Evaluate the integral $\iint_De^{-x^2}\,dx\,dy$

Question

Find $\iint_De^{-x^2}\,dx\,dy$, where $D$ is the triangle formed by points $O(0,0), A(1,0), B(1,1)$

I'm totally lost on this one. I only have experience with circle functions, and converting them to polar and finding the bounds in $π$ and $r$, but here its just points.

Can anyone help me understand this? Do i just bound $x$ to $[0,1]$ and $y$ to $[0.1]$?

Answer 1

Note that $D=\{(x,y)\in\mathbb{R}^2:0\leq x,y\leq1\text{ and }x\leq y\}$. Therefore $$\iint_De^{-x^2}dxdy=\int_0^1\left(e^{-x^2}\int_0^{x}dy\right) \;dx$$ We have that $\int_0^{x}dy=x$ and therefore $$\iint_De^{-x^2}\,dx\,dy=\int_0^1xe^{-x^2}\,dx=\frac12\left(1-\frac1e\right)$$

Answer 2

$$\int_0^x\int_0^1e^{-x^2}\,dx\,dy$$

$$\int_0^1\int_0^xe^{-x^2}\,dy\,dx=\int_0^1(e^{-x^2}y)|_0^x$$ $$\int_0^1xe^{-x^2}\,dx$$

$u=-x^2$

$du=-2x\,dx$

$dx=\frac{du}{-2x}$

$$\int_0^1xe^u\frac{du}{-2x}=-\frac{1}{2}\int_0^{-1}e^u\,du$$

$$-\frac{1}{2}(e^u|_0^{-1})=-\frac{1}{2}(e^{-1}-1)=-\frac{1}{2e}+\frac{1}{2}$$

Answer 3

$$ \iint\limits_{x,y\,:\,0\,<\,y\,<\,x\,<\,1} e^{-x^2} \,d(x,y) $$ Here I have written $d(x,y)$ rather than $dx\,dy$ or $dy\,dx$ to indicate this is a double integral rather than an iterated integral and there is no particular order of integration.

The fact that $e^{-x^2}$ is everywhere nonnegative is enough to imply the double integral equals this iterated integral, whose evaluation is easy: $$ \iint\limits_{x,y\,:\,0\,<\,y\,<\,x\,<\,1} e^{-x^2} \,d(x,y) = \int_0^1 \left( \int_0^x e^{-x^2} \, dy\right) \,dx $$ (The only time a double integral is not equal to an iterated integral in this way is when the integrals of the positive and negative parts of the function are both infinite, in which case the double integral is not well defined and the iterated integral is in some cases still well defined. In that case the two iterated integrals taken in opposite orders ($dx\,dy$ versus $dy\,dx$ sometimes evaluate to different numbers.)