Let $M=\{(x,y,u,v) \in \mathbb{R}^4 : x^2 + y^2 = u^2 + v^2 = 1\}$.
Show that $M$ is an orientable subvariety of $\mathbb{R}^4$, explaining the induced orientation.
Consider the $2$-form $\omega = a(x,y,u,v)dx \wedge du$ with $a \in C^\infty(\mathbb{R}^4)$. Compute $\int_M \omega$ in terms of $a$.
Showing that $M$ is a subvariety of $\mathbb{R}^4$ of dimension $2$ is not a problem.
So for 1., by definition, we need to find an everywhere (on $M$) non-vanishing $2$-form $\omega$.
How can I do that? (The same for induced orientation)
For 2. I have some ideas:
$$\int_M \omega = \int_{S^1 \times S^1} a(x,y,u,v)dx \wedge du = \int_0^{2\pi} \int_0^{2\pi} a(cosu,sinu,cosv,sinv) dudv$$
Is this true?
Thanks in advance!
Let $d\theta =x ~dy - y ~ dx$ and $d\phi = u~dv - v~du$. Then $\omega = d\theta \wedge d\phi$ is such a 2-form.
For the second part, you're not quite on the right track: the parameterization by sine and cosines of two arguments -- $s$ and $t$ might have been better names -- is fine, but you can't just turn "dx" into "ds". Details:
Let $$ X(s, t) = (\cos s, \sin s, \cos t, \sin t) $$ so that $$ x(s, t) = \cos s\\ y(s, t) = \sin s\\ u(s, t) = \cos t\\ v(s, t) = \sin t. $$ Then observe that $$ dx = -\sin s ~ ds\\ du = -\sin t ~dt $$ so that $$ dx \wedge du = \sin s \sin t ~ds \wedge dt. $$ so your integral becomes $$ \int_{[0, 2\pi] \times [0, 2\pi]} a(\cos s, \sin s, \cos t, \sin t) \sin s \sin t ~ ds \wedge dt. $$
I can never remember what to do at this last step, and always have to look up an example in Spivak's Calculus on Manifolds -- do I need a factor of 1/2? Do I have to integrate ds dt and then integrate dt ds or ... well, as I say, I can't recall. I think that in practice you just drop the wedge and integrate, but I could easily be wrong. :(