Evaluate integral of $x^3dy \wedge dz+y^3dz \wedge dx+z^3 dx \wedge dy$ over the upper hemisphere.

Question

Calculate the integral of the differential form $\omega$ over the half-sphere $S^{+}$, where:

• $\omega = x^3 dy \wedge dz + y^3 dz \wedge dx + z^3 dx \wedge dy $
• $S^{+} = \Big\{ (x,y,z) \in \mathbb{R}^3 \big| x^2 + y^2 + z^2 = 1, z \geq 0 \Big\}$

We have such orientation of $S^{+}$ that vector $[0,0,1]$ is normal in point $(0,0,1)$.

We can use Stokes' theorem to calculate the integral of the differential form over half-sphere with circle on the bottom (because we need to calculate a closed solid), we can call it $A$. Then we can subtract the bottom, let's call it $B$ (it should be easy to calculate the bottom - it's just a circle). We have that:

$$\int_{\partial A} \omega = \int_{A} d\omega = \int_{S^{+}} \omega + \int_{B} \omega$$

\begin{align}d \omega &= d(x^3 dy \wedge dz + y^3 dz \wedge dx + z^3 dx \wedge dy)\\ &= d x^3 dy \wedge dz + d y^3 dz \wedge dx + d z^3 dx \wedge dy\\ &= 3x^2 dx \wedge dy \wedge dz + 3y^2 dy \wedge dz \wedge dx + 3z^2 dz \wedge dx \wedge dy\\ &= 3(x^2 + y^2 + z^2) dx \wedge dy \wedge dz\end{align}

First I calculate the whole solid:

\begin{align}\int_{A} d\omega &= \int_{A} 3(x^2 + y^2 + z^2) dx \wedge dy \wedge dz\\ &= 3 \int_{A} x^2 + y^2 + z^2 d^3(x,y,z)\\ &= 3 \int_0^1 \int_{( x^2 + y^2 < 1 - z^2 )} x^2 + y^2 + z^2 d^2(x,y) dz\\ &= 3 \int_0^1 \frac{1}{2}(1 - z^2)^2 + z^2(1 - z^2) dz\\ &= 3 \int_0^1 1 - 2z^2 + z^4 + z^2 - z^4 dz\\ &= 3 \int_0^1 1 - z^2 dz\\ &= \frac{1}{2} - \frac{1}{3}= \frac{1}{6}\end{align}

Then I would like to calculate the circle - I know that $z = 0 \implies dz = 0$, but then I get:

$\int_{B} \omega = \int_{(x^2 + y^2 < 1)} 0 + 0 + z^3 dx \wedge dy = 0$

Why is that? I thought that it should be just $\pi$ (since I evaluate area of a circle with radius $= 1$) but I don't get anything.

Having $\int_{B} \omega = 0$ I would get: $$\int_{\partial A} \omega = \int_{S^{+}} \omega = \frac{1}{6}$$

Any idea what do I do wrong?

Answer 1

From \begin{align}\tag{1} &\iiint\limits_{x^2+y^2+z^2\le 1}(x^2+y^2+z^2)\,dx\,dy\,dz =\int_0^1\int_0^\pi\int_0^{2\pi} r^2\,r^2\sin\theta\,d\theta\,d\varphi\,dr\\ &\quad\quad =\Big(\int_0^1r^4\,dr\Big)\,(2\pi)\,2=\frac{4\pi}{5} \end{align} we see $$\tag{2} \int_Ad\omega=\iiint\limits_{\scriptstyle x^2+y^2+z^2\le 1\atop \scriptstyle z\ge 0}3(x^2+y^2+z^2)\,dx\,dy\,dz=\frac{6\pi}{5} $$ because this integral must be half of $4\pi/5$ times $3\,.$

Proof. Use the symmetry of the integrand in (1) and the integration region under $z\mapsto -z\,.$

To get $$ \int_{S^+}\omega $$ you use Gauss' theorem (not Stokes') by which $$\tag{3} \int_{\partial A}\omega =\int_{S^+}\omega\color{red}{-}\underbrace{\int_B\omega}_{(*)}=\int_A\,d\omega=\frac{6\pi}{5}. $$ The missing piece is (*) over $B=\{x^2+y^2\le 1\}$ which should not be that hard to calculate. The minus is important because we must orient $B$ such that the normal vector points out of $A\,.$

In this case however you are correct that $$ \int_B\omega=0 $$ because only the $dx\wedge dy$ term contributes and there $z^3$ is zero.

There is nothing wrong with this. We have a vector field $$ F=\begin{pmatrix}x^3\\y^3\\z^3\end{pmatrix} $$ that has no flux through $B\,.$ Therefore, all the flux created by the divergence must get out of $A$ through the upper hemisphere.

To further address the comments:

• Orientation again: in $$ \int_B\dots dx\wedge dy $$ the normal vector at points of $B$ points upward (positive $z$ direction) but Gauss wants that we let it point downward. Hence the minus sign in (3). To memorize this you take for $\partial A$ a cube and a constant (hence divergence free) vector field. If you had no minus sign between the integrals of two opposite faces your fluxes would add up instead of becoming zero together.

• There is no "right" coordinate system. Due to the change of variables formula all of them must give the same result. The choice of the coordinate system is pure convenience. Being a good circus horse I shall do it now in cylindrical coordinates: \begin{align} \int_Ad\omega&=\int_0^1\int_0^{\sqrt{1-z^2}}\int_0^{2\pi} 3(\rho^2+z^2)\,\rho\,d\varphi\,d\rho\,dz\\&=6\pi\int_0^1\int_0^{\sqrt{1-z^2}}(\rho^3+z^2\rho)\,d\varphi\,d\rho\,dz\\ &=6\pi\int_0^1\frac{(1-z^2)^2}{4}+\frac{z^2(1-z^2)}{2}\,dz=\frac{6\pi}{5}\tag{4} \end{align} as it must.

Answer 2

At least its easy to integrate the form in spherical coordinates

$$x \to \cos \phi \sin \theta , \ y \to \sin \phi \sin \theta, \ z \to \cos \theta$$

$$\{\text{dx}\to \text{d$\theta $} \ \cos (\theta )\ \cos (\phi )-\text{d$\phi $} \ \sin (\theta ) \ \sin (\phi ),\ \ \text{dy}\to \text{d$\theta $} \ \cos (\theta ) \ \sin (\phi )+\text{d$\phi $} \ \sin (\theta ) \ \cos (\phi ),\ \ \text{dz}\to -\text{d$\theta $} \ \sin (\theta )\}$$

The resulting 2-form is (typo corrected)

$$\sin (\theta ) \text{d$\theta $}\wedge \text{d$\phi $} \left(\frac{1}{4} \sin ^4(\theta ) (\cos (4 \phi )+3)+\cos ^4(\theta )\right)$$

My result is $\frac{6 \pi }{5}$