The problem reads:

Provided the region $ C = \{ \left( x, y \right) | x^2 + y^2 \le 1 \}$, find the mass and center of mass of $C$ if the density $ \rho \left( x, y \right) $ of each point in $C$ is proportional to the distance of the point from the $y$-axis.

The region described by $C$ is a circle with radius 1 centered at the origin and containing all points within it.

In order to find the mass of $C$ with respect to the given $ \rho \left( x, y \right) $, I used two sets of double integrals in an attempt to determine the mass: one in rectangular coordinates and one in polar coordinates.

For rectangular coordinates, $ \rho \left( x, y \right) = kx $, where $ k \gt 0 $.

For polar coordinates, $ \rho \left( r, \theta \right) = kr \cos{\theta} $ , where $ k \gt 0 $. $ (x \to r \cos{\theta}) $

\begin{align} \text{mass of C} & = \iint_C \, kx \, dA \\ & = 2 \left( \int_{-1}^1 \int_0^\sqrt{1-x^2} kx \, dy \, dx \right) \\ & = 2 \left( \int_{0}^{2\pi} \int_0^1 kr^2 \cos{\theta} \, dr \, d\theta \right) \end{align}

Solving both double integrals yields a mass of $0$.

However, in order to yield the center of mass of $C$, I would have to determine the first moment of the mass about the $y$-axis $ M_y = \iint_C \, x \rho \left( x, y \right) \, dA $ and the first moment of the mass about the $x$-axis $ M_x = \iint_C \, y \rho \left( x, y \right) \, dA $.

The coordinates for the center of mass are

$$ \left( \frac{M_y}{mass}, \frac{M_x}{mass} \right) $$

Attempting to solve for the $x$- and $y$-coordinates cannot be done since the mass is $0$.

Is there something wrong with how I set up the problem? Is it preferable to use one coordinate system over another?

  • 1
    $\begingroup$ Your density function gives negative values for points left of the $y$-axis. $\endgroup$ – Umberto P. Mar 22 '17 at 19:36

Your error lies on the inside of the double integral. Because distance should always be a non-negative number, the density function wouldn't be $kx$, but rather $k |x|$. You would then need to set up two separate double integrals and then your mass wouldn't be zero


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