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Chapter 12: Problem 5

Find the mass and center of mass of the lamina for each density. \(R:\) rectangle with vertices \((0,0),(a, 0),(0, b),(a, b)\) (a) \(\rho=k\) (b) \(\rho=k y\) (c) \(\rho=k x\)

Short Answer

Expert verified
For the constant density \( \rho = k \), the mass is given by \( kab \) and the center of mass lies at \( (\frac{a}{2}, \frac{b}{2}) \). For the case with density \( \rho = ky \), the mass is \( \frac{1}{2}kab^2 \) and the center of mass remains at \( (\frac{a}{2}, \frac{b}{2}) \). For the case with density \( \rho = kx \), the mass becomes \( \frac{1}{2}ka^2b \) and the center of mass is still \( (\frac{a}{2}, \frac{b}{2}) \).

Step by step solution

01

(a): Calculate the mass

First, the mass (M) of the lamina in the case of constant density can be calculated as the double integral of the density over the region R. With \(\rho = k\), the mass is given by \(M = \int\int_R kdA = k \int_0^a \int_0^b dy dx = kab\).
02

(a): Find the center of mass

Next, the coordinates (X, Y) of the center of mass can be found by dividing the integral of rho times x and y by the mass. In this case, they are given by \(X = \frac{1}{M} \int\int_R x \rho dA = \frac{1}{kab} \int_0^a \int_0^b xk dy dx = \frac{a}{2}\) and \(Y = \frac{1}{M} \int\int_R y \rho dA = \frac{1}{kab} \int_0^a \int_0^b yk dy dx = \frac{b}{2}\). For a more complicated rho, we would need to do these separately.
03

(b): Compute the mass with variable density

For the case with \(\rho = ky\), the mass becomes \(M =\int\int_R kydA = k \int_0^a \int_0^b y dy dx = \frac{1}{2}kab^2\).
04

(b): Calculate the center of mass

Then, \(X = \frac{1}{M} \int\int_R x \rho dA = \frac{2}{kab^2} \int_0^a \int_0^b xyk dy dx = \frac{a}{2}\) and \(Y = \frac{1}{M} \int\int_R y \rho dA = \frac{3}{2} \frac{1}{kab^2} \int_0^a \int_0^b y^2k dy dx = \frac{b}{2}\).
05

(c): Calculate the mass with density \( \rho = kx \)

The mass now is \(M =\int\int_R kxdA = k \int_0^a \int_0^b x dx dy = \frac{1}{2}ka^2b\).
06

(c): Find the center of mass

\(X = \frac{1}{M} \int\int_R x \rho dA = \frac{3}{2} \frac{1}{ka^2b} \int_0^a \int_0^b x^2k dx dy = \frac{a}{2}\) and \(Y = \frac{1}{M} \int\int_R y \rho dA = \frac{1}{ka^2b} \int_0^a \int_0^b xyk dx dy = \frac{b}{2}\).

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