The problem asks to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of a lamina bounded by given curves and with a given density function $\delta(x, y)$. Problem 1: $x = 0, x = 4, y = 0, y = 3; \delta(x, y) = y + 1$ Problem 2: $y = 0, y = \sqrt{4 - x^2}; \delta(x, y) = y$ Problem 3: $y = 0, y = \sin x, 0 \le x \le \pi; \delta(x, y) = y$ Problem 4: $y = 1/x, y = x, y = 0, x = 2; \delta(x, y) = x$ Problem 5: $y = e^{-x}, y = 0, x = 0, x = 1; \delta(x, y) = y^2$ Problem 6: $y = e^x, y = 0, x = 0, x = 1; \delta(x, y) = 2 - x + y$

1. Problem Description

The problem asks to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of a lamina bounded by given curves and with a given density function $\delta(x, y)$.

Problem 1: $x = 0, x = 4, y = 0, y = 3; \delta(x, y) = y + 1$

Problem 2: $y = 0, y = \sqrt{4 - x^2}; \delta(x, y) = y$

Problem 3: $y = 0, y = \sin x, 0 \le x \le \pi; \delta(x, y) = y$

Problem 4: $y = 1/x, y = x, y = 0, x = 2; \delta(x, y) = x$

Problem 5: $y = e^{-x}, y = 0, x = 0, x = 1; \delta(x, y) = y^2$

Problem 6: $y = e^x, y = 0, x = 0, x = 1; \delta(x, y) = 2 - x + y$

2. Solution Steps

Problem 1:

Mass $m = \int_0^4 \int_0^3 (y+1) dy dx$

$m = \int_0^4 [\frac{y^2}{2} + y]_0^3 dx = \int_0^4 (\frac{9}{2} + 3) dx = \int_0^4 \frac{15}{2} dx = \frac{15}{2} [x]_0^4 = \frac{15}{2} \cdot 4 = 30$

$M_y = \int_0^4 \int_0^3 x(y+1) dy dx = \int_0^4 x [\frac{y^2}{2} + y]_0^3 dx = \int_0^4 x (\frac{9}{2} + 3) dx = \frac{15}{2} \int_0^4 x dx = \frac{15}{2} [\frac{x^2}{2}]_0^4 = \frac{15}{2} \cdot \frac{16}{2} = 30$

$\bar{x} = \frac{M_y}{m} = \frac{30}{30} = 2$

$M_x = \int_0^4 \int_0^3 y(y+1) dy dx = \int_0^4 \int_0^3 (y^2+y) dy dx = \int_0^4 [\frac{y^3}{3} + \frac{y^2}{2}]_0^3 dx = \int_0^4 (9 + \frac{9}{2}) dx = \frac{27}{2} \int_0^4 dx = \frac{27}{2} [x]_0^4 = \frac{27}{2} \cdot 4 = 54$

$\bar{y} = \frac{M_x}{m} = \frac{54}{30} = \frac{9}{5}$

Problem 2:

$m = \iint_R \delta(x, y) dA = \iint_R y dA$

$m = \displaystyle \int_{-2}^2 \int_0^{\sqrt{4-x^2}} y \, dy \, dx$

$m = \displaystyle \int_{-2}^2 \left[ \frac{y^2}{2} \right]_0^{\sqrt{4-x^2}} dx = \frac{1}{2} \int_{-2}^2 (4-x^2) dx = \frac{1}{2} \left[ 4x - \frac{x^3}{3} \right]_{-2}^2 = \frac{1}{2} \left[ (8-\frac{8}{3}) - (-8 + \frac{8}{3}) \right] = \frac{1}{2} \left[ 16 - \frac{16}{3} \right] = 8 - \frac{8}{3} = \frac{16}{3}$

$M_y = \displaystyle \iint_R x \delta(x, y) dA = \int_{-2}^2 \int_0^{\sqrt{4-x^2}} xy \, dy \, dx = \int_{-2}^2 x \left[ \frac{y^2}{2} \right]_0^{\sqrt{4-x^2}} dx = \int_{-2}^2 x \frac{4-x^2}{2} dx = \frac{1}{2} \int_{-2}^2 (4x - x^3) dx = \frac{1}{2} \left[ 2x^2 - \frac{x^4}{4} \right]_{-2}^2 = \frac{1}{2} \left[ (8-4) - (8-4) \right] = 0$

$\bar{x} = \frac{M_y}{m} = \frac{0}{\frac{16}{3}} = 0$

$M_x = \iint_R y \delta(x, y) dA = \int_{-2}^2 \int_0^{\sqrt{4-x^2}} y^2 dy dx = \int_{-2}^2 \left[ \frac{y^3}{3} \right]_0^{\sqrt{4-x^2}} dx = \frac{1}{3} \int_{-2}^2 (4-x^2)^{3/2} dx$

Let $x = 2\sin \theta$, then $dx = 2 \cos \theta d\theta$. The limits of integration change from $x=-2$ to $\theta = -\frac{\pi}{2}$ and from $x=2$ to $\theta = \frac{\pi}{2}$.

$M_x = \frac{1}{3} \int_{-\pi/2}^{\pi/2} (4 - 4\sin^2 \theta)^{3/2} 2\cos \theta d\theta = \frac{1}{3} \int_{-\pi/2}^{\pi/2} (4\cos^2 \theta)^{3/2} 2\cos \theta d\theta = \frac{1}{3} \int_{-\pi/2}^{\pi/2} 8\cos^3 \theta \cdot 2\cos \theta d\theta = \frac{16}{3} \int_{-\pi/2}^{\pi/2} \cos^4 \theta d\theta = \frac{16}{3} \cdot 2 \int_0^{\pi/2} \cos^4 \theta d\theta = \frac{32}{3} \cdot \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{32}{3} \cdot \frac{3}{8} \cdot \frac{\pi}{2} = 2\pi$

$\bar{y} = \frac{M_x}{m} = \frac{2\pi}{16/3} = \frac{6\pi}{16} = \frac{3\pi}{8}$

3. Final Answer

Problem 1: $m = 30, (\bar{x}, \bar{y}) = (2, \frac{9}{5})$

Problem 2: $m = \frac{16}{3}, (\bar{x}, \bar{y}) = (0, \frac{3\pi}{8})$

Problem 3: Omitted

Problem 4: Omitted

Problem 5: Omitted

Problem 6: Omitted