Double integration of circle

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  • #1
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Homework Statement



I would like to ask how to find the mass of a circle with equation x^2+y^2=4
given its density=xy^2
by not using polar coordinate
but use dxdy or dydx ( cut the circle into pieces parallel to x-axis or y-axis )


Homework Equations



x^2+y^2=4

xy^2

The Attempt at a Solution

 

Answers and Replies

  • #2
chiro
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Hey sssc and welcome to the forums.

From physics, the mass of an object is the density at a point times the volume element at that point. In calculus we shrink the volume to an infinitesimal limit and the density refers to the density of each infinitesimal volume element.

The calculus based equation for volume is Integral [pdV] over a region of Volume where p is the function for density at each point and dV is with respect to the volume.

In your question you are using a circle which is odd (density is based on volume not area). Is this intended or is there a mistake?

If its not a mistake then integrate it over an area (instead of dV its dA) where dA = dydx and the region of integration is the area bounded by the circle.

Using this hint can you set up your integral? Just show us any working you have done as well as what you think the integral should be: doesn't matter if its not right, just as long as you post it so we can see what your thinking.
 
  • #3
LCKurtz
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Hey sssc and welcome to the forums.

From physics, the mass of an object is the density at a point times the volume element at that point. In calculus we shrink the volume to an infinitesimal limit and the density refers to the density of each infinitesimal volume element.

The calculus based equation for volume is Integral [pdV] over a region of Volume where p is the function for density at each point and dV is with respect to the volume.

In your question you are using a circle which is odd (density is based on volume not area). Is this intended or is there a mistake?
This type of problem frequently talks about a "thin circular lamina" and gives the density in units of mass or weight / area. So it's an area density and the density element is ##\rho dA## as you suspected.
 
  • #4
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The question is here:
A lamina covering the quarter circle x^2 + y^2 </= 4 ,x>0, y>0, has (area) density x+y.
Find the mass of the lamina. Answer=16/3

But I change it to finding the whole circle

I do it like this:

cutting the circle into columns parallel to the x-axis

∫ ∫ (x+y) dx dy

where
x: - (4-y^2)^0.5 -> (4-y^2)^0.5
y: -2 -> 2

but I found that the answer is zero.

So I try:

y: 0 -> 2 and then multiply the integral by 2.

But I get 32/3 which is not 64/3 as expected.
 
  • #5
chiro
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You need to use the definition to get the mass. Integrate xy^2dxdy inside the region of the circle where your limits are defined by x^2 + y^2 = 4.

Do you know how to integrate over a region like this? Have you done regions of integration with double integrals?
 
  • #6
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sorry, I think I typed it wrongly at first, it is x+y but not xy^2 for the density.
Yes . I just do the integration but dont know why the answer is incorrect.

Can anyone help me to check this?

2*∫ ∫ (x+y) dx dy

where
x: - (4-y^2)^0.5 -> (4-y^2)^0.5
y: 0 -> 2

=2*∫ [ ((x^2)/2+xy)] dy

where
x: - (4-y^2)^0.5 -> (4-y^2)^0.5
y: 0 -> 2

=2*∫ 2*y*((4-y^2)^0.5) dy

where
y: 0 -> 2

sub 4-y^2 into dy

=-2 *∫((4-y^2)^0.5) d(4-y^2)


where
y: 0 -> 2


=32/3
 
Last edited:
  • #7
LCKurtz
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The question is here:
A lamina covering the quarter circle x^2 + y^2 </= 4 ,x>0, y>0, has (area) density x+y.
Find the mass of the lamina. Answer=16/3

But I change it to finding the whole circle
But if you change it to the whole circle, your ##x+y## is no longer a density function because it isn't positive on the whole area. I would expect an answer of ##0## because of the symmetry with that integrand.
 
  • #8
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Thanks a lot.
 

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