Double integration with a little bit difficult interval

Question

Let $T$ be the area between $x=y^2 + 1$ and $x=2$. Find the value of

$$\int\int_{T} [35xy^2 + 7e^x y^3] dx dy$$

I'll show what I've done: my first step was to find $y=\pm 1$. So, the limits of the integral would be $-1\leq y\leq 1$ and $2\leq x \leq y^2+1$, obtaining $$ \int_{-1}^{1} \int_{2}^{y^2+1}[35xy^2 + 7e^x y^3] dx dy.$$ Integrating with respect to $x$, I got: $$7 \int_{-1}^{1} [\frac{5}{2}y^6 + 5y^4 + \frac{5}{2}y^2 + y^3 (e^{(y^2+1)}) -10y^2 -e^2y^3] dy$$ I kept on trying several times but I've lost myself in computations, obtaining strange results. Probably, there are just few steps to the correct result, but I can't "view" the idea.

I also foolishly tried to compute the limits of the integral in a second (wrong) way, obtaining $0\leq x\leq 2$ and $ -\sqrt{x-1} \leq y \leq \sqrt{x-1}$, but I didn't go furtherly, because I noticed by myself $0\leq x< 1$ is out of domain with respect to $y$.

Can someone help me? Thanks in advance.

Answer 1

Hint. Note that the domain of integration is symmetric with respect to the line $y=0$, hence your integral is just $$\iint_{T} [35xy^2 + 7e^x y^3] dx dy=\iint_{T} [35xy^2] dx dy= 2\cdot 35\int_{y=0}^1 y^2\left(\int_{x=y^2+1}^{2}x\, dx\right) dy$$ because $7e^x y^3$ is an odd function in the variable $y$ and $35xy^2$ is an even one.

Answer 2

Note that your method is correct. The issue is with setting the limits for $x$, which must be $y^2+1\le x\le 2$. You calculated the inner definite integral correctly (just reverse the signs due to the reversed limits). And then, as Robert Z noted, note that $y^3e^{y^2+1}, e^2y^3$ are odd, hence their integrals at $[-1,1]$ are $0$. The rest are polynomial, easy to integrate and find the final answer $16$.