Double integral of $e^{x^2+y^2}dydx$?

Question

I'd like to know the detailed solution of $$\int_{x=0}^{x=1} \int_{y=0}^{y=2} e^{x^2+y^2} dy dx.$$

I know I must find $\int_{y=0}^{y=2} e^{x^2+y^2}dy$, first but I don't even know how to start it.

Answer 1

Without context it's hard to suggest the tools, but for the exact problem you are posing notice that $$ \int_{y=0}^{y=2} e^{x^2+y^2} dy = \int_{y=0}^{y=2} e^{x^2} e^{y^2} dy = e^{x^2} \int_{y=0}^{y=2} e^{y^2} dy $$ So if we define $\int_0^x e^{u^2} du = F(x)$ then your double integral becomes $$ \int_{x=0}^{x=1} \int_{y=0}^{y=2} e^{x^2+y^2} dy dx = \left( \int_{x=0}^{x=1} e^{x^2} dx \right) \cdot \left( \int_{y=0}^{y=2} e^{y^2} dy \right) = F(2) \cdot F(1). $$ How you find $F(x)$ is a more complex problem.

Here is another approach. Note that you can switch to polar coordinates, and then $$ \int \int e^{x^2+y^2} dxdy = \int \int e^{r^2} r dr d\theta $$ and $$ \int r e^{r^2} dr $$ can be integrated using the substituion $u = r^2$. The trick is to map the region correctly, which for your square is not an easy thing to do either.

Hence me asking, in what context is your question?

Answer 2

$\int_0^1\int_0^2e^{x^2+y^2}~dy~dx$

$=\int_0^1\int_0^2e^{x^2}e^{y^2}~dy~dx$

$=\int_0^1e^{x^2}~dx\int_0^2e^{y^2}~dy$

$=\left[\dfrac{\sqrt\pi~\text{erfi}(x)}{2}\right]_0^1\left[\dfrac{\sqrt\pi~\text{erfi}(y)}{2}\right]_0^2$

$=\dfrac{\pi~\text{erfi}(1)~\text{erfi}(2)}{4}$