Define a differential 1-form $\omega:\mathbb R^2 \to \Lambda_1(\mathbb R^2)$ by $\omega(a,b)=adx+bdy.$ Compute: $\int_{\Phi|_{[0,1]}}\omega$

Question

Let $\Phi : \mathbb R \to \mathbb R^2$ and $\Phi(t)=(t^3,t^4)$. Define a differential 1-form $\omega:\mathbb R^2 \to \Lambda_1(\mathbb R^2)$ by $\omega(a,b)=adx+bdy.$

Compute: $\int_{\Phi|_{[0,1]}}\omega$

My idea:

$ \frac{\partial\Phi}{\partial t}(a,b)=(3t^2,4t^3)$

$J_Φ(a, b)= \begin{pmatrix} 3t^2 \\ 4t^3 \end{pmatrix}$

$ω(Φ(a, b))(J_Φ(a, b))=adx\begin{pmatrix} 3t^2 \\ 4t^3 \end{pmatrix}+bdy\begin{pmatrix} 3t^2 \\ 4t^3 \end{pmatrix}=a3t^2+b4t^3$

Then $\int_{\Phi|_{[0,1]}}\omega=\int_{0}^{1}a3t^2+b4t^3$ ?

Is this correct so far? If not how do I proceed?

Answer 1

Your approach is correct (aside from the missing $dt$), but it's computationally more natural to make the symbolic substitutions $$ \begin{aligned} x &= t^{3}, \\ dx &= 3t^{2}\, dt; \end{aligned} \qquad \begin{aligned} y &= t^{4}, \\ dy &= 4t^{3}\, dt. \end{aligned} $$ Thus $$ \int_{\Phi[0,1]} \omega = \int_{0}^{1} (3at^{2} + 4bt^{3})\, dt = a + b. $$