Let $\Omega\subset\mathbb{R}^3$ be the upper half ball with radius $a$, i.e. the region bounded below by $z=0$ and above by $x^2+y^2+z^2=a^2$. Compute $$ \int_{\partial\Omega} xz\,dy\wedge dz + yz\,dz\wedge dx + (x^2+y^2+z^2)\,dx\wedge dy. $$
I have computed this integral directly and by Stokes' theorem, but I got different results.
Let $V=\{(r,\theta):0<r<a, 0<\theta<2\pi\}$ and $U=\{(\phi, \theta):0<\phi<\pi/2,0<\theta<2\pi\}$.
$$\begin{align*} &\int_{\partial\Omega} xz\,dy\wedge dz + yz\,dz\wedge dx + (x^2+y^2+z^2)\,dx\wedge dy \\ =& \int_{V} r^2\,d(r\cos\theta)\wedge d(r\sin\theta) \\ &+ \int_{U} (a\sin\phi\cos\theta)(a\cos\phi)\, d(a\sin\phi\sin\theta)\wedge d(a\cos\phi) \\ &+ \int_{U} (a\sin\phi\sin\theta)(a\cos\phi)\, d(a\cos\phi)\wedge d(a\sin\phi\cos\theta) \\ &+\int_{U} a^2\, d(a\sin\phi\cos\theta)\wedge d(a\sin\phi\sin\theta) \\ =& \int_{V} r^3\,dr\wedge d\theta \\ &+ \int_{U} (a\sin\phi\cos\theta)(a\cos\phi)(a^2\sin^2\phi\cos\theta)\, d\phi\wedge d\theta \\ &+ \int_{U} (a\sin\phi\sin\theta)(a\cos\phi)(a^2\sin^2\phi\sin\theta)\, d\phi\wedge d\theta \\ &+\int_{U} a^2(a^2\cos\phi\sin\phi)\, d\phi\wedge d\theta \\ =& \int_0^{2\pi}\int_0^a r^3\,dr\,d\theta \\ &+ a^4\int_0^{\frac{\pi}{2}}\int_0^{2\pi} (\sin^3\phi\cos\phi\cos^2\theta+\sin^3\phi\cos\phi\sin^2\theta+\cos\phi\sin\phi)\,d\theta\,d\phi \\ =&\ 2\pi a^4. \end{align*}$$
By Stokes' theorem, $$ \begin{align*} &\int_{\partial\Omega} xz\,dy\wedge dz + yz\,dz\wedge dx + (x^2+y^2+z^2)\,dx\wedge dy \\ =& \int_{\Omega} (z+z+2z)\,dx\wedge dy\wedge dz \\ =& \int_0^a\int_0^{\frac{\pi}{2}}\int_0^{2\pi} (4\rho\cos\phi) (\rho^2\sin\phi)\, d\theta\,d\phi\,d\rho \\ =&\ \pi a^4. \end{align*} $$
I have been debugging for hours, but I'm still struggling to find a mistake.