# integration in five dimensions space part two

I am following the discussion here: 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 the result from the previous discussion, and I still have some problems about the following work.

We showed that the integration can be evaluated by$$\int_{\partial S}a\wedge da$$(using Stokes's theorem), where $$\partial S:=A\cup B:=p_1^2+p_2^2+q_1^2+q_2^2=1\cup p_1^2+p_2^2+q_1^2+q_2^2=2$$ and $$a\wedge da=p_1dq_1\wedge dp_2\wedge dq_2+p_2dp_2\wedge dp_1\wedge dq_1+terms\quad with\quad dt.$$

So I just start to calculate the above integral(if $dp_1\wedge dp_2\wedge dq_1\wedge dq_2\wedge dt>0$): $$\int_{A\cup B}p_1dq_1\wedge dp_2\wedge dq_2+p_2dp_2\wedge dp_1\wedge dq_1$$

Firstly, I use the parametric form for $A$ (with radius 1): $p_1=\cos\theta_1$, $p_2=\sin\theta_1\cos\theta_2$, $q_1=\sin\theta_1\sin\theta_2\cos\theta_3$, $q_2=\sin\theta_1\sin\theta_2\sin\theta_3$, where $\theta_{1,2}\in[0,\pi], \theta_3\in[0,2\pi].$

Secondly, I calculate the Jacobi $$\left. \begin{array}{c} p_1\\ p_2\\ q_1\\ q_2\\ \end{array} \right. \left( \begin{array}{ccc} -\sin\theta_1&0&0\\ \cos\theta_1\cos\theta_2&-\sin\theta_1\sin\theta_2&0\\ \cos\theta_1\sin\theta_2\cos\theta_3&\sin\theta_1\cos\theta_2\cos\theta_3&-\sin\theta_1\sin\theta_2\sin\theta_3\\ \cos\theta_1\sin\theta_2\sin\theta_3&\sin\theta_1\cos\theta_2\sin\theta_3&-\sin\theta_1\sin\theta_2\cos\theta_3\\ \end{array} \right)$$

Thirdly, I do the integration of the first part on $A$: $$\int_{A}-p_1dp_2\wedge dq_1\wedge dq_2.$$ This is equal to(plug in the parameters form):

$$\int_0^{\pi}\int_0^{\pi}\int_0^{2\pi}(-\cos\theta_1)\space \det(2,3,4\,\text{rows of Jacobi})\space d\theta_1d\theta_2d\theta_3.$$

Am I doing the right way to calculate the integral?(So that I could continue with the rest three terms)

Another question is how to determine the order of the parameters wedge product?Is it $d\theta_1d\theta_2d\theta_3$, or some other order? What determine this order? I need to plug the parameters into $dp_2\wedge dq_1\wedge dq_2$ to see the result whether it is $d\theta_1d\theta_2d\theta_3$ or not, right?

Thank you very much!

• Goodness. This is painful. Why on Earth would someone assign this? This is an exercise in tedium if I've ever seen it... – Cameron Williams Apr 14 '14 at 23:41
• This is qualify exam. I am not very familiar with this topic, so I am here to ask some some tricks or other small points in practice that I need to know. – breezeintopl Apr 15 '14 at 0:47

Let $H=p_1^2+p_2^2+q_1^2+q_2^2.$ Let $D$ be the subset of $\mathbb R^4$ given by $1\leq H\leq 2$. Then $S$ is the graph of the function $t=H$ over $D$. Now change variables in the integral to integrate over $D$ instead of $S$ and you get the volume of $D$ (since $da\wedge da$ pulls back to the volume form on $D$).
Let $c$ be the volume of the unit 4 ball. Then the volume of the 4 ball of radius $r$ is $cr^4$. So the required integral is $4c-c=3c$.
To calculate $c$ use induction. Consult any vector calculus textbook.
• Thank you for your answer! So you do not use Stokes formula and calculate directly?(I actually don't understand pull back) what about the first two terms $p_1dq_1+p_2dq_2$? – breezeintopl Apr 16 '14 at 17:35
• Pull back just means change of variable in a differential form. Just make the change of variable $t=H(p_1,...)$ and you get that $da\wedge da$ changes to the volume form in $\mathbb R^4$. – Gil Bor Apr 16 '14 at 18:12