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!
[Math] total mass of lamina and center of mass
integration
Related Solutions
To integrate this region in polar coordinates, it is advisable to break up the integral into two parts, as shown in the figures below:
The two parts of the integral are divided by the diagonal line through the upper right corner of the rectangle. Since the sides of the rectangle are $a$ and $b$, this diagonal line is at the angle $\arctan \frac ba.$
For $0 \leq \theta \leq \arctan \frac ba,$ you would integrate over $0 \leq r \leq a \sec\theta,$ and for $\arctan \frac ba \leq \theta \leq \frac\pi2,$ you would integrate over $0 \leq r \leq b \csc\theta.$
If you actually try this, I think you'll find that it is no easier than doing the integration in rectangular coordinates. It may even be worse.
An alternative approach, rather than combining $x^2+y^2$ into $r^2$, is to integrate the terms separately:
$$\begin{eqnarray} m &=& \int_0^a\int_0^b (1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b dy\,dx +\int_0^a\int_0^b x^2 \,dy\,dx +\int_0^a\int_0^b y^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{x} &=& \int_0^a\int_0^b x(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b x \,dy\,dx +\int_0^a\int_0^b x^3 \,dy\,dx +\int_0^a\int_0^b xy^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{y} &=& \int_0^a\int_0^b y(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b y \,dy\,dx +\int_0^a\int_0^b x^2y \,dy\,dx +\int_0^a\int_0^b y^3 \,dy\,dx \end{eqnarray}$$
Now you have nine integrals to solve, but they're all quite simple.
Actually, the factor of $2$ is unimportant. The center of mass in the $y$ direction is given by
$$\bar{y} = \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)(r \sin{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)}$$
Note that the weight of the $y$ coordinate is expressed in the $r \sin{\theta}$ term in the numerator. Note also the boundary of the lamina is expressed in its polar form, $r=\sin{\theta}$. Evaluating the radial integrals, we get
$$\bar{y} = \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \sin^5{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}$$
or
$$\bar{y} = \frac{6}{5}$$
For $\bar{x}$, we do a similar calculation:
$$\begin{align}\bar{x} &= \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)(r \cos{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)}\\ &= \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \cos{\theta} \sin^4{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}\\ &= \frac{9}{20}\end{align}$$
Best Answer
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
$$ M \ = \ A \ \cdot \ \rho(x_m, y_m) \ = \ (b-a) \ (d-c) \cdot \ \rho(\frac{a+b}{2},\frac{c+d}{2}) $$
$$ = \ 7 \ \cdot \ 2 \ \cdot \ \rho(\frac{7}{2},1) \ = \ 14 \ \cdot (3 \cdot \frac{7}{2} \ + \ 4 \cdot 1 \ + \ 5) \ = \ 14 \ \cdot \ 19.5 \ = \ 273\ \ . $$