Suppose that the surface $S$ has projection $R$ on the $x y$ plane (see figure of Prob.17). Show that
$$
\int_{S} \int \mathbf{A} \cdot \mathbf{n} d S=\int_{\mathbb{R}} \int A \cdot \mathbf{n} \frac{d x d y}{|\mathbf{n} \cdot \mathbf{k}|}
$$
By Problem 17 , the surface integral is the limit of the sum
$$
\sum_{p=1}^{N} \mathbf{A}_{p} \cdot \mathbf{n}_{p} \Delta s_{p}
$$
The projection of $\Delta S_{\phi}$ on the $x y$ plane is $\left|\left(\mathbf{n}_{p} \Delta S_{p}\right) \cdot \mathbf{k}\right|$ or $\left|\mathbf{n}_{p} \cdot \mathbf{k}\right| \Delta S_{p}$ which is equal to $\Delta x_{p} \Delta y_{p}$ ' so that $\Delta S_{p}=\frac{\Delta x_{p} \Delta y_{p}}{\left|\mathbf{n}_{\hat{p}} \cdot \mathbf{k}\right|}$. Thus the sum $(1)$ becomes
$$
\sum_{p=1}^{M} \mathrm{~A}_{\hat{p}} \cdot \mathbf{n}_{p} \frac{\Delta x_{p} \Delta_{y_{p}}}{\left|\mathbf{n}_{\hat{p}} \cdot \mathbf{k}\right|}
$$
By the fundamental theorem of integral calculus the limit of this sum as $M \rightarrow \infty$ in such a manner that the largest $\Delta x_{p}$ and $\Delta y_{p}$ approach zero is
$$
\int_{R} \int A \cdot n \frac{d x d y}{|\mathbf{n} \cdot k|}
$$
and so the required result follows.
Strictly speaking, the result $\Delta S_{p}=\frac{\Delta_{x}_{p} \Delta y_{p}}{\left|\mathbf{n}_{\hat{p}} \cdot \mathbf{k}\right|}$ is only approximately true but it can be shown on closer examination that they differ from each other by infinitesimals of order higher than $\Delta x_{p} \Delta y_{p}$, and using this the limits of $(1)$ and $(2)$ can in fact be shown equal.