Calculating the surface integral $\int_{S_1(0)}y_j \ d\sigma(y)$ I'm learning how to take surface integrals on the surface of spheres in $\mathbb{R}^n$. 

Definition 22.4.
  Suppose $\varSigma \colon D \subset_0 \mathbb{R}^{n-1} \to M \subset \mathbb{R}^n$ is a $C^1$-parameterized hypersurface of $\mathbb{R}^n$ and $f \in C_c(M, \mathbb{R})$.
  Then the surface integral of $f$ over $M$, $\int_M f \,\mathrm{d}\sigma$, is defined by
  \begin{align*}
     \int_M f \,\mathrm{d}\sigma
  &= \int_D f \circ \varSigma(y)
            \left|
              \det
              \left[
                \frac{\partial \varSigma(y)}{\partial y_1}
                \middle|
                \dotsb
                \middle|
                \frac{\partial \varSigma(y)}{\partial y_{n-1}}
                \middle|
                n(y)
               \right]
            \right|
            \,\mathrm{d}y  \\
  &= \int_D f \circ \varSigma(y)
            \left|
              \det
              \left[
                \varSigma'(y) e_1
                \middle|
                \dotsb
                \middle|
                \varSigma'(y) e_{n-1}
                \middle|
                n(y)
               \right]
            \right|
            \,\mathrm{d}y
\end{align*}
  where $n(y) \in \mathbb{R}^n$ is a unit normal vector perpenticular of $\operatorname{ran}(\varSigma'(y))$ for each $y \in D$.
  We will abbreviate this formula by writing
  $$
    \mathrm{d}\sigma
  = \left|
      \det
      \left[
        \frac{\partial \varSigma(y)}{\partial y_1}
        \middle|
        \dotsb
        \middle|
        \frac{\partial \varSigma(y)}{\partial y_{n-1}}
        \middle|
        n(y)
      \right]
    \right|
    \,\mathrm{d}y.
$$
(Original image here.)

I'm asked to compute
$$\int_{S_1(0)}y_j \ d\sigma(y)$$ 
According to the definition above, all I have to do is to find a parametrization $\Sigma$ for the unit sphere with center $0$ that goes from the 'circumference' to the 'sphere' analogous in $n$ dimensions. I can only find such parametrization locally, not globally. So I guess the integral is going to have to be broken in two hemispheres. Am I rigth?
Let's imagine the north hemisphere first, through the parametrization $\Sigma(y) = \left(y,\sqrt{1-|y|^2}\right)$:
$$\int_{S_1(0)^+}y_j \ d\sigma(y) = \int_{C_1(0)+}(0,\cdots,y_j,\cdots,0)\circ\Sigma(y)\det [ye_1\ \cdots \  ye_{n-1} \ n(y)]\ dy$$ 
where $C_1(0)$ is just the 'circunference' on which the 'sphere' is parametrized. We can think of it as the region $|y|<1$ for $y\in\mathbb{R}^{n-1}$
Remember that the unit normal $n(y)$ in the unit sphere is just $y$. So the determinant above would give us $y_1\cdots y_{n-1}\sqrt{1-(y_1+\cdots y_{n-1})^2}$ because these are the diagonal terms multiplied and the rest of the elements are $0$ (except for the normal column but it gets multiplied by the other $0$s)
So we end up with
$$\int_{S_1(0)^+}y_j \ d\sigma(y) = \int_{C_1(0)+} y_1\cdots y_j^2 \cdots \ y_{n-1}\sqrt{1-(y_1^2+\cdots +  y_{n-1}^2)}\  dy$$
Now, breaking these onto the region $C_1(0)$ we get:
$$\int_{S_1(0)^+}y_j \ d\sigma(y) = \int_{-1}^1\cdots\int_{-1}^1 y_1\cdots y_j^2 \cdots \ y_{n-1}\sqrt{1-(y_1^2+\cdots +y_{n-1}^2)} \ dy_1\cdots dy_j \cdots dy_{n-1}$$
And for the south hemisphere:
$$\int_{S_1(0)^-}y_j \ d\sigma(y) = \int_{-1}^1\cdots\int_{-1}^1 y_1\cdots y_j^2 \cdots \ y_{n-1}\sqrt{-1+(y_1^2+\cdots + y_{n-1}^2)} \ dy_1\cdots dy_j \cdots dy_{n-1}$$
I think it helps if I do the two cases of integration for the north hemisphere first. I have to integrate $y_k\sqrt(\cdots)$ or $y_k^2\sqrt(\cdots)$. Let's do the first one first:
$$\frac{1}{2}\int_{-1}^12y_k\sqrt{1-(y_1^2 + \cdots + y_k^2 + \cdots + y_{n-1}^2)}\ dy_k = \\\frac{1}{2}\int_{1}^{1}\sqrt{1-(y_1^2 + \cdots + u + \cdots + y_{n-1}^2)}\ du$$
I'm integrating with the substitution $u = y_k^2$ from $u(-1) = 1$ to $u(1) = 1$ so it should be $0$? This doesn't make sense since everything else would get multiplied by $0$.
So two questions arise: 
what am I doing wrong and is there a better way to solve this? 
I know that there may be some non elementar ways but I'm doing a PDE course and it may be that my teacher wants me to take this integral by hand or maybe he wants me to use some trick. Is there a way to express this integral in terms of the famous $w_n$ number which is the volume or area, something like that, in $\mathbb{R}^n$?
UPDATE:
Can the divergence theorem be used? If I see $y_j$ as the divergence of $\vec{X}(y_1,\cdots,y_j,\cdots,y_n) = (0,\cdots,\frac{y_j^2}{2},\cdots,0)$ then I could see the integral of $\mbox{div} \vec{X}$ over the sphere as being an integral over the boundary of the sphere. I don't know however about $d\sigma$ and how to put that into the divergence theorem. It's just an idea.
UPDATE:
My book says that this integral should indeed be $0$. Did I do everything rigth or did I magically come up with a true answer?
 A: Solution Employing Symmetry
Let $S^{n-1}$ denote the unit hypersphere in $\mathbb{R}^n$ (with coordinate vector $\mathbf{y}=(y_1,y_2,\ldots,y_n)$) centered at the origin $\boldsymbol{0}_n$.  Write $\sigma_{n-1}$ for the hypersurface area measure of $S^{n-1}$.  Declare the northern hemisphere $S^{n-1}_+$ to be the one with $y_j>0$, and the southern hemisphere $S^{n-1}_-$ to be the one with $y_j<0$.  By symmetry, we have
$$\int_{S^{n-1}_-}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y}) = -\int_{S^{n-1}_-}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y})\,.$$
This shows that $$\int_{S^{n-1}}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{S^{n-1}_+}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y})+\int_{S^{n-1}_-}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y})=0\,.$$

