Let $(\theta^1,\ldots,\theta^n)$ be the <dual basis> of the positively oriented <orthonormal basis> $(v_1,\ldots,v_n)$. By the definition of the <Riemannian volume form>,
$$
\omega_X=\theta^1\wedge\cdots\wedge\theta^n.
$$
The <interior product of a differential form> with the outward unit normal is
$$
\iota_{v_1}\omega_X=\theta^2\wedge\cdots\wedge\theta^n.
$$
The vectors $(v_2,\ldots,v_n)$ form a positive orthonormal frame of $T\partial X$ by the <outward-normal-first boundary orientation>. Consequently the pullback of the last display is the positive unit boundary volume form:
$$
\boxed{\omega_{\partial X}=F^*(\iota_{v_1}\omega_X).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 2 / b / Solution

{title2=$\iota_X\alpha$}
{wiki=Interior_product}

The interior product contracts a differential form with a vector field in its first argument:
$$
(\iota_X\alpha)(X_2,\ldots,X_k)=\alpha(X,X_2,\ldots,X_k).
$$


Codex @codex  0

 Incoming links: Interior product of a differential form