I would like to find the centroid $(x_c^A,y_c^A)$ of the orange area $A$, marked in the attached picture, i.e. of the area between $$ f(x)=\ln(x)\qquad\textrm{ and }\qquad g(x)=1, 0\leqslant x\leqslant e. $$
To this end, I first determined that $A=e-1$, because $B=\int_{1}^e\ln(x)\, dx=1$.
Hence $C=A+B=e$.
Now, the $x$-coordinate $x_c^C$ of the centroid of $C$, which is a rectangle satisfies $$ x_c^C=\frac{(x_c^A\cdot A)+(x_c^B\cdot B)}{C}=\frac{e}{2}. $$
Moreover, $$ x_c^B=\frac{1}{B}\int_1^e x\ln x\, dx=\frac{1}{4}(e^2+1), $$ so that $$ x_c^A=\frac{e^2-1}{4(e-1)}. $$
Equivalently, I get $$ y_c^A=\frac{1}{e-1}. $$
Would like to know if I am correct.
If $A$ is bounded also by $y=0$ then the $x$-coordinate of the centroid is correct, but the $y$-coordinate is not. A more direct way: $$|A|=\int_{y=0}^1\int_{x=0}^{e^y}dxdy=[e^y]_0^1=e-1$$ and $$x_A=\frac{1}{e-1}\int_{y=0}^1\int_{x=0}^{e^y}xdxdy=\frac{[\frac{e^{2y}}{4}]_0^1}{e-1}=\frac{e^{2}-1}{4(e-1)},$$ $$y_A=\frac{1}{e-1}\int_{y=0}^1\int_{x=0}^{e^y}ydxdy=\frac{[e^{y}(y-1)]_0^1}{e-1}=\frac{1}{e-1}.$$
If we consider just the given bounds (without $y=0$) then the domain $A$ is not bounded but it has finite area: $$|A|=\int_{y=-\infty}^1\int_{x=0}^{e^y}dxdy=[e^y]_{-\infty}^1=e$$ and $$x_A=\frac{1}{e}\int_{y=-\infty}^1\int_{x=0}^{e^y}xdxdy=\frac{[\frac{e^{2y}}{4}]_{-\infty}^1}{e}=\frac{e}{4},$$ $$y_A=\frac{1}{e}\int_{y=-\infty}^1\int_{x=0}^{e^y}ydxdy=\frac{[e^{y}(y-1)]_{-\infty}^1}{e}=0.$$