Volume of a Special Pyramid 
Let $P$ be a plane in $\mathbb{R}^3$ parallel to the $xy$-plane. Let $\Omega$ be a closed, bounded set in the $xy$-plane with $2$-volume $B$. Pick a point $Q$ in $P$ and make a pyramid by joining each point in $\Omega$ to $Q$ with a straight line segment. Find the $3$-volume of this pyramid.

I know that the volume with be dependent on $B$ and the distance from $\Omega$ to $P$, and the solution probably involves multiple integration, but beyond that I don't know where to start.
Any help would be appreciated.
N.b.: Although tagged as "homework," this problem is no longer "live," i.e. the homework has already been turned in.
 A: Let's answer to a bit more ambitious question: Let $\Omega$ be an open, bounded and connected subset of an $(n-1)$-dimensional hyperplane $P$ of $\mathbb{R}^n$, and let $x_0\in\mathbb{R}^n$ be a point not contained in $P$. What is the volume of the pyramid with basis $\Omega$ and vertex $x_0$?
By a rotation, we may assume $P=\{x\in\mathbb{R}^n|x_n=0\}$. By a reflection, we may also assume $(x_0)_n>0$. Our pyramid is given by
$$\Pi = \{tx_0+(1-t)x|t\in[0,1],\ x\in\Omega\}$$
and its volume is
$$V_\Pi=\int_\Pi dV$$
where $dV=dx_1\ldots dx_n$. We make the following change of variables:
$$\begin{align}
x_n = & t(x_0)_n\\
x_i = & (1-t)y_i + t(x_0)_i
\end{align}$$
where $i=1,\ldots,n-1$. With this we have $dV = (x_0)_n(1-t)^{n-1}dtdy_1\ldots dy_{n-1}$ and after the change of variables our integral becomes
$$\begin{align}
V_\Pi = & \int_\Omega dy_1\ldots dy_{n-1}\int_0^1 dt\ \bigg((x_0)_n(1-t)^{n-1}\bigg)\\
= & \frac{1}{n}(x_0)_nV_\Omega
\end{align}$$
where $V_\Omega$ is the $(n-1)$-dimensional volume of $\Omega$.
Notice how this reduces to the following known formulas


*

*$n=2$: $\mathrm{area\ triangle} = \frac{1}{2}\mathrm{basis}\times\mathrm{height}$

*$n=3$: $\mathrm{volume\ pyramid} = \frac{1}{3}\mathrm{basis\ area}\times\mathrm{height}$

A: Euclid solves this by dividing a triangular prism into three pyramids of equal volume.
