Find the integral $\iint\limits_{S}x^2dy\wedge dz+y^2dz\wedge dx+z^2dx\wedge dy$

Question

Find the integral $$\iint\limits_{S}x^2dy\wedge dz+y^2dz\wedge dx+z^2dx\wedge dy,$$ where $S$ is the lower side of the hemisphere $x^2+y^2+z^2=R$, $z\geq 0$

$\wedge$ denotes the exterior product. If I understand correctly, this is just a more correct notation for $dx dy$. It is nothing more than the notation here.

I tried to use the formula for the oriented surface $S$ $$\int\limits_{S}\omega = \iint\limits_{D}\left(\omega_1\frac{\partial(y,z)}{\partial(u,v)}+\omega_2\frac{\partial(z,x)}{\partial(u,v)}+\omega_3\frac{\partial(x,y)}{\partial(u,v)}\right)dudv,$$ where $D$ is the projection of the surface $S$ onto the plane $Oxy$ in the parametric form $(x(u,v), y(u,v), z(u,v))$.

Since $x^2+y^2+z^2=R$, $z\geq 0$, I decided to go to spherical coordinates $(\rho, \theta, \phi)$, where $$x = \rho\sin\theta\cos\phi,\quad y = \rho\sin\theta\sin\phi,\quad z = \rho\cos\theta$$

Then the Jacobian would be $$\frac{\partial(x, y, z)}{\partial(\rho, \theta, \phi)} = \rho^2\sin\theta$$

But then I started looking at cases, but I couldn't come up with anything. Does anyone have any ideas?

Answer 1

The wedge product $dx\wedge dy$ is a differential form, and these give you an oriented version of integration on manifolds. In that sense it is not "more correct" notation for $dxdy$, rather it emphasizes a choice that can be made when defining the integral which is already present in the $1$-dimensional setting, although there it is so straight-forward to address that the issue is rarely dwelt upon: In the one-variable case, there is a choice to be made when defining the (say Riemann) integral of a suitably well-behaved function $f(t)$ on an interval $[a,b]$.

Should $\int_a^b f(t)$ be:

(i) the limit of Riemann sums over the interval, and hence we might just as well denoted it by $\int_{[a,b]} f(t)$,

(ii) or should we view $\int_a^b$ as the operation of integrating f from a to b, so that we take the same limit of Riemann sums, but now adjusted by $\frac{b-a}{|b-a|}\in \{\pm1\}$, i.e. oriented so that $\int_{a}^b f(t) = -\int_{b}^{a} f(t)$. This has the advantage the change of variables formula no longer requires one to take the absolute value of $\phi'$ where $\phi\colon [a,b] \to [c,d]$ is a diffeomorphism between intervals $[a,b]$ and $[c,d]$, indeed $\int_{\phi(a)}^{\phi(b)} f(s)ds = \int_{a}^b f(\phi(t))\phi'(t)dt$.

Option (i) is natural when, for example, you first learn a definition of the Riemann integral on an interval, and the integral is usually defined using that definition (at least initially). Option (ii) however becomes quite natural in the context of the fundamental theorem of calculus, where you want to write $f(x)-f(a) = \int_{a}^{x} \frac{df}{dt}dt$, since the right-hand side changes sign if I swap $a$ and $x$. Notice that I wrote "$dt$" for the first time here, because in the oriented definition there needs to be something to indicate which orientation one is taking.

The signed version of the integral that the fundamental theorem of calculus leads you too is the 1-dimensional theory of differential forms, and if you generalise to higher dimensions the natural statement is Stokes theorem for differential forms: this says that if you have a $k$-form $\omega$ on a manifold with boundary $S=\partial M$, then $\int_{S} \omega = \int_{M} d\omega$, where $d\colon \Omega^k(M) \to \Omega^{k+1}(M)$ is the differential on forms extending the map sending a function $f$ to $df$ its differential (that is, a section of $T^*M$) and satisfies the signed version of the Leibniz rule.

Here it may be worth thinking about Stokes theorem, because if $$ \omega = x^2dy\wedge dz + y^2dz\wedge dx + z^2 dx\wedge dy, $$ and we write $d\nu = dx\wedge dy \wedge dz$, then $d\omega =2(x+y+z)d\nu$.

