A lamina occupies the part of the rectangle , $ \ 0 \le y \le 2 \ $ , and the density at each point is given by the function
What is the total mass?
Where is the center of mass?
My problem is setting the integral!
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$\begingroup$The mass of a rectangular lamina with boundaries $ \ a \le x \le b \ $ and $ \ c \le y \le d \ $ is given by
$$ M \ = \ \int_c^d \ \int_a^b \ \ \rho (x,y) \ \ dx \ dy \ . $$
The coordinates of the centroid ("center-of-mass" in physical applications) are found from
$$ \overline{x} \ = \ \frac{\int_c^d \ \int_a^b \ \ x \cdot \rho (x,y) \ \ dx \ dy}{M} \ \ , $$
$$ \overline{y} \ = \ \frac{\int_c^d \ \int_a^b \ \ y \cdot \rho (x,y) \ \ dx \ dy}{M} \ \ . $$
Note that since the density gets larger as we go toward the "upper-right" corner, $ \ (7,2) \ , $ we should expect the centroid to be "above and to the right" of the center of the rectangle at $ \ ( \frac{7}{2} , 1 ) \ . $ [Later, I'll mention a handy way to get the mass without integration when the density function is linear in each dimension...]
For the sake of guidance, I get $ \ M \ = \ 273 \ , \ \overline{x} \ = \ \frac{1127}{273} \ , \ \overline{y} \ = \ \frac{875}{3 \ \cdot \ 273} \ . $ (No one said the centroid coordinates would be pretty -- just rational...)
EDIT: The "short-cut" I was holding off mentioning is that when linear functions are involved, some calculations can be greatly simplified because of the way some integrals work. The density function in this problem, $ \ \rho(x, y) = 3x + 4y + 5 \ $ , is linear in both the $ \ x-$ and $ \ y-$ directions. In this situation, the mass of the lamina is just the density at the geometrical center of the region (not the centroid; the two only coincide for uniform density) times the area. For this problem, we thus find
$$ M \ = \ A \ \cdot \ \rho(x_m, y_m) \ = \ (b-a) \ (d-c) \cdot \ \rho(\frac{a+b}{2},\frac{c+d}{2}) $$
$$ = \ 7 \ \cdot \ 2 \ \cdot \ \rho(\frac{7}{2},1) \ = \ 14 \ \cdot (3 \cdot \frac{7}{2} \ + \ 4 \cdot 1 \ + \ 5) \ = \ 14 \ \cdot \ 19.5 \ = \ 273\ \ . $$
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