Given $D \subseteq \mathbb R^2$ and a scalar function $f: \mathbb R^2 \to \mathbb R$, if we express $D$ in the following two way: $$D = \{(x, y) : a\leq x\leq b \mbox{ and } \phi_1(x)\leq y\leq \phi_2(x)\}= \{(x, y) : \psi_1(y)\leq x\leq \psi_2(y)\mbox{ and } c\leq y\leq d \}$$
We can define the double integral over $D$ with respect to the function $f$ as follows:
$$\iint_D f \, dA = \int^b_a\int^{\phi_2(x)}_{\phi_1(x)}f(x, y) \, dy \, dx =\int^d_c\int^{\psi_2(y)}_{\psi_1(y)}f(x, y) \, dx \, dy $$.
In $\mathbb R^3$, I also came across the notion of a surface integral of scalar field over a region $S$ with respect to another scalar function $g$ , denoted by $\iint_S g \, d\mathbf S$. So I attempt to link the two of them as follows: we can construct $S$ such that the projection of $S$ into the $x$–$y$ plane is $D$.
Specifically, I want all the points in $S$ takes the form of $(x, y, f(x, y))$, so it "captures" the region $D$ with $f$ intuitively. Ignore the problem of the normal because it is only going to differ the final answer by a factor of $-1$ anyways.
Now, if I let $g$ to be identically $1$, the surface integral $\iint_Sd\mathbf S$ calculates the surface area of $S$, isn't this what $\iint_D f \, dA$ is? If yes, is it true that
$$\iint_Sd\mathbf S=\iint_D f \, dA$$
? But I worked out that
$$ \iint_S d \mathbf S=\iint_{D}\left(\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}}\right) d x d y$$
Am I correct? If yes, then it seems like the result is not $\iint_D f \, dA$, why is this so?
Best Answer
The surface integral $\iint_X \mathrm d S$ calculates the surface area of $S$. The integral $\iint_D f \mathrm dA$ calculates the (signed) volume under the graph of $f$.