Calculation of Integral of Pulled-Back $1$-Form. I have the function $F: S^1 \rightarrow S^1 \times S^1 \times  S^1$ given by $F(p) =(p,p,p)$ and a 1-form $\omega = dt_1 + dt_2 + dt_3$ over $S^1$.
I need to do the explicit calculations of 
$\int_{F(S^1 )}\omega$.
I know that $F^{*}\omega = 3\,dt$. But how do i calculate the Integral?
 A: Your first thoughts are correct. To be precise, let's parametrize $S^1\subseteq \mathbb{R}^2$ by $\theta(x)=(\cos x,\sin x)$. Let's give $S^1\times S^1\times S^1$ coordinates on almost every point by $(t_1,t_2,t_3)$ parametrized analogously. Then, your map is $F(\theta)=(\theta,\theta,\theta)$. It follows that the Jacobian for this map is 
$$ 
\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}.$$
As a consequence, the Jacobian for the dual map on forms is 
$$ \begin{bmatrix}
1&1&1
\end{bmatrix}.$$
So, as you predicted $F^*(\omega)=F^*(dt_1+dt_2+dt_3)=3 \, d\theta.$ Now, we can use the usual change of variables formula twice to complete the calculation:
$$ \int_{F(S^1)}\omega=\int_{S^1}F^*(\omega) = \int_{S^1} 3 \, d\theta = \int_0^{2\pi}3 \| \theta'(x)\| \, dx=\int_0^{2\pi}3 \, dx=6\pi.$$
Note: You may object to the fact that my parametrization does not cover the whole manifold, but it misses a mere point. The integral is not perturbed by the absence of a measure zero subset of the manifold and so we are done.
