Solución

Primero calculamos la masa $m$ . Necesitamos describir la región entre la gráfica de $y = x^2$ y las líneas verticales $x = 0$ y $x = 2$ :

\begin{aligned} m &= \iint_Rdm = \iint_R \rho(x,y)dA = \int_{x=0}^{x=2}\int_{y=0}^{y=x^2} (x+y)dydx = \int_{x=0}^{x=2}\bigg[xy + \frac{y^2}{2}\bigg|_{y=0}^{y=x^2}\bigg]dx\\ &= \int_{x=0}^{x=2}\bigg[\frac{x^3}{2}\bigg]dx = \bigg[\frac{x^4}{4} + \frac{x^5}{10}\bigg]\bigg|_{x=0}^{x=2} = \frac{36}{5} \end{aligned}

Ahora calculamos los momentos $M_x$ y $M_y$ :

M_x = \iint_R y\rho(x, y)dA = \int_{x=0}^{x=2}\int_{y=0}^{y=x^2} y(x+y)dydx = \frac{80}{7}

M_y = \iint_R x\rho(x, y)dA = \int_{x=0}^{x=2}\int_{y=0}^{y=x^2} x(x+y)dydx = \frac{176}{15}

Finalmente, evaluamos el dentro de masa,

\bar{x} = \frac{M_y}{m} = \frac{\iint_R x\rho(x, y)dA}{\iint_R \rho(x, y)dA} = \frac{176/15}{36/5} = \frac{44}{27}

\bar{y} = \frac{M_x}{m} = \frac{\iint_R y\rho(x, y)dA}{\iint_R \rho(x, y)dA} = \frac{80/7}{36/5} = \frac{100}{63}

Por lo tanto, el centro de masa es $(\bar{x}, \bar{y}) = \big(\frac{44}{27} , \frac{100}{63} \big)$ .