Finding the Centroid

Question

Find the centroid of the plane area in the second quadrant bounded by the curve $x=y^2-9$. I already computed the value of $x$ but it is $36/5$ which is positive. It should be negative because it's in the second quadrant!

Answer 1

The $x$-coordinate of the centroid of the plane area $$A:=\{(x,y)\in\mathbb{R}^2\;:\; 0\leq y \leq 3, \;y^2-9\leq x\leq 0\}$$ should be $$\overline{x}=\frac{\int_{y=0}^3\int_{x=y^2-9}^{0}x dx dy}{\int_{y=0}^3\int_{x=y^2-9}^{0} dx dy} =-\frac{\frac{1}{2}\int_{y=0}^3(y^2-9)^2 dy}{\int_{y=0}^3 (9-y^2) dy} =-\frac{[\frac{y^5}{5}-6y^3+81y]_0^3}{2[9y-\frac{y^3}{3}]_0^3} =-\frac{18}{5}.$$ Compare your computations with mine.

What is the $y$-coordinate? You should find $\overline{y}=9/8$.