Thus if $D_3(R)$ is the upper half-ball which has boundary $S\cup D_2(R)$ where $$ B_2(R) = \{(x,y,0):x^2+y^2\leq R^2\} $$ is the disk of radius $R$ in the $xy$-plane. Thus Stokes theorem shows (since the question takes the surface to be the "lower side" of the hemisphere) that $$ \int_S \omega = \int_{B_2(R)} \omega - \int_{D_3(R)}2(x+y+z)\mathrm{d}\nu d$$

Proceeding from this equation and paying attention to some obvious symmetries which force some terms to vanish, you can readily reduce the calculation of $\int_S \omega$ to $\int_{D_3(R)} 2z \mathrm{d}\nu$.

Answer 2

The parametrisation $(x(u, v), y(u, v), z(u, v))$ involves two parameters, $u, v$. Your spherical coordinates involves three parameters, $\rho$, $\theta$ and $\phi$. So we have a problem.

To resolve this problem, note that $\rho$ should not be viewed as a parameter, since $\rho$ takes the constant value of $R$ on the surface $S$.

The two other spherical coordinates $\theta$ and $\phi$ are the true parameters that parametrise our surface $S$. $\theta$ and $\phi$ play the role of $u$ and $v$. To spell it out, the parametrisation of $S$ is \begin{align} x(\theta, \phi) & = R \sin \theta \cos \phi \\ y(\theta, \phi) & = R \sin \theta \sin \phi \\ z(\theta, \phi) & = R \cos \theta \end{align} with $\theta$ and $\phi$ lying in the range $$ 0 \leq \theta \leq \frac{\pi}{2}, \qquad 0 \leq \phi < 2\pi. $$

So you need to compute $$ \int_{\theta = 0}^{\theta = \frac{\pi}2} \int_{\phi = 0}^{\phi = 2\pi} \left( \omega_1(\theta, \phi) \frac{\partial (y, z)}{\partial(\theta, \phi)} + \omega_2(\theta, \phi) \frac{\partial (z, x)}{\partial(\theta, \phi)} + \omega_3(\theta, \phi) \frac{\partial (x, y)}{\partial(\theta, \phi)} \right) d\theta d\phi.$$

You'll have to compute three Jacobians: \begin{align} \frac{\partial (y, z)}{\partial(\theta, \phi)} &= \frac{\partial y}{\partial \theta}\frac{\partial z}{\partial \phi} - \frac{\partial z}{\partial \theta}\frac{\partial y}{\partial \phi} \\ \frac{\partial (z, x)}{\partial(\theta, \phi)} &= \frac{\partial z}{\partial \theta}\frac{\partial x}{\partial \phi} - \frac{\partial x}{\partial \theta}\frac{\partial z}{\partial \phi} \\ \frac{\partial (x, y)}{\partial(\theta, \phi)} &= \frac{\partial x}{\partial \theta}\frac{\partial y}{\partial \phi} - \frac{\partial y}{\partial \theta}\frac{\partial x}{\partial \phi} \end{align} You'll have to compute them separately.

[By the way, the question states that $S$ is the lower side of the hemisphere. I presume this is telling us that our choice of orientation means that we have to add an overall minus sign in front of the integral that I wrote down.]

Answer 3

Using a nineteenth century notation, $$ \iint\limits_Sx^2dy\wedge dz+y^2dz\wedge dx+z^2dx\wedge dy=\iint\limits_S\left(\begin{smallmatrix}x^2\\y^2\\z^2\end{smallmatrix}\right)\cdot\boldsymbol{n}\,dS $$ where $\boldsymbol{n}$ is the unit normal to the sphere which we know is $$ \boldsymbol{n}=\textstyle\frac{1}{R}\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)\,. $$ Therefore, the integral over the upper hemisphere becomes \begin{align} &\frac{1}{R}\iint\limits_S x^3+y^3+z^3\,dS\\ &= \frac{1}{R}\int_0^{2\pi}\int_0^{\pi/2} R^3\Big(\cos^3\varphi\sin^3\theta+\sin^3\varphi\sin^3\theta+\cos^3\theta\Big) R^2\sin\theta\,d\theta\,d\varphi\\[2mm] &=\frac{R^4\,\pi}{2}\,. \end{align} To see this, notice that $\int_0^{2\pi}\cos^3\varphi\,d\varphi=\int_0^{2\pi}\sin^3\varphi\,d\varphi=0 $ and we only have to calculate

$$ \int_0^{2\pi}\int_0^{\pi/2}\cos^3\theta\sin\theta\,d\theta\,d\varphi=\frac{\pi}{2}\,. $$