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Let $A \subseteq \Bbb R^{n-1}$ be a $J$-measurable set, let $p = (p_j)_{j=1}^n \in \Bbb R^n$ with $p_n > 0$, and define $$S = \{ x \in \Bbb R^n \mid x = (1-t)a+tp , \text{for }a\in A \times \{0\} \text{ and }t\in ]0,1[\}.$$Then we must find a formula for the volume of $S$ in terms of the volume of $A$. I did it, but I'm a bit insecure about my work and I'd like some feedback - if there are any mistakes, how can the solution be improved, etc. This is a question from an old advanced calculus test, here.

Define $f\colon A \times ]0,1[ \to S$ by $$\begin{align} f(a_1,\ldots,a_{n-1},t) &= (1-t)(a_1,\ldots,a_{n-1},0) + tp \\ &= ((1-t)a_1+tp_1,\ldots,(1-t)a_{n-1}+tp_{n-1},tp_n) \end{align}$$

It is easy to see that $f$ is a diffeomorphism. We have: $$Df(a_1,\ldots,a_{n-1},t) = \begin{pmatrix} 1-t & 0 & \cdots & 0 & p_1-a_1 \\ 0 & 1-t & \cdots & 0 & p_2-a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1-t & p_{n-1}-a_{n-1} \\ 0 & 0 & \cdots & 0 & p_n\end{pmatrix},$$so $\det Df(a_1,\cdots,a_{n-1},t) = p_n (1-t)^{n-1}>0$. Finally: $$\begin{align} {\rm Vol}(S) &= \int_S 1 = \int_{f(A\times ]0,1[)} 1 \\ &= \int_{A \times ]0,1[} p_n(1-t)^{n-1}\,{\rm d}a_1\,{\rm d}a_2\,\ldots\,{\rm d}a_{n-1}\,{\rm d}t \\ &= \int_0^1 \int_A p_n(1-t)^{n-1}\,{\rm d}a_1\,{\rm d}a_2\,\ldots\,{\rm d}a_{n-1}\,{\rm d}t \\ &= p_n\,{\rm Vol}(A) \int_0^1(1-t)^{n-1}\,{\rm d}t \\ &= \frac{p_n}{n}\,{\rm Vol}(A).\end{align}$$

Thanks.

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I asked the professor and this was the solution expected.

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