optimize with respect to domain shape

Let $\Gamma$ be the set of all closed $C^2$ curves in the plane which enclose unit area and let $\Omega$ be the set of all subsets of $\mathbb{R}^2$ that are enclosed by some curve in $\Gamma$. Now let $f: \mathbb{R}^2\rightarrow\mathbb{R}$ be a real-valued function on the plane. How can we find the set $\omega\in\Omega$ with boundary $\gamma\in\Gamma$ such that $\int_\omega\ f\$ is minimized?

A related question (with a more physics-style interpretation): Let $\Omega$ and $\Gamma$ be as before. Let $u(\vec{x})$ be the real-valued function on the plane that solves the PDE $$\Delta u(\vec{x}) = 0\ \text{for } \vec{x}\in\omega$$

$$u(\vec{x}) = 1\ \text{for } \vec{x}\in \partial\omega,$$

where $\omega$ is some set in $\Omega$. How can we find the $\omega$ that minimizes the quantity $\int_{\gamma}|du/dn|^2$ where $\gamma$ is the boundary of $\omega$ with unit outward normal $n$ ?

I'm more interested in finding out which branch of mathematics studies questions like this and what concepts/tools are important to approach questions like this. Any references or suggestions to similar problems are greatly appreciated. (A friend suggested I tag this as geometric measure theory, but I don't know how appropriate that is)

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I'd say these belongs to the family of shape optimization problems in the Calculus of Variations (of whom the most famous example is the isoperimetric problem). As to the first one, you should make some assumption on $f$ to ensure existence (for e.g. $f(x):=\frac{1}{1+|x|^2}$ gives a non attained infimum, equal to 0). Natural assumptions are e.g.: $f$ is smooth, coercive, and its only critical point is its global minimum. This should give as minimizer a level set of $f$ -the last assumption is just to get a regular simple curve. (The latter problem seems less easy and should require at least some small computation. I guess the solution is a circle). PS: Say, what about any $\omega$ as minimizer, as $u=1$ constant in $\omega$ ? :-) The latter problem was a joke, wasn't it?