Solution Implementing the Divergence Theorem
Let $\lambda_n$ be the Lebesgue measure on $\mathbb{R}^n$ and $\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n$ the standard basis vectors of $\mathbb{R}^n$.  The normal vector to the hypersurface $S^{n-1}$ is given by $$\mathbf{n}=y_1\mathbf{e}_1+y_2\mathbf{e}_2+\ldots+y_n\mathbf{e}_n\,.$$  Write $B^n_r(\mathbf{x})$ for the open ball of radius $r>0$ centered at $\mathbf{x}\in\mathbb{R}^n$.  Note from the Divergence Theorem that $$\int_{\partial B^n_1(\boldsymbol{0}_n)}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{\partial B^n_1(\boldsymbol{0}_n)}\,\mathbf{e}_j\cdot\mathbf{n}\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{B^n_1(\boldsymbol{0}_n)}\,\big(\boldsymbol{\nabla}\cdot\mathbf{e}_j\big)\,\text{d}\lambda_n(\mathbf{y})\,.$$   Obviously, $\boldsymbol{\nabla}\cdot\mathbf{e}_j=0$.  Consequently, the integral $\displaystyle\int_{S^{n-1}}\,y_j\,\text{d}\sigma_{n-1}(\mathbf{y})$ equals $0$.

Solution Following the OP's Idea
Without loss of generality, suppose that $j=n$.  First, it is easy to prove that 
$$\text{d}\sigma_{n-1}(\mathbf{y})=\frac{1}{\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}}\,\text{d}y_1\,\text{d}y_2\,\cdots \,\text{d}y_{n-1}\,.$$
For $i=1,2,\ldots,n-1$, write $$b_{i}:=\sqrt{1-y_1^2-y_2^2-\ldots-y_{i-1}^2}\,.$$  Observe that there are two values of $y_n$ for a given value of $(y_1,y_2,\ldots,y_{n-1})$: $$\pm\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}\,.$$  Hence, we have
$$\int_{-b_{n-1}}^{+b_{n-1}}\,\frac{\Big(\pm\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}\Big)}{\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}}\,\text{d}y_{n-1}=\pm\,\int_{-b_{n-1}}^{+b_{n-1}}\,\text{d}y_{n-1}\,.$$
Observe that
$$\int_{S^{n-1}}\,y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{S_+^{n-1}}\,y_n\,\text{d}\sigma_{n-1}(\mathbf{y})+\int_{S^{n-1}_-}\,y_n\,\text{d}\sigma_{n-1}(\mathbf{y})$$
and
$$\int_{S^{n-1}_\pm}\,y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=\small\int_{-b_1}^{+b_1}\,\int_{-b_2}^{+b_2}\,\cdots\int_{-b_{n-2}}^{+b_{n-2}}\,\left(\int_{-b_{n-1}}^{+b_{n-1}}\,{\tiny\frac{\Big(\pm\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}\Big)}{\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}}}\,\text{d}y_{n-1}\right)\,\text{d}y_{n-2}\,\cdots\,\text{d}y_2\,\text{d}y_1\,.$$
This proves that
$$\int_{S^{n-1}_\pm}\,y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=\pm\,\int_{B^{n-1}_1(\mathbf{0}_{n-1})}\,\text{d}\lambda_{n-1}(y_1,y_2,\ldots,y_{n-1})\,,$$
and so we conclude that $$\int_{S^{n-1}}\,y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{B^{n-1}_1(\mathbf{0}_{n-1})}\,\text{d}\lambda_{n-1}(y_1,y_2,\ldots,y_{n-1})-\int_{B^{n-1}_1(\mathbf{0}_{n-1})}\,\text{d}\lambda_{n-1}(y_1,y_2,\ldots,y_{n-1})=0\,.$$
