= Solution
{c}
The original <double integral> covers the <triangle> $0\leq y\leq x\leq1$. By <Fubini's theorem>, after reversing the order, $y$ runs from $0$ to $1$ and $x$ from $y$ to $1$. Since a <continuous function> is integrable on this bounded region,
$$
\int_0^1\int_0^x f(y)\,dy\,dx
=\int_0^1\int_y^1 f(y)\,dx\,dy
=\boxed{\int_0^1(1-y)f(y)\,dy}.
$$
For the <square pyramid>, a horizontal section at height $z$ is a <square> of side $2(1-z)$: each vertex moves linearly from the base towards $(1,1,1)$. Hence its <cross-sectional area> is $4(1-z)^2$, and its <volume> is $4\int_0^1(1-z)^2\,dz$. To use the identity just proved, choose $f(y)=1-y$:
$$
\begin{aligned}
\operatorname{Vol}
&=4\int_0^1\int_0^x(1-y)\,dy\,dx\\
&=4\int_0^1\left(x-\frac{x^2}{2}\right)dx
=4\left(\frac12-\frac16\right)
=\boxed{\frac43}.
\end{aligned}
$$
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