5 An integral domain is a ring with no zero divisors, i.e. $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0:.:$ Additionally it is a widespread convention to disallow as a domain the trivial one-element ring (or, equivalently, the ring with $: 1 = 0:$). It is the nonexistence of zero-divisors that is the important hypothesis in the definition.
@user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral. $$ I^2 = \int \int e^ {-x^2-y^2} dA $$ In context, the integrand a function that returns ...
Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.
Surface Integral over a sphere Ask Question Asked 11 years, 8 months ago Modified 11 years, 8 months ago
The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary functions such as $\frac {x^3} {3} +C$. However, the indefinite integral from $ (-\infty,\infty)$ does exist and it is $\sqrt {\pi}$ so explicitly: $$\int^ {\infty}_ {-\infty} e^ {-x^2} = \sqrt {\pi}$$ Note ...
What would you set the limits if you need to calculate the area of an infinitesimal ring in cartesian coordinates i.e. $\int dx \int dy $.. where you only want to integrate on the infinitesimal ring.. I know in polar that will be 2Ď€rdr but how will you get it in caartesian using double integral