Find the center of mass of a plate

[Davidllerenav]

You can run the integral from 0 if you wish, but the constraint 0<x²<y-1 only applies for y>1. So you need to break the integral into separate ranges 0 to 1 and 1 to 5. The 0 to 1 range is trivially null.

So I just need to integrate from 1 to 5 on both integrals?

 

[haruspex]

So I just need to integrate from 1 to 5 on both integrals?

The numerator and denominator integrals for x>0, yes.

 

[Davidllerenav]

The numerator and denominator integrals for x>0, yes.

Ok, so just to check, the two integrals on the left must be from 1 to 5? So I will have four integrals from 1 to 5?

 

[haruspex]

Ok, so just to check, the two integrals on the left must be from 1 to 5? So I will have four integrals from 1 to 5?
View attachment 240464

No, some confusion here.
You have integrals that relate to the 7x5 rectangle on the left (x<0) and integrals that relate to the curved flange on the right (x from 0 to 2).
Those for the right involve the constraint x²>y-1 and therefore only apply for y=1 to 5.
Those for the left apply for y=0 to 5, but since they do not involve the constraint x²<y-1 they should not contain any square roots so should not give imaginary nunbers.

 

[Davidllerenav]

No, some confusion here.
You have integrals that relate to the 7x5 rectangle on the left (x<0) and integrals that relate to the curved flange on the right (x from 0 to 2).
Those for the right involve the constraint x²>y-1 and therefore only apply for y=1 to 5.
Those for the left apply for y=0 to 5, but since they do not involve the constraint x²<y-1 they should not contain any square roots so should not give imaginary nunbers.

But as you can see in here, I ended up with square roots on the left:

 

[haruspex]

But as you can see in here, I ended up with square roots on the left:
View attachment 240465

Consider finding the mass of some shape. The general form is ∫σ.dA. In Cartesian, that's ∫σ.dxdy.
In the present problem, σ=Axy, so ∫∫Axy.dxdy. For the rectangle (x<0) the bounds are easy: ∫_x=-7^x=0_y=0^y=5Axy.dxdy.
See if you can solve that and do the right hand portion and the numerator integrals the same way.

 

[Davidllerenav]

Consider finding the mass of some shape. The general form is ∫σ.dA. In Cartesian, that's ∫σ.dxdy.
In the present problem, σ=Axy, so ∫∫Axy.dxdy. For the rectangle (x<0) the bounds are easy: ∫_x=-7^x=0_y=0^y=5Axy.dxdy.
See if you can solve that and do the right hand portion and the numerator integrals the same way.

I'm new to the doble integral notation, but I think that it would be $\int_{0}^2\int_{1}^5 Axy\cdot dydx$.

 

[haruspex]

I'm new to the doble integral notation, but I think that it would be $\int_{0}^2\int_{1}^5 Axy\cdot dydx$.

No, that would be a rectangle. You need to relate the bounds on the inner integral to the value of the dummy variable in the outer integral.
E.g. suppose we integrate wrt y first. That means for the purpose of that integral x is a constant. What is the range of y in terms of that x?

 

[Davidllerenav]

No, that would be a rectangle. You need to relate the bounds on the inner integral to the value of the dummy variable in the outer integral.
E.g. suppose we integrate wrt y first. That means for the purpose of that integral x is a constant. What is the range of y in terms of that x?

Sorry, but I don't understand what you mean.

 

[haruspex]

Sorry, but I don't understand what you mean.

Look at the top right diagram in your post #1. You show a vertical band width dx and a horizontal band height dy in the rectangle and another pair in the x>0 portion.
Consider the vertical band in the x>0 area. What is the range of values of y in that band?

 

[Davidllerenav]

Look at the top right diagram in your post #1. You show a vertical band width dx and a horizontal band height dy in the rectangle and another pair in the x>0 portion.
Consider the vertical band in the x>0 area. What is the range of values of y in that band?

The range of the y values are from 5-y to 5.

 

[haruspex]

The range of the y values are from 5-y to 5.

No, the range of y can't depend on y. It depends on x.

