Taking the surface integral $\int_{S_1(0)}y_jy_k \ d\sigma(y)$

Question

I'm learning how to take surface integrals on the surface of spheres in $\mathbb{R}^n$. This question is related to Calculating the surface integral $\int_{S_1(0)}y_j \ d\sigma(y)$ where I try to compute a similar integral but I don't know if I did everything ok. (read that answer to have a better understanding of how to integrate over a surface if you need)

I'm asked to compute

$$\int_{S_1(0)}y_jy_k \ d\sigma(y)$$

Let's imagine the north hemisphere first, through the parametrization $\Sigma(y) = \left(y,\sqrt{1-|y|^2}\right)$:

$$\int_{S_1(0)^+}y_jy_k \ d\sigma(y) = \int_{C_1(0)+}y_jy_k\det [ye_1\ \cdots \ ye_{n-1} \ n(y)]\ dy$$

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

$$\int_{S_1(0)^+}y_jy_k \ d\sigma(y) = \int_{C_1(0)+} y_1\cdots y_j^2 \cdots y_k^2 \cdots \ y_{n-1}\sqrt{1-(y_1^2+\cdots + y_{n-1}^2)}\ dy$$

Now, breaking these onto the region $C_1(0)$ we get:

$$\int_{S_1(0)^+}y_jy_k \ d\sigma(y) = \\ \int_{-1}^1\cdots\int_{-1}^1 y_1\cdots y_j^2\cdots y_k^2 \cdots \ y_{n-1}\sqrt{1-(y_1^2+\cdots +y_{n-1}^2)} \ dy_1\cdots dy_j \cdots dy_{n-1}$$

And for the south hemisphere:

$$\int_{S_1(0)^-}y_jy_k \ d\sigma(y) =\\ \int_{-1}^1\cdots\int_{-1}^1 y_1\cdots y_j^2 \cdots y_k^2 \cdots \ y_{n-1}\sqrt{-1+(y_1^2+\cdots + y_{n-1}^2)} \ dy_1\cdots dy_j \cdots dy_{n-1}$$

I think it helps if I do the two cases of integration for the north hemisphere first. If $k\neq j$ then I have to integrate $y_i\sqrt(\cdots)$ or $y_i^2\sqrt(\cdots)$. If $k=j$ then it's a matter of integrating $y_i\sqrt(\cdots)$ and $y_i^3\sqrt(\cdots)$.

So for $k\neq j$, if we integrate $y_i\sqrt(\cdots)$:

$$\frac{1}{2}\int_{-1}^12y_i\sqrt{1-(y_1^2 + \cdots + y_i^2 + \cdots + y_{n-1}^2)}\ dy_i = \\\frac{1}{2}\int_{1}^{1}\sqrt{1-(y_1^2 + \cdots + u + \cdots + y_{n-1}^2)}\ du$$

I'm integrating with the substitution $u = y_i^2$ from $u(-1) = 1$ to $u(1) = 1$ so it should be $0$. This gets multiplied with every other integral so the entire integral is $0$ which is ok according to my book.

Now for $k=j$, if we integrate $y_i^4\sqrt(\cdots)$ we would get something, but it would also be multiplied by the integral of $y_i\sqrt(\cdots)$, so it should be $0$ too. But my book says it is $n^{-1}\int{S_1(0)}1$.

Batominovski's Comment: In the paragraph above, the claim that "it would also be multiplied by the integral of $y_i\sqrt(\cdots)$" is wrong. We do not have $\int\,(fg) =\left(\int\,f\right)\,\left(\int\,g\right)$. So, $\int\,g=0$ does not imply $\int\,(fg)=0$.

What is wrong?

Answer 1

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 $y_j<0$. If $j\neq k$, then by symmetry, we have $$\int_{S^{n-1}_-}\,y_jy_k\,\text{d}\sigma_{n-1}(\mathbf{y}) = -\int_{S^{n-1}_-}\,y_jy_k\,\text{d}\sigma_{n-1}(\mathbf{y})\,.$$ This shows that $$\int_{S^{n-1}}\,y_jy_k\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{S^{n-1}_+}\,y_jy_k\,\text{d}\sigma_{n-1}(\mathbf{y})+\int_{S^{n-1}_-}\,y_jy_k\,\text{d}\sigma_{n-1}(\mathbf{y})=0\,.$$

If $j=k$, then we note from symmetry that $$\int_{S^{n-1}}\,y_1^2\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{S^{n-1}}\,y_2^2\,\text{d}\sigma_{n-1}(\mathbf{y})=\ldots=\int_{S^{n-1}}\,y_n^2\,\text{d}\sigma_{n-1}(\mathbf{y})\,.$$ Since $\sum\limits_{i=1}^n\,y_i^2=1$ on $S^{n-1}$, $$\int_{S^{n-1}}\,y_j^2\,\text{d}\sigma_{n-1}(\mathbf{y})=\frac{1}{n}\,\int_{S^{n-1}}\,\sum_{i=1}^n\,y_i^2\,\text{d}\sigma_{n-1}(\mathbf{y})=\frac{1}{n}\,\int_{S^{n-1}}\,\text{d}\sigma_{n-1}(\mathbf{y})=\frac{1}{n}\,\Sigma_{n-1}\,,$$ where $\Sigma_{n-1}:=\displaystyle \int_{S^{n-1}}\,\text{d}\sigma_{n-1}(\mathbf{y})=\dfrac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}$ is the hypersurface area of $S^{n-1}$. Here, $\Gamma$ is the usual gamma function. That is, $$\frac{1}{n}\,\Sigma_{n-1}=\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)}\,,$$ which equals the volume of a unit $n$-dimensional hypersphere.

