Evaluation of $I=\iint_R e^{-(x^2+y^2)} \,dx\,dy$ by change of variable

Question

I am asked to evaluate: $$I=\iint_R e^{-(x^2+y^2)} \,dx\,dy$$ where R denotes the quadrant $x, y \ge 0$, by converting to polar coordinates, ie. $$x=r\cdot \cos(t) \ ,y = r \cdot \sin(t)$$ I have worked this out to $I=\dfrac{\pi}{4}$.

My issue is with the following...I also tried the change of variables as:$$x=\cos(t) \ ,y = \sin(t)$$ taking t in the range of: $t \in [\pi/4, \pi/2]$. These were taken this way since for those values of t: $x, y \ge 0$.

Differentiating we get: $$dx = -\sin(t)dt \\ dy = \cos(t)dt$$

However this does not work out to $I=\dfrac{\pi}{4}$, which is the correct answer. Any reasons why? Is the range of $t$ incorrect? Or is the change of variable incorrect? If so, why?

My working:

$$\iint_R e^{-(x^2+y^2)} \,dx\,dy$$ $$=\int_{\pi/4}^{\pi/2}\int_{\pi/4}^{\pi/2} e^{-(\cos(t)^2+\sin(t)^2)} \,(-\sin(t)dt)\,(\cos(t)dt)$$ $$=-e^{-1}\int_{\pi/4}^{\pi/2}\int_{\pi/4}^{\pi/2} \sin(t)\cos(t)dt\,dt$$ $$=-\dfrac{1}{2e}\int_{\pi/4}^{\pi/2}\int_{\pi/4}^{\pi/2} \sin(2t)dt\,dt$$ $$=-\dfrac{e \pi}{16}$$

Answer 1

Interpret the original double integral as volume under the surface (surface being defined by $f(x,y) = e^{-(x^2 + y^2)}$, the region of integration being the whole x-y plane).

So when you change the variables using just $x = \cos{t}$ and $y = \sin{t}$, you would be integrating just over the unit circle. What you are trying to do is tantamount to reducing a surface integral to a line integral in a very weird way. (Weird in the sense that you changed the variables, brought it down to a simple curve, yet the integral is still a double integral). Thus your double integral needs to remain a double integral if you want to span the whole x-y plane. In short, it's conceptually wrong to reduce a surface integral to a line integral in the way you did it. (I'm trying not to invoke Green's theorem here, which does reduce surface integral to line integral).

Also, when you change the variables in a double integral using a function $g: \mathbb{R}^2 \to \mathbb{R}^2$, you must take into consideration the change in the differential surface element $dx dy$, which is caused by the transformation. The factor by which the differential surface changes is given by the determinant of the Jacobian.

Answer 2

Note that

$$x^2+y^2=\rho^2$$

$$dx \,dy = \rho \,d\rho \,d \theta$$

thus

$$I=\iint_R e^{-(x^2+y^2)} \,dx\,dy=\iint_{R'} \rho e^{-\rho^2} \,d\rho\,d\theta=\int_0^{\frac{\pi}{2}} d\theta \quad \int_0^{+\infty} \rho e^{-\rho^2} \,d\rho$$