 

[Davidllerenav]

No, the range of y can't depend on y. It depends on x.

Then the range would be from x=0 to X=2, right?

 

[haruspex]

Then the range would be from x=0 to X=2, right?

No, you are looking for a range of y as a function of x.
Look at your diagram. For that shaded vertical band, what is the value of y at the top? What is it at the bottom?

 

[Davidllerenav]

No, you are looking for a range of y as a function of x.
Look at your diagram. For that shaded vertical band, what is the value of y at the top? What is it at the bottom?

For y at the top the value is 7, and for y at the bottom it is 7-y. Right?

 

[haruspex]

For y at the top the value is 7, and for y at the bottom it is 7-y. Right?

If y equals 7-y then y=3.5. We don't want y as a function of itself, that's not helpful. We want y as a function of x.

The bottom of the shaded strip lies on the curve y=x²+1. If the X coordinate is x, what is its Y coordinate in terms of x?

 

[Davidllerenav]

If y equals 7-y then y=3.5. We don't want y as a function of itself, that's not helpful. We want y as a function of x.

The bottom of the shaded strip lies on the curve y=x²+1. If the X coordinate is x, what is its Y coordinate in terms of x?

Y in terms of x would be function, right? $y=x^2+1$?

 

[haruspex]

Y in terms of x would be function, right? $y=x^2+1$?

Right, so that is your lower bound for y in that strip. The y integral is $\int_{y=x^2+1}^5Axy.dy$.

 

[Davidllerenav]

Right, so that is your lower bound for y in that strip. The y integral is $\int_{y=x^2+1}^5Axy.dy$.

Ok, and should I change that on every integral? Even on part A?

 

[haruspex]

Ok, and should I change that on every integral? Even on part A?

For part A it was not necessary to do a double integral because the density was constant. You could just write down that the mass of the vertical strip on the right was σ(5-(x²+1))dx. That was instead of the integral (∫_y=x²+1^y=5σ.dy).dx

 

[Davidllerenav]

For part A it was not necessary to do a double integral because the density was constant. You could just write down that the mass of the vertical strip on the right was σ(5-(x²+1))dx. That was instead of the integral (∫_y=x²+1^y=5σ.dy).dx

So I am correct on part A, I just need to correct part B, right?

 

[haruspex]

So I am correct on part A, I just need to correct part B, right?

Your part A is ok.

 

[Davidllerenav]

Your part A is ok.

Ok, thanks. And for part B I just need to correct $Y_{cm}$? Or $X_{cm}$ too?

 

[haruspex]

Ok, thanks. And for part B I just need to correct $Y_{cm}$? Or $X_{cm}$ too?

You need to use double integrals everywhere in part B.
By always doing the y integral first you may avoid the square roots.

 

[Davidllerenav]

You need to use double integrals everywhere in part B.
By always doing the y integral first you may avoid the square roots.

What do you mean by doing the y integral first? Why would that make me avoid square roots?

 

[haruspex]

What do you mean by doing the y integral first? Why would that make me avoid square roots?

You have a double integral in y and x. You can do either first.
If you do x first it will have a bound that depends on y and involves a square root, so the result of the integral may involve a square root, which then becomes part of the integrand for the second integral.
If you do the y integral first it has bounds that involve x but no square root. The result will be a function of x, which you then need to integrate, but it will not have a square root in the integrand.
If you are feeling energetic you can try it both ways and compare.

 

[Davidllerenav]

You have a double integral in y and x. You can do either first.
If you do x first it will have a bound that depends on y and involves a square root, so the result of the integral may involve a square root, which then becomes part of the integrand for the second integral.
If you do the y integral first it has bounds that involve x but no square root. The result will be a function of x, which you then need to integrate, but it will not have a square root in the integrand.
If you are feeling energetic you can try it both ways and compare.

I didn't know I was able to iintegrate double integrals in any order. It's like partial derivatives, right?

 

[haruspex]

I didn't know I was able to iintegrate double integrals in any order. It's like partial derivatives, right?

Yes.