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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 $$\int_{\partial B^n_1(\boldsymbol{0}_n)}\,y_j\,y_k\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{\partial B^n_1(\boldsymbol{0}_n)}\,y_k\mathbf{e}_j\cdot\mathbf{n}\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{B^n_1(\boldsymbol{0}_n)}\,\big(\boldsymbol{\nabla}\cdot(y_k\mathbf{e}_j)\big)\,\text{d}\lambda_n(\mathbf{y})\,.$$ Obviously, $\boldsymbol{\nabla}\cdot(y_k\mathbf{e}_j)=0$ for $j\neq k$ and $\boldsymbol{\nabla}\cdot(y_j\mathbf{e}_j)=1$.

Consequently, the integral $\displaystyle\int_{S^{n-1}}\,y_jy_k\,\text{d}\sigma_{n-1}(\mathbf{y})$ equals $\lambda_n\big(B_1^n(\textbf{0}_n)\big)\,\delta_{j,k}$. Here, $\delta$ is the Kronecker delta. Since $\lambda_n\big(B^n_1(\textbf{0}_n)\big)=\dfrac{1}{n}\,\Sigma_{n-1}$, we get the same result as the first solution.

Answer 2

You have made some errors in setting up the integral, both here and in your previous post. The parameterization of the upper hemisphere we're using is

$$\Sigma (y) = (y,(1-|y|^2)^{1/2},$$

for $y$ in the open unit ball $B_{n-1}$ in $\mathbb R^{n-1}.$ We calculate

$$\tag 1\frac{\partial \Sigma(y)}{\partial y_k} = (e_k,-y_k(1-|y|^2)^{-1/2}), \,\, k = 1,\dots , n-1.$$

Here $e_k$ is the standard basis vector in $\mathbb R^{n-1}.$

We want to think of the $n\times n$ matrix with $(1)$ giving the first $n-1$ column vectors, and $n(y),$ the unit normal vector, as the last column. But $n(y)$ is not $y,$ it is $\Sigma (y).$ ($\Sigma (y)$ is on the sphere; $y$ lives in $B_{n-1}.$) The determinant of this matrix is

$$\tag 2\frac{1}{(1-|y|^2)^{1/2}}.$$

Thus $d\sigma(\Sigma (y)) = dy/(1-|y|^2)^{1/2}.$ This is true in all dimensions, nicely enough.

To verify $(2),$ let's take advantage of the fact that $\Sigma$ is a graph. To review: Suppose $U\subset \mathbb R^{n-1}$ is open and $f:U\to \mathbb R$ is smooth. Then the graph $\{(y,f(y)):y\in U\}$ has surface area measure given by

$$d \sigma(y,f(y))= \left[(1+(D_1f(y))^2 + \cdots + (D_{n-1}f(y))^2\right]^{1/2}\,dy.$$

This follows from the general surface area formula you cited in your previous post. You should try to verify this. In the case of the upper hemisphere, $f(y) = (1-|y|^2)^{1/2}.$ A straightforward computation then gives $(2).$

Hopefully this will help with the trouble you are having with your specific integrals. Give it a try. I'll stop here for now.

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Added later, another approach: Your integrals can be easily done if you accept one property of the surface measure $\sigma$ on the sphere: Rotation invariance. More precisely, if $T$ is an orthogal transformation of $\mathbb R^n$ and, say, $f$ is continuous on the unit sphere $S,$ then

$$\int_S f(y)\,d\sigma(y) = \int_S f(T(y))\,d\sigma (y).$$

So assume $i\in \{1,2\dots, n\},$ and $T$ is the orthogonal transformation that sends each standard basis vector $e_j$ to itself except for $j=i,$ where $T(e_i) = -e_i.$ Suppose $j\ne i,$ and $f(y) = y_iy_j.$ Then $f(T(y))=-f(y)$ for $y\in S.$ Thus

$$ \int_S y_iy_j\,d\sigma (y) = \int_S f(y)\,d\sigma(y) = \int_S f(T(y))\,d\sigma f(y) = -\int_S f(y)\,d\sigma(y).$$

Thus this integral is $0.$

For the other integral, suppose $i\ne j.$ Define $T$ to be the orthogonal transformation that switches $e_i$ with $e_j,$ and leaves the other basis vectors alone. Let $f(y)= y_i^2.$ Then

$$\int_S y_i^2\,d\sigma (y) = \int_S f(y)\,d\sigma (y) = \int_S f(T(y))\,d\sigma(y) = \int_S y_j^2\,d\sigma (y).$$

From this we get, for fixed $i,$

$$\int_S y_i^2d\sigma (y) = \frac{1}{n}\int_S \left (\sum_{j=1}^{n}y_j^2\right )\,d\sigma (y) = \frac{1}{n}\int_S 1\,d\sigma (y) = \frac{\sigma (S)}{n}.$$

Answer 3

MathJax is killing my slow browser, so I have to write a separate answer. All notations can be found in my first answer.

