The boundary of a lamina consists of the semicircles $y = \sqrt{1 - x^2}$ and $y = \sqrt{9 - x^2}$ together with the portions of the $x$-axis that join them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.

Neetesh Kumar | November 27, 2024

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This is the solution to Math 1D
Assignment: 15.4 Question Number 8
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Step-by-step solution:

Step 1: Understand the region and density

The lamina lies between two semicircles:

• The inner semicircle is $y = \sqrt{1 - x^2}$ with radius $1$,
• The outer semicircle is $y = \sqrt{9 - x^2}$ with radius $3$,
• The lamina is symmetric about the $y$-axis and includes the portions of the $x$-axis connecting the semicircles.

The density $\rho(x, y)$ is inversely proportional to the distance from the origin. The distance from the origin to any point $(x, y)$ is $r = \sqrt{x^2 + y^2}$, so:

$\rho(x, y) = \frac{k}{\sqrt{x^2 + y^2}}$

for some constant $k$.

Step 2: Center of mass formulas

The center of mass $(\bar{x}, \bar{y})$ is given by:

$\bar{x} = \frac{1}{m} \iint_D x \rho(x, y) \, dA, \quad \bar{y} = \frac{1}{m} \iint_D y \rho(x, y) \, dA$

where the mass $m$ is:

$m = \iint_D \rho(x, y) \, dA$

Due to symmetry, $\bar{x} = 0$, as the region and density are symmetric about the $y$-axis. We only need to compute $\bar{y}$.

Step 3: Convert to polar coordinates

In polar coordinates:

• $x = r\cos\theta$, $y = r\sin\theta$,
• $x^2 + y^2 = r^2$,
• $dA = r \, dr \, d\theta$.

The bounds for the region are:

• $r \in [1, 3]$ (between the two semicircles),
• $\theta \in [0, \pi]$ (semicircles span the upper half-plane).

Substitute $\rho(x, y) = \frac{k}{r}$ into the mass integral:

$m = \int_0^{\pi} \int_1^3 \frac{k}{r} \cdot r \, dr \, d\theta = \int_0^{\pi} \int_1^3 k \, dr \, d\theta$

Step 4: Compute the mass

Evaluate the $r$-integral:

$\int_1^3 k \, dr = k \int_1^3 1 \, dr = k [r]_1^3 = k(3 - 1) = 2k$

Now, integrate with respect to $\theta$:

$m = \int_0^{\pi} 2k \, d\theta = 2k \int_0^{\pi} 1 \, d\theta = 2k \cdot \pi = 2k\pi$

Step 5: Compute $\bar{y}$

The formula for $\bar{y}$ is:

$\bar{y} = \frac{1}{m} \iint_D y \rho(x, y) \, dA$

In polar coordinates, $y = r\sin\theta$ and $\rho(x, y) = \frac{k}{r}$, so:

$\bar{y} = \frac{1}{m} \int_0^{\pi} \int_1^3 r\sin\theta \cdot \frac{k}{r} \cdot r \, dr \, d\theta$

Simplify:

$\bar{y} = \frac{1}{m} \int_0^{\pi} \int_1^3 k r \sin\theta \, dr \, d\theta$

Separate the integrals:

$\bar{y} = \frac{1}{m} \cdot k \int_0^{\pi} \sin\theta \, d\theta \cdot \int_1^3 r \, dr$

Step 6: Evaluate the integrals

First term ($\theta$-integral):

$\int_0^{\pi} \sin\theta \, d\theta = -\cos\theta \Big|_0^{\pi} = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2$

Second term ($r$-integral):

$\int_1^3 r \, dr = \left[\frac{r^2}{2}\right]_1^3 = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4$

Step 7: Compute $\bar{y}$

Substitute the results into the formula for $\bar{y}$:

$\bar{y} = \frac{1}{m} \cdot k \cdot 2 \cdot 4$

Substitute $m = 2k\pi$:

$\bar{y} = \frac{1}{2k\pi} \cdot k \cdot 8 = \frac{8}{2\pi} = \frac{4}{\pi}$

Final Answer:

The center of mass of the lamina is: $(\bar{x}, \bar{y}) = \boxed{\left(0, \frac{4}{\pi}\right)}$

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