centroid of a right triangle

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I'm asked to find the $M_x, M_y$ and the centroid of the shape created by the functions $5x/6$ and $x=6$ that has a density of $5$.

I find $$M_y \int_0^5 \frac{5}{6} x^2dx = \left. \frac{5}{18} x^3 \right|_0^5 = \frac{625}{18}.$$ With density it's $M_y = 3125/18$.

I find $$ M_x \int_0^5 \frac{\rho}{2} \times \frac{25}{36} x^2 dx = \left. \frac{25\rho}{216} x^3 \right|_0^5 = \frac{3125}{216} \times \rho = \frac{15625}{216}. $$ Now $$\bar{X} = \frac{M_y}{15\rho} = \frac{125}{54}$$ and $$\bar{Y} = \frac{M_x}{15\rho} = \frac{625}{648}.$$

The only problem is that none of those answers are right. What am I missing?

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Total mass is given by $$ \int_0^6 \frac{5x}{6} \cdot 5 dx = 75 $$

Now for the centroid $$ M_y = \int_0^6 \frac{5x^2}{6} \cdot 5 dx = 300 $$ so $\bar{X} = 300/75 = 4$. Can you take it from here?