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Solution Following the OP's Idea

Without loss of generality, suppose that $j=n$. First, it is easy to prove that $$\text{d}\sigma_{n-1}(\mathbf{y})=\frac{1}{\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}}\,\text{d}y_1\,\text{d}y_2\,\cdots \,\text{d}y_{n-1}\,.$$ If $k\neq j$, then we may assume without loss of generality that $k=n-1$. 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 $$\int_{-b_{n-1}}^{+b_{n-1}}\,\frac{y_{n-1}\,\Big(\pm\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}\Big)}{\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}}\,\text{d}y_{n-1}=\pm\,\int_{-b_{n-1}}^{+b_{n-1}}\,y_{n-1}\,\text{d}y_{n-1}=0\,.$$ Observe that $$\int_{S^{n-1}}\,y_{n-1}y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=\int_{S_+^{n-1}}\,y_{n-1}y_n\,\text{d}\sigma_{n-1}(\mathbf{y})+\int_{S^{n-1}_-}\,y_{n-1}y_n\,\text{d}\sigma_{n-1}(\mathbf{y})$$ and $$\small\int_{S^{n-1}_\pm}\,y_{n-1}y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=\small\int_{-b_1}^{+b_1}\,\int_{-b_2}^{+b_2}\,\cdots\int_{-b_{n-2}}^{+b_{n-2}}\,\left(\int_{-b_{n-1}}^{+b_{n-1}}\,{\tiny\frac{y_{n-1}\,\Big(\pm\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}\Big)}{\sqrt{1-y_1^2-y_2^2-\ldots-y_{n-1}^2}}}\,\text{d}y_{n-1}\right)\,\text{d}y_{n-2}\,\cdots\,\text{d}y_2\,\text{d}y_1\,.$$ This proves that $$\int_{S^{n-1}_\pm}\,y_{n-1}y_n\,\text{d}\sigma_{n-1}(\mathbf{y})=0\,,$$ and so we conclude that $$\int_{S^{n-1}}\,y_j\,y_k\,\text{d}\sigma_{n-1}(\mathbf{y})=0$$ if $j\neq k$.

Now, we tackle the case $k=j=n$. On any hemisphere, we have $$y_j\,y_k=y_n^2=1-y_1^2-y_2^2-\ldots-y_{n-1}^2=1-r_{n-1}^2\,,$$ where $r_{n-1}:=\sqrt{y_1^2+y_2^2+\ldots+y_{n-1}^2}$. Thus, combining the two hemispheres, we get $$\int_{S^{n-1}}\,y_n^2\,\text{d}\sigma_{n-1}(\textbf{y})=2\,\int_{B^{n-1}_1(\mathbf{0}_{n-1})}\,\sqrt{1-r_{n-1}^2}\,\text{d}\lambda_{n-1}(y_1,y_2,\ldots,y_{n-1})\,.$$ We have $$\sqrt{1-r_{n-1}^2}\,\text{d}\lambda_{n-1}(y_1,y_2,\ldots,y_{n-1})=r_{n-1}^{n-2}\,\sqrt{1-r_{n-1}^2}\,\text{d}r_{n-1}\,\text{d}\sigma_{n-2}(y_1,y_2,\ldots,y_{n-1})$$ and it can be easily shown that $$\int_0^1\,r_{n-1}^{n-2}\,\sqrt{1-r_{n-1}^2}\,\text{d}r_{n-1}=\frac{1}{n}\,\int_0^1\,\frac{r_{n-1}^{n-2}}{\sqrt{1-r_{n-1}^2}}\,\text{d}r_{n-1}\,.$$ Ergo, $$\int_{S^{n-1}}\,y_n^2\,\text{d}\sigma_{n-1}(\textbf{y})={\small\frac{2}{n}\,\int_{B^{n-1}_1(\mathbf{0}_{n-1})}\,\frac{1}{\sqrt{1-r_{n-1}^2}}\,\text{d}\lambda_{n-1}(y_1,y_2,\ldots,y_{n-1})}={\small\frac1n\,\int_{S^{n-1}}\,\text{d}\sigma_{n-1}(\textbf{y})}=\frac{1}{n}\,\Sigma_{n-1}\,.$$

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In general, let $a_1,a_2,\ldots,a_n$ be nonnegative integers such that at least one of the $a_i$'s is an odd integer. Then, $$\int_{S^{n-1}}\,y_1^{a_1}y_2^{a_2}\cdots y_n^{a_n}\,\text{d}\sigma_{n-1}(\textbf{y})=0\,.$$
On the other hand, if each $a_i$ is even, then $$\int_{S^{n-1}}\,y_1^{a_1}y_2^{a_2}\cdots y_n^{a_n}\,\text{d}\sigma_{n-1}(\textbf{y})>0\,.$$