integration in five dimensions space I am doing this problem: Consider the differential form
$$a=p_1 \, dq_1+p_2 \, dq_2-(p_1^2+p_2^2+q_1^2+q_2^2) \, dt\text{ in }\mathbb R^5=(p_1,p_2,q_1,q_2,t).$$
(a) Compute the differential $da$ and the form $a\wedge da$.
(b) Evaluate the integral $\int_S da\wedge da$ where $S$ is the 4-dim surface in $R^5$ defined by the equation $p_1^2+p_2^2+q_1^2+q_2^2=t$ and the inequality $1\le t\le 2$.
Here is part of my solution:
(a) $da=dp_1\wedge dq_1+dp_2\wedge dq_2-2p_1dp_1\wedge dt-2p_2dp_2\wedge dt-2q_1dq_1\wedge dt-2q_2dq_2\wedge dt$
$a\wedge da=p_1dq_1\wedge(dp_2\wedge dq_2-2p_1dp_1\wedge dt-2p_2dp_2\wedge dt-2q_2dq_2\wedge dt)$
${}+p_2dp_2\wedge(dp_1\wedge dq_1-2p_1dp_1\wedge dt-2p_2dp_2\wedge dt-2q_1dq_1\wedge dt)$ 
${}-(p_1^2+p_2^2+q_1^2+q_2^2)dt\wedge(dp_1\wedge dq_1+dp_2\wedge dq_2)$
I stopped here, since I think this expression is too long. I cannot imagine this problem is so complicated. 
So my first question is that is there any quick way to calculate $a\wedge da$? Or the only way is to continue the above calculation, and leave so many terms there(I guess only two pairs can be collected)?
(b) I realized that the integrand $$da\wedge da=d(a\wedge da)$$, so I think this is a good indicator for us to use the Stokes's theorem $$\int_Sda\wedge da=\int_{\partial S}a\wedge da$$. And $\partial S$ is just $p_1^2+p_2^2+q_1^2+q_2^2=1$ and $p_1^2+p_2^2+q_1^2+q_2^2=2$, which I guess is what the people who made this problem want us to do. 
So my second question is how do we calculate $\int_{\partial S}a\wedge da$. By this I mean:


*

*Do I really need to calculate the integral one by one(I guess there are at least eight terms in the expression of $a\wedge da$ after collecting the terms)?

*Any orientation issues I need to be careful about the two "balls" $p_1^2+p_2^2+q_1^2+q_2^2=1$ and $p_1^2+p_2^2+q_1^2+q_2^2=2$? I think I need to use the parametric form to calculate the integral, right? By the way, do I really need to use the polar coordinate, which is four layers of $sin, cos$? or there is some other way to calculate the final integral?
Thank you very much!
 A: I'm not sure why you were asked to "compute" those forms in all their lengthy glory. The key thing to realize is this: As you suggested, $\partial S$ is the union of those two $3$-spheres (appropriately oriented—I'll get to that in a moment). Since $dt=0$ on $\partial S$ (as each component is contained in a slice with $t=\text{constant}$), any term in $a\wedge da$ involving $dt$ can be ignored. So, note that we have
$$a\wedge da = p_1\,dq_1\wedge dp_2\wedge dq_2 + p_2\,dq_2\wedge dp_1\wedge dq_1 + \text{terms involving } dt\,.$$
To integrate this $3$-form over a sphere $p_1^2+p_2^2+q_1^2+q_2^2=R^2$ it is easiest to apply Stokes's Theorem again, since this sphere does bound a $4$-dimensional ball of radius $R$. You should easily convince yourself that you get twice the volume of that ball when you integrate. Well, not so quickly. We may get the negative.
So far as orientation is concerned, we first need to orient the $4$-manifold with boundary $S\subset\Bbb R^5$. Let's agree (since $S$ can be parametrized globally by $(p_1,p_2,q_1,q_2)$) to orient $S$ by declaring $dp_1\wedge dp_2\wedge dq_1\wedge dq_2>0$ on $S$. (Thus $da\wedge da$ gives the negative of the volume form with this convention.) Now think about how you orient a cylinder $S^1\times [1,2]\subset \Bbb R^3$. The two boundary components have opposite orientation, and since boundary orientation is determined by putting the outward-pointing normal first in the list of tangent vectors, we see that $\partial (S^1\times [1,2]) = S^1\times \{1\} - S^1\times\{2\}$. You should now think this through, analogously, with the surface $M$ given by $z=x^2+y^2$, $1\le z\le 2$. Orienting it by $dx\wedge dy = r\,dr\wedge d\theta$ we see that $\partial M$ has the top circle oriented positively and the bottom circle oriented negatively.
I now leave it to you to sort out all the signs in your problem in $\Bbb R^5$. You might want to double-check with your instructor on how $S$ is oriented. I made an arbitrary choice.
