I am doing some practice exams for multivariable calculus and I am having some trouble solving the follwing integral. $$\int_{0}^{1}\left(\int_{x^2}^{1} 4xe^{y^2} dy\right)dx$$
I know that the integration region for is the area enclosed by the curves $y_1 =1$, $x_1=0$ and $y_2 = x^2$. This region can be described by polar coordinates like this i think: for $0 \leq \theta \leq \frac{\pi}{4}: 0 \leq r \leq 1/Cos^2(\theta)$ and for the region $\frac{pi}{4} \leq \theta \leq \pi:0 \leq r = \sqrt{cos^2(\theta)+1}$, I'm going to call this region E henceforth. If not for the 4x part i think this would be enough to solve the integral, i belive i should simply be able to square it, convert it to polar coordinates and then take the square root of it:
$$ \int_{0}^{1}(\int_{x^2}^{1} 4xe^{y^2} dy)dx= \int_{0}^{1}4x\sqrt {\int_{x^2}^{1}e^{y^2} dy \int_{x^2}^{1}e^{z^2} dz } \ dx =\int_{0}^{1}4x\sqrt {\int_{x^2}^{1} \int_{x^2}^{1}e^{y^2} e^{z^2} dydz } \ dx = \int_{0}^{1}x\sqrt{\iint\limits_{E}2re^{r^2}drd\theta} \ dx$$
But changing to polar coordinates wothout changing that x doesn't really feel allowed...
what at first seemed correct was not to square the thing rather i jjust tried converting the whole thing to polar coordinates:
$$\iint\limits_{E}r^2cos(\theta)e^{r^2sin^2(\theta)}drd\theta$$
But since there's no $2 sin(\theta)$ in this expression i'm having a hard time figuring out what the antiderivative should be.
I know what the answer is supposed to be, I'm just not entirely sure how they arrived there. And the way's ive tried solving it so far just seem like they're a bit overly complex.
Anybody have any suggestions
Flipping the order of integration should do it. If you plot the region over which the integration is carried out, and reverse the order of the integration, you'll get
$ I = \displaystyle \int_{x=0}^1 \int_{y = x^2}^1 4 x e^{y^2} dy dx = \int_{y=0}^1 \int_{x = 0}^{\sqrt{y}} 4 x e^{y^2} dx dy $
Now, by carrying out the inner integral, this reduces to
$I = \displaystyle \int_{y = 0}^1 2 y e^{y^2} dy $
which reduces further to,
$ I = \left[ e^{y^2} \right]_0^1 = e - 1 $