(x)}\rho(x,y) dy dx$

 

[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 doesn't have an x value

 

[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
$$M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx$$

 

[cloudboy]

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
$$M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx$$

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
how bad is it?

 

[haruspex]

$$M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx$$

That's M_y, right?

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

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

 

[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!

 

[haruspex]

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!

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.

 

[cloudboy]

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.

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

 

[haruspex]

That's strange. I thought xbar = My/M ?

No. ∫∫ydydx/∫∫dydx must be computing an average value of y.

Maybe I am getting my limits messed up, I evaluated My from 0->3, because from 3->4 is symmetric.

What is from 0 to 3? x? The x range is determined by where the two curves meet.