Is it possible to determine the actual distance it would take to walk from one point to another on a map (including hills)? 
I have just read about line integration and how we can determine the area along a curve (the blue area in between the red curve and the blue curve in the image linked). In the image linked above, I was wondering if it is possible to determine the equation of the blue curve by knowing the equation of the red curve and blue area between the red curve and blue curve.
I was thinking this could be useful because say we had a map and wanted to determine the distance between two points with hills in between them. By knowing the equation of the curve drawn on a 2D map (which would represent sea level and the red curve in the image), we can determine the distance of that path without considering the hills using arc length integration. And if we also know the function that gives the surface of the earth (the surface including the hills), I was wondering if this is enough information to determine the actual distance of the path (the blue curve in the image).
 A: Setup
Suppose the path in the $xy$-plane (the red curve) is given by a known parametric function $\mathbf r(t)=\langle x(t),y(t)\rangle$ for $a\lt t\lt b$. You have asked two questions:

*

*Is it possible to determine the equation of the blue curve (the path in 3d space) by knowing the blue area (the signed area counting area below the $xy$ plane as negative) between the red curve and the blue curve?

*If we know the function that gives the surface, can we determine the length of the blue curve?

The short answers are "maybe, depending on what you mean" and "yes", respectively.
Path from Area
If you just know the total signed area, that's not nearly enough information. But if you know that signed area for the portion of the curve from $t=a$ to $t=T$ for any number $T$, then under nice conditions you can indeed recover the blue curve.
Suppose the area up until time $T$ is given by $A(T)$, and is known for $a\le T\le b$. And let $h(t)$ denote the height of the blue curve at time $t$. Then by the definition of line integral, we know $\int_a^T h(t)|\mathbf r'(t)|\,\mathrm dt=A(T)$. Differentiating both sides with respect to $T$ and applying the Fundamental Theorem of Calculus, we have $h(T)|\mathbf r'(T)|=A'(T)$. This allows is to solve for $h$: $h(T)=A'(T)/|\mathbf r'(T)|$.
The entire parametrization of the blue curve combines the height information with the red curve, so it's $\langle x(t),y(t),h(t)\rangle$, or $\left\langle x(t),y(t),\dfrac{A'(t)}{\sqrt{\left(x'(t)\right)^2+\left(y'(t)\right)^2}}\right\rangle$.
Length from Surface
Suppose the height of the surface is given everywhere by $f(x,y)$. Then the blue curve is automatically parametrized by $\mathbf b(t)=\langle x(t),y(t),f(x(t),y(t))\rangle$. The length of the blue curve is then given by a 3d line integral of the scalar function that is constantly $1$.
The length is $\int_a^b|\mathbf b'(t)|\,\mathrm dt$, which expands to $$\int_a^b\sqrt{\left(x'(t)\right)^2+\left(y'(t)\right)^2+\left(\dfrac{\mathrm{d}}{\mathrm dt}f(x(t),y(t))\right)^2}\,\mathrm dt\text{.}$$ Using the multivariate chain rule, we can rewrite this as follows, where $f_x$ and $f_y$ are the partial derivatives of $f$: $$\int_a^b\sqrt{\left(x'(t)\right)^2+\left(y'(t)\right)^2+\left(f_x(x(t),y(t))*x'(t)+f_y(x(t),y(t))*y'(t)\right)^2}\,\mathrm dt\text{.}$$
If you like keeping the $\left(x'(t)\right)^2$ terms together, and abbreviate $x(t)$ as $x$, etc., you could also rewrite this as:
$$\int_a^b\sqrt{(x')^2\left(1+(f_x(x,y))^2\right)+(y')^2\left(1+(f_y(x,y))^2\right)+2f_x(x,y)f_y(x,y)x'y'}\,\mathrm dt\text{.}$$
