integration
The following masses mi are located at the given points $P_i: m_1=6, P_1(1,5), m_2=5,P_2(3,−2), m_3=10,P_3(−2,−1)$. Find the moments $M_x$, $M_y$, and the $x$ and $y$-coordinates of the center of mass of the system.
$M_x= ?$
$M_y= ?$
$x$-coordinate of the center of mass: $x= ?$
$y$-coordinate of the center of mass: $y= ?$
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.
See the picture below.
As you can see, $\cos x$ is above $\sin x$ or $\cos x>\sin x$ for $\left[0,\dfrac\pi4\right]$. Therefore $$ \begin{align} M&=\int_0^\frac\pi4(\cos x-\sin x)\ dx=\sqrt{2}-1,\\\\ M_y&=\int_0^\frac\pi4x(\cos x-\sin x)\ dx=\frac14(\sqrt{2}\pi-4),\quad\Rightarrow\quad\text{use IBP} \end{align} $$ and $$ \begin{align} M_x&=\frac12\int_0^\frac\pi4(\cos x+\sin x)(\cos x-\sin x)\ dx\\ &=\frac12\int_0^\frac\pi4(\cos^2 x-\sin^2 x)\ dx\\ &=\frac12\int_0^\frac\pi4\cos2x\ dx\\ &=\frac14. \end{align} $$ I think you can handle it from here. I hope this helps.
Best Answer
Since formulas for $M_x$ and $M_y$ are $M_x =\displaystyle{\sum_{k=1}^{3}{m_ky_k}}$ and $M_y =\displaystyle{\sum_{k=1}^{3}{m_ky_k}}$, we have
$M_x=6(5)+5(-2)+10(-1)=30-10-10=10$
$M_y=6(1)+5(3)+10(-2)=6+15-20=1$
We set $M=m_1+m_2+m_3=6+5+10=21$
Then, the coordinates of the center of mass are $x=\dfrac{M_y}{M}=\dfrac{1}{21}$ and $y=\dfrac{M_x}{M}=\dfrac{20}{21}$.