I think my prof wants me to convert it to polar coordinates but I can't find the bounds for θ, I found π/4 but then I got arcsin(1/3) and arcos(1/3) maybe I'm just stuck and blind.
#Can't find the bounds for θ
winter basin
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I think my prof wants me to convert it to polar coordinates but I can't find the...
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winter basin
Kinda messy sorry
warm scarab
I think my prof wants me to convert it to polar coordinates but I can't find the...
Try drawing the region first. Then you'll see that it's easier to split it into two pieces.
And since the given (what I assume is) density is radially symmetric, you can just consider one half of the region.
winter basin
Yeah it's for finding the centroid or center of mass
That's what I have to do
I've thought of something, red dot is what I thought of yesterday, purple dot is something i thought of today
warm scarab
I see.
We might be able to simplify the problem a bit by rotating the region a bit, I think.
The density is pretty symmetric, after all.
winter basin
Hmm rotating?
warm scarab
So, here's the region, right?
winter basin
Yep need to find the centeroid
warm scarab
What if we do this, instead?
So, just rotate it a bit.
We can do this, since the density is radially symmetric.
Now it's obvious that the center of mass lies on the x-axis.
warm scarab
Well, as usual.
Suppose the initial region is D and the region in polar coordinates is D'. Then:
m = ∫∫(σ dxdy, D) = ∫∫(σr drdθ, D')
m_x = ∫∫(σx dxdy, D) = ∫∫(σr^2 cos(θ) drdθ, D')
m_y = ∫∫(σy dxdy, D) = ∫∫(σr^2 sin(θ) drdθ, D'); no need to find it in our case.
Then r_c = {m_x/m, m_y/m}.
In our case in polar coordinates σ = r.
winter basin
The bounds looks complicated
warm scarab
Well, the upper one is easy.
The lower one you can derive from the equation of the line.
I don't think this can be made any easier.
Translating the region isn't going to help, since the density isn't translationally symmetric.
winter basin
So for r, it's 1≤r≤3?, for θ hmm
winter basin
Then √2
warm scarab
No. The lower bound isn't a circle.
Let's look at the top line.
Its equation is y = x - √(2).
Substitute polar coordinates and try expressing r.
winter basin
r=(-√2)/sin(θ)-cos(θ)
warm scarab
No, not quite right.
warm scarab
So for r, it's 1≤r≤3?, for θ hmm
As for θ, you can just find the intersection point of one of the lines and the circle, then find θ from there.
winter basin
I feel like I'm wrong, I got
3π/8 ≤θ≤ 5π/8
Honestly I'm new to radians
We had to work with radians because of those polar coords, and it's kinda my first time doing them for calculus
warm scarab
I feel like I'm wrong, I got
3π/8 ≤θ≤ 5π/8
Oh, you switched to the initial region again?
winter basin
That wasn't my intension but damn
warm scarab
Well, let's work with the final region first.
Let's find the intersection point. We have a system:
x^2 + y^2 = 9
y = x - √(2)
Taking x = 3cos(θ), y = 3sin(θ), we get:
cos(θ) - sin(θ) = √(2)/3
cos(θ + π/4) = 1/3
θ + π/4 = ±arccos(1/3) + 2πn, n ∈ ℤ
θ = -π/4 ± arccos(1/3) + 2πn, n ∈ ℤ
Our angle of interest is in the first quadrant. It's easy to see that it will be θ = arccos(1/3) - π/4.
And it is indeed that angle, as you can see.
winter basin
Oh
warm scarab
So, now figure out the lower limit corresponding to the straight line, and it's just straight calculation from there.
You only need to compute two integrals, since we obviously know that the y-coordinate of the center of mass is 0.
warm scarab
Oh, moreover, you can only look at the top half of this piece for the integrals.
Its center of mass won't lie on the x-axis, but since we're looking for its x-coordinate, it will be the same as for the larger piece.
If it's not obvious why that is so, I can elaborate.
winter basin
Oh, moreover, you can only look at the top half of this piece for the integrals....
It's kinda confusing; center of mass is not on x axis but we are looking for it's point in x
warm scarab
Alright, let me explain.
Here's a rough sketch of our region.
Here it is with the centers of mass of each half.
Notice that they lie on the same vertical line. Do you see why that's true?
winter basin
Yeah
warm scarab
Right.
And now recall the theorem about the center of mass of a compound body:
r(c) = (m(1) r(c, 1) + m(2) r(c, 2))/(m(1) + m(2))
So, the center of mass of a body made from several others is just the arithmetic mean of the centers of mass of each piece, weighted by their mass.
In our case it's pretty obvious that m(1) = m(2). So:
r(c) = (r(c, 1) + r(c, 2))/2
So, this is the center of mass of the whole piece.
But notice how now all three points lie on the same vertical line.
winter basin
Ohhh I see qhere your going now
warm scarab
Yeah. Essentially, we're simplifying the problem by doing three things:
- Rotating the piece (which is allowed due to radially symmetric density) so that it's aligned with the x-axis.
- Noticing that due to symmetry the y-coordinate of the resulting piece's center of mass is 0.
- Noticing that, again, due to symmetry the x-coordinate of the resulting piece's center of mass is the same as that of its top half. It's easier to work with, since it has the same lower and upper limits in polar coordinates.
So, now all that's left is to calculate two integrals - ∫(dS) and ∫(xdS) - over the top half, which will allow us to find the x-coordinate of the center of mass.
And if you specifically want the coordinates of the center of mass of the piece as it is given initially, just rotate the resulting point back by π/4.
winter basin
Idk if its me but sounds like a bunch of extra steps
warm scarab
Well, these three steps don't really require you to do anything in writing, but they simplify the integrals and reduce their amount by one.
You can do it without this, of course, but I think that will take longer.
winter basin
Anyway I passed my homework with
arctan(1/2√2)≤ θ ≤ arctan(2√2) so now I feel like I got the right answer. I hope
winter basin
You can do it without this, of course, but I think that will take longer.
Yeah it is kinda long
warm scarab
Anyway I passed my homework with
arctan(1/2√2)≤ θ ≤ arctan(2√2) so now I feel li...
Oh, for the initial shape? Let me check...
winter basin
Yeah
winter basin
Well it's long so; but I'm sure my prof will be OK with it
warm scarab
Well, I got to this, and now I'm too lazy to continue 😅
And yeah, considering the substitution, I think it is indeed better not to rotate.
Or not... Not sure.
Anyway, it's not very hard, just takes a while.
Nah, it's still better to rotate, I think, since it's easier to exploit symmetry that way.
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@winter basin
winter basin
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