The question asks for the transformation of the double integral $\int_D e^{x^2 + y^2} \, dx\,dy$ into polar coordinates, where the domain $D$ is defined as $D = \{(x, y) \in \mathbb{R}^2 \mid 0 \leq x \leq y \text{ and } 0 \leq y \leq 1\}$.

Key steps to solve:

1. Interpret the region $D$:

   • $0 \leq x \leq y$: In Cartesian coordinates, this describes the region below the line $x = y$.
   • $0 \leq y \leq 1$: This bounds $y$ between 0 and 1.

   Together, $D$ is the triangular region in the first quadrant bounded by $x = 0$, $y = 1$, and $x = y$.

2. Convert the integral to polar coordinates:

   • In polar coordinates: $x = r\cos\theta, \quad y = r\sin\theta, \quad dx\,dy = r\,dr\,d\theta.$ Also, $x^2 + y^2 = r^2$, so $e^{x^2 + y^2} = e^{r^2}$.

   • The line $x = y$ translates to $\theta = \frac{\pi}{4}$ in polar coordinates.

   • The line $y = 1$ translates to $r\sin\theta = 1 \implies r = \frac{1}{\sin\theta}$.

   • The line $x = 0$ translates to $\theta = 0$, and the lower bound $y = 0$ corresponds to $r = 0$.

   Therefore, the bounds in polar coordinates are: $0 \leq r \leq \frac{1}{\sin\theta}, \quad 0 \leq \theta \leq \frac{\pi}{4}.$

3. Set up the transformed integral: The integral in polar coordinates becomes: $\int_0^{\pi/4} \int_0^{1/\sin\theta} e^{r^2} r \, dr\,d\theta.$

4. Match the correct option:

   • Option 1 matches exactly: $\int_0^{\pi/4} \int_0^{1/\sin\theta} r e^{r^2} \, dr\,d\theta$.

Final Answer:

Option 1 is correct.

Do you want further explanation or detailed steps?

Here are five related questions to deepen understanding:

1. How do polar coordinates relate to Cartesian coordinates?
2. Why does $x = y$ translate to $\theta = \pi/4$?
3. What is the role of the Jacobian ($r$) in changing to polar coordinates?
4. How do you determine the upper bound of $r$ for polar transformations?
5. Could this integral be solved directly in Cartesian coordinates?

Tip: Always sketch the region $D$ to verify bounds before transforming integrals!