# Calculate the integral $\iint_D e^{\frac{y}{x}}\, dx \,dy$

Question: Calculate the integral

$$\iint_D e^{\frac{y}{x}} \,dx \,dy$$

where $D$ is the area defined by the inequality $x^2 \leq y \leq x^3 \leq 8$.

My problem is that I have a hard time understanding how to write the inequality in terms of $x$ and $y$ (and what limits to use when calculating the integral).

Looking at the graphs of $x^2$ and $x^3$ you could make the case that

$$D = \left\{ (x,y) : 1 \leq x \leq 2,\ x^2 \leq y \leq x^3 \right\},$$

but then we are missing the area from $0 \leq x \leq 1$.

• $y \geq x^2$ implies that $y\geq 0$; since $y\leq x^3$, this also implies that $x\geq 0$. $x^2\leq x^3$ then implies that $x\geq 1$. There is no area from $0\leq x\leq 1$ in the actual domain. Apr 21 '17 at 4:35
• Note that you'll need to use Fubini's theorem to get an explicit answer, because as it's currently written the inner integral has a non-elementary result. Apr 21 '17 at 4:39
• Couldn't you just integrate the dy and use $x^2,x^3$ as limits of integration, then do the dx part using $x\le 8$. Apr 21 '17 at 4:42

$x^2 \leq 8$ gives $-2\sqrt{2} \leq x \leq 2\sqrt{2}$ and $x^3 \leq 8$ gives $x \leq 2$. There's another inequality which is $x^2 \leq x^3$. This means $x \geq 1$. Combining, $1 \leq x \leq 2$. So your $D$ is correct.