Question

A lamina occupies the part of the disk x2 + y2 ≤ 25 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.

Answer 1

Answer:

The center of mass is located at coordinates $(x_0,y_0) = (\frac{15}{8}, \frac{15\pi }{16})$

Step-by-step explanation:

Consider the disk $x^2+y^2\leq 25$ and its' portion located at the first quadrant. We are being said that the density $\rho$ is proportional to the distance of the point to the x-axis. Given a point of coordinates (x,y), its' distance to the x-axis is y. Therefore, we know that the density function $\rho(x,y) = k\cdoty$ for some constant k, whose value is not relevant. Let (x_0,y_0) be the center of mass and let D be the region occupied by the lamina. The center of mass's coordinates fulfill the following equations:

$x_0 = \frac{\int_{D}x\cdot\rho(x,y)dA}{M}$

$y_0 = \frac{\int_{D}y\cdot\rho(x,y)dA}{M}$

where M is the mass of the region, which in this case is given by $M=\int_D \rho(x,y)dA$.

Let us calculate the mass of the lamina. For this, we will use the polar coordinates. Recall that they are given by the change of coordinates $x=r\cos \theta, y = r\sin \theta$, where r and theta are the new parameters. Given a point (x,y) in the plane, r is the distance from (x,y) to the origin and theta is the angle formed between the line that joins the origin and the point, and the x-axis. We want to describe the region D in terms of the new parameters. Replacing the values of x,y in the given inequality, we get that $(r\cos\theta)^2+(r\sin\theta)^2$. Since $\cos^2\theta + \sin^2\theta = 1$ and r>0 we get that r<=5. On the other side, in order to describe the whole region of the first quadrant, we need to sweep the angle theta from 0 to $\frac{\pi}{2}$. With that, we can calculate the mass of the lamina as follows

M = $\int_D \rho(x,y)dA = \int_D ky dA = \int_{0}^{\frac{\pi}{2}}\int_{0}^5 k r \sin \theta \cdot r drd\theta = k (\frac{5^3}{3}-0)\int_{0}^{\frac{\pi}{2}} \sin \theta d\theta =k \cdot \frac{5^3}{3}$

In here, the extra r appears as the jacobian of the change of coordinates (the explanation of why this factor occurs is beyond the scope of this answer. Please refer to the internet for further explanation).

Then, we need to calculate the following integrals.

$\int_{D}x\cdot\rho(x,y)dA = k\int_D xy dA = k \int_{0}^{\frac{\pi}{2}}\int_{0}^5 r\cos \theta \cdot r\sin \theta r dr d\theta =k(\frac{5^4}{4}-0) \int_0^{\frac{\pi}{2}}\frac{\sin(2\theta)}{2}d\theta = k\cdot \frac{5^4}{8}$ (recall $\cos(\theta)\sin\theta = \frac{\sin(2\theta)}{2}$).

Then, $x_0=\frac{k \cdot \frac{5^4}{8}}{k \cdot \frac{5^3}{3}}= \frac{15}{8}$

On the other hand:

$\int_{D}y\cdot\rho(x,y)dA = k\int_D y^2 dA = k \int_{0}^{\frac{\pi}{2}}\int_{0}^5 r\cdot r^2\sin^2 \theta dr d\theta = k(\frac{5^4}{4}-0) \int_{0}^{\frac{\pi}{2}}\frac{1-\cos(2\theta)}{2}d\theta = k \cdot \frac{5^4\pi}{16}$

(recall $\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}$

Then, $y_0 = \frac{k \cdot \frac{5^4\pi}{16}}{k \cdot \frac{5^3}{3}} = \frac{15\pi }{16}$

To check that the answer makes sense, the center of mass must lie in the disk, that is, it should satisfy the equation. We can easily check that $(\frac{15}{8})^2+ (\frac{15\pi }{16})^2=12.19$