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To evaluate the given integral, we need to approach the problem using spherical coordinates due to the symmetry and the nature of the region \( D \).

### Problem Setup
The integral is:
\[
\iiint_D e^{(x^2 + y^2 + z^2)^{3/2}} \, dV
\]
where \( D \) is the region:
1. Above the cone \( z = \sqrt{x^2 + y^2} \),
2. Below the hemisphere \( z = \sqrt{1 - x^2 - y^2} \).

### Step 1: Transform to Spherical Coordinates
In spherical coordinates:
- \( x = \rho \sin\theta \cos\phi \)
- \( y = \rho \sin\theta \sin\phi \)
- \( z = \rho \cos\theta \)
- The volume element \( dV = \rho^2 \sin\theta \, d\rho \, d\theta \, d\phi \).

Additionally, we have \( r^2 = x^2 + y^2 + z^2 = \rho^2 \), so \( (x^2 + y^2 + z^2)^{3/2} = \rho^3 \).

### Step 2: Define the Limits
The region \( D \) is bounded by:
- The cone \( z = \sqrt{x^2 + y^2} \), which in spherical coordinates corresponds to \( \theta = \frac{\pi}{4} \).
- The hemisphere \( z = \sqrt{1 - x^2 - y^2} \), which means \( \rho \leq 1 \).

Thus, the limits for integration are:
- \( \rho \): from 0 to 1 (bounded by the hemisphere),
- \( \theta \): from 0 to \( \frac{\pi}{4} \) (bounded by the cone),
- \( \phi \): from 0 to \( 2\pi \) (full rotation around the z-axis).

### Step 3: Set Up the Integral
Substituting these bounds and the expression for \( (x^2 + y^2 + z^2)^{3/2} = \rho^3 \), the integral becomes:
\[
\int_0^{2\pi} \int_0^{\pi/4} \int_0^1 e^{\rho^3} \cdot \rho^2 \sin\theta \, d\rho \, d\theta \, d\phi
\]

### Step 4: Solve the Integral
Separate the variables:
\[
\int_0^{2\pi} d\phi \int_0^{\pi/4} \sin\theta \, d\theta \int_0^1 \rho^2 e^{\rho^3} \, d\rho
\]

#### 1. Integral over \( \phi \):
\[
\int_0^{2\pi} d\phi = 2\pi
\]

#### 2. Integral over \( \theta \):
\[
\int_0^{\pi/4} \sin\theta \, d\theta = \left[-\cos\theta\right]_0^{\pi/4} = 1 - \frac{\sqrt{2}}{2}
\]

#### 3. Integral over \( \rho \):
Let \( u = \rho^3 \Rightarrow du = 3\rho^2 d\rho \Rightarrow d\rho = \frac{du}{3\rho^2} \), giving:
\[
\int_0^1 \rho^2 e^{\rho^3} \, d\rho = \frac{1}{3} \int_0^1 e^u \, du = \frac{1}{3} \left( e^1 - e^0 \right) = \frac{1}{3}(e - 1)
\]

### Step 5: Combine Results
Now, putting it all together:
\[
2\pi \cdot \left(1 - \frac{\sqrt{2}}{2}\right) \cdot \frac{e - 1}{3} = \frac{2\pi}{3} \left(1 - \frac{\sqrt{2}}{2}\right)(e - 1)
\]

This matches option B:
\[
\boxed{\frac{2\pi}{3} (2 - \sqrt{2})(e - 1)}
\]

Would you like further details on any of these steps or have any other questions?

### Additional Questions
1. How does the cone's equation affect the integration limits in spherical coordinates?
2. Why is spherical coordinates preferred over Cartesian for this integral?
3. What is the significance of \( e^{\rho^3} \) in the integrand?
4. Can we generalize this approach for other cone-hemisphere intersections?
5. What is the physical interpretation of this integral in a real-world scenario?

### Tip
In spherical coordinates, always verify the bounds by examining each surface’s shape relative to spherical angles to avoid incorrect integration limits.

Evaluate the integral \( \iiint_D e^{(x^2 + y^2 + z^2)^{3/2}} \, dV \) where \( D \) is the region above the cone \( z = \sqrt{x^2 + y^2} \) and below the hemisphere \( z = \sqrt{1 - x^2 - y^2} \) using spherical coordinates.

Learn how to evaluate a triple integral of \( e^{(x^2 + y^2 + z^2)^{3/2}} \) over a region bounded above by a hemisphere and below by a cone using spherical coordinates. This solution includes setting up spherical coordinates, defining limits, and solving the integral step-by-step. Ideal for university-level calculus students.

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