Consider finding the mass of some shape. The general form is ∫σ.dA. In Cartesian, that's ∫σ.dxdy.Davidllerenav said:But as you can see in here, I ended up with square roots on the left:
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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 said: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=-7x=0y=0y=5Axy.dxdy.
See if you can solve that and do the right hand portion and the numerator integrals the same way.
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.Davidllerenav said:I'm new to the doble integral notation, but I think that it would be ##\int_{0}^2\int_{1}^5 Axy\cdot dydx##.
Sorry, but I don't understand what you mean.haruspex said: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?
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.Davidllerenav said:Sorry, but I don't understand what you mean.
The range of the y values are from 5-y to 5.haruspex said: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?
No, the range of y can't depend on y. It depends on x.Davidllerenav said:The range of the y values are from 5-y to 5.
Then the range would be from x=0 to X=2, right?haruspex said:No, the range of y can't depend on y. It depends on x.
No, you are looking for a range of y as a function of x.Davidllerenav said:Then the range would be from x=0 to X=2, right?
For y at the top the value is 7, and for y at the bottom it is 7-y. Right?haruspex said: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?
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.Davidllerenav said:For y at the top the value is 7, and for y at the bottom it is 7-y. Right?
Y in terms of x would be function, right? ##y=x^2+1##?haruspex said: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?
Right, so that is your lower bound for y in that strip. The y integral is ##\int_{y=x^2+1}^5Axy.dy##.Davidllerenav said:Y in terms of x would be function, right? ##y=x^2+1##?
Ok, and should I change that on every integral? Even on part A?haruspex said:Right, so that is your lower bound for y in that strip. The y integral is ##\int_{y=x^2+1}^5Axy.dy##.
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+1y=5σ.dy).dxDavidllerenav said:Ok, and should I change that on every integral? Even on part A?
So I am correct on part A, I just need to correct part B, right?haruspex said: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+1y=5σ.dy).dx
Your part A is ok.Davidllerenav said:So I am correct on part A, I just need to correct part B, right?
Ok, thanks. And for part B I just need to correct ##Y_{cm}##? Or ##X_{cm}## too?haruspex said:Your part A is ok.
You need to use double integrals everywhere in part B.Davidllerenav said:Ok, thanks. And for part B I just need to correct ##Y_{cm}##? Or ##X_{cm}## too?
What do you mean by doing the y integral first? Why would that make me avoid square roots?haruspex said:You need to use double integrals everywhere in part B.
By always doing the y integral first you may avoid the square roots.
You have a double integral in y and x. You can do either first.Davidllerenav said:What do you mean by doing the y integral first? Why would that make me avoid square roots?
I didn't know I was able to iintegrate double integrals in any order. It's like partial derivatives, right?haruspex said: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.
Yes.Davidllerenav said:I didn't know I was able to iintegrate double integrals in any order. It's like partial derivatives, right?