Video Transcript
Find the center of mass of the
region 𝑅 the set of points 𝑥, 𝑦 such that 𝑦 is greater than or equal to zero, 𝑥
is greater than or equal to zero, and one is less than or equal to 𝑥 squared plus
𝑦 squared less than or equal to four with the given density function 𝜌 of 𝑥, 𝑦
equals the square root of 𝑥 squared plus 𝑦 squared.
Recall that to find the center of
mass of a 2D lamina shape, we need to balance the moments of the lamina about two
perpendicular axes. Usually we pick the 𝑥- and 𝑦-axes
for simplicity. Specifically, if we imagine that
the total mass of the lamina is concentrated at a single point, the object’s center
of mass, then the moment of this single point is equal to the sum of all the
individual moments of the area elements of the lamina. Algebraically, about the 𝑥-axis,
for instance, the total mass 𝑀 times the 𝑥-coordinate of the center of mass, 𝑥
naught, is equal to the sum of all of the individual products of the masses of the
area elements, 𝛿𝑚, times their 𝑥-coordinates, 𝑥.
If we let the size of these area
elements tend to zero, then this sum becomes an integral over the region 𝑅 of 𝑥
times d𝑚. And dividing through by the total
mass on both sides, this gives us the 𝑥-coordinate of the center of mass, 𝑥
naught. However, we cannot currently
evaluate this since we don’t know what the total mass is, and we cannot integrate 𝑥
with respect to 𝑀. Since we have a density function,
d𝑚 is just the density at this point times the size of the area element d𝐴.
Similarly, the total mass is just
the integral over the region 𝑅 of all of the mass elements d𝑚. And we can once again reexpress d𝑚
as 𝜌 d𝐴. This gives us the formula for the
𝑥-coordinate of the center of mass, 𝑥 naught, equal to the double integral over
the region 𝑅 of 𝜌𝑥 d𝐴 divided by the double integral over the region 𝑅 of 𝜌
d𝐴. We can do the same thing for the
𝑦-coordinate just with 𝑦 replacing 𝑥. Let’s place these formulae up here
to give us some space.
We now need to consider our region
𝑅 and our density function 𝜌. So first of all, 𝑅 is restricted
to 𝑦 greater than or equal to zero and 𝑥 greater than or equal to zero. So this is restricted to the upper
right quadrant of the 𝑥𝑦-plane. And secondly, 𝑅 is restricted to
one less than or equal to 𝑥 squared plus 𝑦 squared less than or equal to four. 𝑥 squared plus 𝑦 squared is the
square of the distance from the origin. So the square of the distance from
the origin must be between one and four. Therefore, the distance from the
origin must be between one and two. So the region 𝑅 looks something
like this, a quarter annulus in the upper right quadrant of the 𝑥𝑦-plane with
inner radius one and outer radius two.
Now, we could proceed with the
integration in the Cartesian system. But it is in fact easier to convert
everything into polar coordinates first, especially because the density function 𝜌
is defined very simply in polar coordinates. In polar coordinates, 𝑟 is the
distance of a point from the origin. And 𝜃 is the argument the point
makes with the positive 𝑥-axis. So 𝑥 is equal to 𝑟 cos 𝜃 and 𝑦
is equal to 𝑟 sin 𝜃, which in turn means that 𝑟 squared is equal to 𝑥 squared
plus 𝑦 squared. Since 𝜌 is equal to the square
root of 𝑥 squared plus 𝑦 squared, in polar coordinates, 𝜌 is simply equal to
𝑟. Similarly, defining the limits of
the integration will be much simpler in polar coordinates. The limits of both 𝑟 and 𝜃 will
be constant and independent. So 𝑟 one equals one, 𝑟 two equals
two, 𝜃 one equals zero, and 𝜃 two equals 𝜋 by two.
Lastly, the area element d𝐴 is
given in polar coordinates by 𝑟 d𝑟 d𝜃. So starting with the 𝑥-coordinate,
substituting these expressions into our formula for 𝑥 naught, we get 𝑥 naught
equals the integral between 𝜃 equals zero and 𝜃 equals 𝜋 by two and between 𝑟
equals one and 𝑟 equals two of 𝑟𝑥 times the area element 𝑟 d𝑟 d𝜃 all divided
by the same integral between 𝜃 equals zero and 𝜋 by two and 𝑟 equals one and two
but just 𝑟 times the area element 𝑟 d𝑟 d𝜃.
Now we cannot currently integrate
this since we cannot integrate 𝑥 with respect to 𝑟 and 𝜃. But remember that 𝑥 is equal to 𝑟
cos 𝜃, so we can substitute in this expression for 𝑥. Doing this and tying up both double
integrals gives us the integral between zero and 𝜋 by two of cos 𝜃 d𝜃 times the
integral between one and two of 𝑟 cubed d𝑟 all over the integral between zero and
𝜋 by two of d𝜃 times the integral between one and two of 𝑟 squared d𝑟. Integrating with respect to 𝑟 on
the numerator and denominator gives us one-quarter times 𝑟 to the fourth evaluated
between one and two and one-third times 𝑟 cubed evaluated between one and two.
Evaluating and taking these
constants outside of the integration gives us 15 over four on the numerator and
seven over three on the denominator. Now integrating with respect to 𝜃
gives us sin 𝜃 evaluated between zero and 𝜋 by two on the numerator and just 𝜃
evaluated between zero and 𝜋 by two on the denominator. Evaluating gives us 15 over four
times one minus zero over seven over three times 𝜋 by two minus zero. And simplifying gives us the
𝑥-coordinate of the center of mass, 45 over 14𝜋.
Now we could use the exact same
procedure to find the 𝑦-coordinate of the center of mass. But notice that this lamina is
entirely symmetric about the 𝑥- and 𝑦-axes. There is nothing to distinguish the
𝑥-coordinate from the 𝑦-coordinate. Even the density function is given
by the square root of 𝑥 squared plus 𝑦 squared. So the dependence on 𝑦 is exactly
the same as the dependence on 𝑥. Therefore, the 𝑦-coordinate of the
center of mass, 𝑦 naught, must be exactly the same as the 𝑥-coordinate, 45 over
14𝜋. Therefore, the center of mass of
this shape is at 45 over 14𝜋, 45 over 14𝜋.
