I am interested in finding the centroid of the surface enclosed by the lines $y=x^n$ and $y=x^{1/n}$ where $n \in \mathbb N, n >1$. The exercise, gives the following sketch.
I know that in order to find the centroid, we need to calculate its $x$ and $y$ coordinates, namingly :
$$x_c = \frac{\iint_A x \mathrm{d}A}{A} , \; y_c = \frac{\iint_A y \mathrm{d}A}{A}$$
Now, since $n>1$ I think that the only two points that the two lines meet are at $0$ and $1$. Thus, the area should be :
$$A = \int_0^1 ( x^{1/n} - x^n)\mathrm{d}x = \left[ \frac{x^{1/n + 1}}{1/n + 1}- \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1}{\frac{1}{n}+1} - \frac{1}{n+1} $$
I am having a bit of a struggle now, though, calculating the numerator quantities. Please, correct me if I am wrong in my approach for them as follows :
$$\iint_A x\mathrm{d}A = \int_0^1 \left( \int_0^{y^{1/n}} x \mathrm{d}x \right) \mathrm{d}y, \; \iint_A y\mathrm{d}A = \int_0^1 \left( \int_0^{x^{1/n}}y\mathrm{d}y\right)\mathrm{d}x$$
Thanks in advance.
$$\iint_A x\mathrm{d}A = \int_0^1 \left( \int_{y^n}^{y^{1/n}} x \mathrm{d}x \right) \mathrm{d}y$$
$$= \int_0^1 \left[\frac {x^2}2 \right]_{y^n}^{y^{1/n}} \mathrm{d}y$$
$$= \frac 12 \int_0^1 \left[y^{\frac 2n}-y^{2n} \right] \mathrm{d}y$$
$$= \frac 12 \left[\frac{y^{\frac 2n+1}}{\frac 2n+1}-\frac{y^{2n+1}}{2n+1} \right]_0^1 $$
$$= \frac 12 \left[\frac{y^{\frac {2+n}{n}}}{\frac {2+n}{n}}-\frac{y^{2n+1}}{2n+1} \right]_0^1 $$
$$= \frac 12 \left[\frac{1}{\frac {2+n}{n}}-\frac{1}{2n+1} \right] $$
$$= \frac{n}{2(2+n)}-\frac{1}{2(2n+1)}$$
$$= \frac{n(2n+1)-(2+n)}{2(2+n)(2n+1)}$$
$$= \frac{2n^2-2}{2(2+n)(2n+1)}$$
$$= \frac{n^2-1}{(2+n)(2n+1)}$$