Find the center of mass of a plate

[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.