Find the volume enclosed between ﻿f left parenthesis x comma y right parenthesis equals space e to the power of x squared plus y squared end exponent﻿ and the region defined by a circle with a radius of 1.6 and centred at the origin. Round to 1 decimal place. 

To find the volume enclosed between the surface \( f(x, y) = e^{x^2 + y^2} \) and the region defined by a circle with a radius of 1.6 centered at the origin, we need to integrate the function \( f(x, y) \) over the given circular region.

### Step-by-Step Solution

1. **Convert to Polar Coordinates:**

   Since the region is a circle, polar coordinates will simplify the integration. Recall that in polar coordinates:
   \[
   x = r \cos(\theta), \quad y = r \sin(\theta), \quad \text{and} \quad dx \, dy = r \, dr \, d\theta.
   \]

   The function in polar coordinates becomes:
   \[
   f(x, y) = e^{x^2 + y^2} = e^{r^2}.
   \]

2. **Set Up the Integral:**

   The circular region has radius 1.6, so \( r \) ranges from 0 to 1.6, and \( \theta \) ranges from 0 to \( 2\pi \). The volume \( V \) is given by:
   \[
   V = \int_{0}^{2\pi} \int_{0}^{1.6} e^{r^2} \, r \, dr \, d\theta.
   \]

3. **Integrate with Respect to \( r \):**

   The integral with respect to \( r \) is:
   \[
   \int_{0}^{1.6} e^{r^2} \, r \, dr.
   \]
   Let \( u = r^2 \), hence \( du = 2r \, dr \) or \( \frac{du}{2} = r \, dr \). The limits for \( u \) will change from 0 to \( 1.6^2 = 2.56 \). The integral becomes:
   \[
   \int_{0}^{2.56} e^u \, \frac{du}{2} = \frac{1}{2} \int_{0}^{2.56} e^u \, du.
   \]
   Integrating with respect to \( u \):
   \[
   \frac{1}{2} \left[ e^u \right]_{0}^{2.56} = \frac{1}{2} \left( e^{2.56} - e^0 \right) = \frac{1}{2} \left( e^{2.56} - 1 \right).
   \]

4. **Integrate with Respect to \( \theta \):**

   Now integrate with respect to \( \theta \):
   \[
   V = \int_{0}^{2\pi} \frac{1}{2} \left( e^{2.56} - 1 \right) \, d\theta = \frac{1}{2} \left( e^{2.56} - 1 \right) \int_{0}^{2\pi} d\theta.
   \]
   The integral of \( d\theta \) from 0 to \( 2\pi \) is \( 2\pi \), so:
   \[
   V = \frac{1}{2} \left( e^{2.56} - 1 \right) \cdot 2\pi = \pi \left( e^{2.56} - 1 \right).
   \]

5. **Compute the Numerical Value:**

   Now, let's calculate the numerical value of \( V \), rounding to 1 decimal place.

\[
e^{2.56} \approx 12.930.
\]

Thus,

\[
V \approx \pi \left( 12.930 - 1 \right) = \pi \times 11.930.
\]

Calculating the final volume:

\[
V \approx 37.5.
\]

### Final Answer

The volume enclosed between the surface \( e^{x^2 + y^2} \) and the region defined by the circle with a radius of 1.6 is approximately **37.5 cubic units** (rounded to one decimal place).

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Would you like more details, or do you have any questions?

Here are some related questions to explore further:
1. How would the volume change if the function were \( e^{-(x^2 + y^2)} \)?
2. What if the region was an ellipse instead of a circle?
3. How do you convert between Cartesian and polar coordinates?
4. How does changing the radius of the region affect the volume?
5. What are other examples where polar coordinates simplify integration?

**Tip:** When dealing with circular regions, polar coordinates often simplify the integration significantly.

Learn how to calculate the volume enclosed between the surface e^(x^2 + y^2) and a circle with a radius of 1.6 centered at the origin. Detailed step-by-step solution using integration in polar coordinates. Suitable for advanced undergraduate and graduate-level mathematics.

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