To evaluate the given triple integral using spherical coordinates, let’s transform the integral step by step:

The integral is:

$\iiint_E \frac{e^{-(x^2 + y^2 + z^2)}}{x^2 + y^2 + z^2} \, dV$

1. Transformation to Spherical Coordinates

In spherical coordinates:

• $x = \rho \sin\phi \cos\theta$,
• $y = \rho \sin\phi \sin\theta$,
• $z = \rho \cos\phi$,
• $x^2 + y^2 + z^2 = \rho^2$,
• Volume element $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$.

Thus, the integral becomes: $\iiint_E \frac{e^{-\rho^2}}{\rho^2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$

Simplify: $\iiint_E e^{-\rho^2} \sin\phi \, d\rho \, d\phi \, d\theta$

2. Define the Bounds of Integration

The region $E$ is bounded by the spheres $\rho = 2$ and $\rho = 5$:

• $\rho \in [2, 5]$,
• $\phi \in [0, \pi]$ (entire range of polar angle),
• $\theta \in [0, 2\pi]$ (full revolution in the azimuthal angle).

3. Separate the Variables

The integral can now be written as: $\int_{0}^{2\pi} \int_{0}^\pi \int_{2}^5 e^{-\rho^2} \sin\phi \, d\rho \, d\phi \, d\theta$

Step 1: Integrate over $\theta$

The $\theta$-integral is straightforward: $\int_{0}^{2\pi} 1 \, d\theta = 2\pi$

Step 2: Integrate over $\phi$

The $\phi$-integral involves only $\sin\phi$: $\int_{0}^\pi \sin\phi \, d\phi = \left[-\cos\phi\right]_0^\pi = -\cos(\pi) + \cos(0) = 2$

Step 3: Integrate over $\rho$

The $\rho$-integral is: $\int_{2}^{5} e^{-\rho^2} \, d\rho$

Let $u = \rho^2$, so $du = 2\rho \, d\rho$. When $\rho = 2$, $u = 4$, and when $\rho = 5$, $u = 25$. Thus: $\int_{2}^{5} e^{-\rho^2} \, d\rho = \frac{1}{2} \int_{4}^{25} e^{-u} \, du$

The integral of $e^{-u}$ is: $\int e^{-u} \, du = -e^{-u}$

Apply the limits: $\frac{1}{2} \int_{4}^{25} e^{-u} \, du = \frac{1}{2} \left[-e^{-u}\right]_4^{25} = \frac{1}{2} \left(-e^{-25} + e^{-4}\right)$

4. Combine the Results

The total integral is: $2\pi \cdot 2 \cdot \frac{1}{2} \left(e^{-4} - e^{-25}\right)$

Simplify: $\text{Result} = 2\pi \left(e^{-4} - e^{-25}\right)$

Final Answer:

$\boxed{2\pi \left(e^{-4} - e^{-25}\right)}$

This is the value of the given triple integral.

---

Let me know if you'd like further details or assistance!

Expanding Knowledge

1. How do spherical coordinates simplify triple integrals for spherical regions?
2. What are other examples where spherical coordinates are useful?
3. How does one handle regions bounded by cones in spherical coordinates?
4. Why is the Jacobian ($\rho^2 \sin\phi$) critical in transformations?
5. What is the significance of exponential decay functions in integrals?

Tip

When dealing with spherical coordinates, always ensure the bounds for $\rho$, $\phi$, and $\theta$ are correctly aligned to the geometry of the region!