Final answer:
To find the center of mass of a rectangular object with a non-uniform density ρ(x, y) = ρ0xy, we integrate to find both the total mass and the first moments about the x and y axes. The x-coordinate and y-coordinate of the center of mass are found to be a/3 and b/3, respectively, making the correct answer (b) (a/3, b/3).
Step-by-step explanation:
The question is asking for the center of mass of a rectangular object with non-uniform density ρ(x, y) = ρ0xy when placed in the xy-plane. To find the center of mass, we use the formulas for the x-coordinate and y-coordinate of the center of mass, which are given by:
\[\bar{x} = \frac{\int{xdm}}{M}\]
\[\bar{y} = \frac{\int{ydm}}{M}\]
Here, dm = ρ(x, y) dA is the infinitesimal mass element where dA is the differential area element and M is the total mass of the rectangle. The density function ρ(x, y) = ρ0xy suggests that the mass varies linearly along both the x and y axes.
Let's calculate the total mass M first:
\[M = \int_0^b{\int_0^a{ρ0xy dxdy}}\]
\[M = ρ0 \cdot \left(\frac{a^2}{2}\right) \cdot \left(\frac{b^2}{2}\right)\]
Then, the x-coordinate of the center of mass will be:
\[\bar{x} = \frac{1}{M} \int_0^b{\int_0^a{x ρ0xy dxdy}}\]
\[\bar{x} = \frac{1}{M} ρ0 \cdot \left(\frac{a^3}{3}\right) \cdot \left(\frac{b^2}{2}\right)\]
\[\bar{x} = \frac {a}{3}\]
Similarly, for y-coordinate of the center of mass:
\[\bar{y} = \frac{1}{M} \int_0^b{\int_0^a{y ρ0xy dxdy}}\]
\[\bar{y} = \frac{1}{M} ρ0 \cdot \left(\frac{a^2}{2}\right) \cdot \left(\frac{b^3}{3}\right)\]
\[\bar{y} = \frac{b}{3}\]
Therefore, the correct answer is (b) (a/3, b/3).