How do I find the area of a general closed curve? (And then generalisation to multiple dimensions) 
*

*Let us say I have a general equation where $f$ is a general function:
$$f(x,y)=0$$
If the function in the Cartesian plane forms a closed curve, is there a general way to find the area bounded by the curve? Or is it only possible for some forms like circles/ellipses, for example.

*In addition, what would happen if I were to ask the same question, but with
$$f(x_1,x_1,x_3,...,x_n)=0$$
Would there be a general way to find the (don't quite know what the name is for this) 'hypervolume' if the function forms a closed shape in $\Bbb{R}^n$?

*Finally, would it be always possible to find the area/hypervolume of the intersecting place of 2 intersecting closed curves (here I ask this question for $\Bbb{R}^2$ and then for $\Bbb{R}^n$)?
EDIT: Unfortunately, some of the answers I'm getting here seem to say that $f(x, y)=0$ is not enough and I need to impose restrictions on it. I was aiming more for a calculus/integration kind of answer.
I don't quite know the terminology for this, but I'm assuming that $f$ can be expressed in terms of $x$, $y$, etc using 'basic' functions like addition, multiplication, exponentiation, etc. Things you'd typically allow in calculus which you can integrate/differentiate.
 A: As to question 1, if you can solve $f(x,y)=0$ for one variable in terms of the other or express the bounds of integration as functions of a single variable, then you can of course use the methods you already know to find the area.  
Another way is to look at the (signed) area “swept out” by the radius vector as you traverse the curve. If you can transform the equation of the curve into polar form $r=g(\theta)$, then this area can be found by integrating $\frac12r^2\,d\theta = \frac12g(\theta)^2\,d\theta$. (This integrand is just the area of a right triangle with sides $r$ and $r\,d\theta$.) More generally, the signed area enclosed by a curve is given in Cartesian coordinates by the line integral $$\frac12\int_\Gamma x\,dy-y\,dx.$$ To evaluate this, you’ll generally need to parametrize the curve, which turns this into an ordinary integral.
A: You have to impose some restrictions on $f$ if your questions is to be meaningful. You also need to clarify what you mean by "closed curve", and by its area. Perhaps "closed curve" is clear enough, you are likely  talking about a continuous Jordan closed curve, i.e. a subset of the plane which is homeomorphic to the circle. (If you don't assume continuity, then 
every plane set is the set of solutions $f(x,y)=0$ where $f$ is the characteristic function of the complement of the set.) 
More importantly, I assume that when you say "area of this shape" you mean the area of the region bounded by the curve, rather that the area of the curve itself. This distinction may seem superficial since the area of most curves (or most nice curves, e.g. differentiable ones) is $0$, but this is not true for every continuous curve and should be taken into account. 
An example of a (simple closed) curve with positive area (the curve itself) was constructed by Osgood. See https://en.wikipedia.org/wiki/Osgood_curve where a reference to his original paper (available online) is given plus more references. 
The first answer posted (by @amd) already tells you how to use an integral along the curve to find the enclosed area (for nice curves). I am personally not aware of something easier, in general. There is a nice connection to the so-called Inscribed square problem by Toeplitz, I won't describe it in detail, but a good survey (on this problem in general) is 
Matschke, Benjamin (2014), "A survey on the square peg problem", Notices of the American Mathematical Society 61 (4): 346–253, doi:10.1090/noti1100, and an interesting relation to area bounded by curves is in the papers by Karasev(and coauthors) and Makeev, referenced there. 
A: This isn't a definitive answer, just a couple of pessimistic observations based on the assumption that you want to start with a closed curve
$$
C = \{(x, y) : f(x, y) = 0\}
$$
and perform some computational procedure on $f$ to obtain the area enclosed by $C$:


*

*The fact that $C$ is the locus of an equation $f(x, y) = 0$ is no restriction: Every closed subset (in the sense of topology) of the plane is the zero set of a continuous function. If $C$ is "smooth" (i.e. is a $1$-dimensional embedded submanifold), you can choose $f$ to be infinitely differentiable with non-vanishing gradient along the curve.

*For any particular curve $C$, there is an infinite-dimensional space of functions defining $C$. For example, if $\phi$ is an arbitrary real-valued function of one variable that vanishes only at $0$, then $g = \phi \circ f$ also defines $C$. Any successful procedure must return the same numerical value for $f$ and $g$.
