Pessimist Singularitarian

#3
February 12th, 2018,
22:18
The first thing I would do is sketch the bounded area:

Now, let's compute the mass (noting that the curves $y=x$ and $y=2-x^2$ intersect at $x=1$ in QI):

$ \displaystyle m=\rho A=3\int_{0}^{1}\int_{x}^{2-x^2}\,dy\,dx=3\int_{0}^{1}2-x-x^2\,dx=3\left(2-\frac{1}{2}-\frac{1}{3}\right)=\frac{7}{2}$

Next, let's compute the moments of the lamina:

$ \displaystyle M_x=3\int_{0}^{1}\int_{x}^{2-x^2}y\,dy\,dx=\frac{3}{2}\int_{0}^{1}\left(2-x^2\right)^2-x^2\,dx=\frac{3}{2}\int_{0}^{1}x^4-5x^2+4\,dx=\frac{19}{5}$

$ \displaystyle M_y=3\int_{0}^{1}x\int_{x}^{2-x^2}\,dy\,dx=3\int_{0}^{1}2x-x^2-x^3\,dx=\frac{5}{4}$

Hence:

$ \displaystyle \overline{x}=\frac{M_y}{m}=\frac{\frac{5}{4}}{\frac{7}{2}}=\frac{5}{14}$

$ \displaystyle \overline{y}=\frac{M_x}{m}=\frac{\frac{19}{5}}{\frac{7}{2}}=\frac{38}{35}$