To simplify the expression down to a single integral in terms of $x$, let's go through the given integral step by step.

The original expression for the volume is:

$\int_{-4}^{4} \int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} \int_{2x^2 + 2z^2}^{80 - 3x^2 - 3z^2} \, dy \, dz \, dx$

Here, the integrand is 1, so we’re essentially integrating over the volume bounded by the given limits.

Step 1: Integrate with respect to $y$

The innermost integral is: $\int_{2x^2 + 2z^2}^{80 - 3x^2 - 3z^2} \, dy$

Since this is an integral with respect to $y$ with integrand 1, the result is simply the length of the interval in $y$: $80 - 3x^2 - 3z^2 - (2x^2 + 2z^2) = 80 - 5x^2 - 5z^2$

Thus, we have:

$\int_{-4}^{4} \int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} (80 - 5x^2 - 5z^2) \, dz \, dx$

Step 2: Integrate with respect to $z$

Now, we integrate with respect to $z$. This integral becomes: $\int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} (80 - 5x^2 - 5z^2) \, dz$

This can be split as:

$\int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} 80 \, dz - \int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} 5x^2 \, dz - \int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} 5z^2 \, dz$

Let's compute each part individually:

1. First part: $\int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} 80 \, dz = 80 \cdot 2\sqrt{16 - x^2} = 160\sqrt{16 - x^2}$

2. Second part: $\int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} 5x^2 \, dz = 5x^2 \cdot 2\sqrt{16 - x^2} = 10x^2\sqrt{16 - x^2}$

3. Third part: $\int_{-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} 5z^2 \, dz$

   Since $z^2$ is an even function, we can compute this as: $2 \int_{0}^{\sqrt{16 - x^2}} 5z^2 \, dz = 10 \int_{0}^{\sqrt{16 - x^2}} z^2 \, dz$

   Substituting $z^3 / 3$ for $z^2$, we get: $10 \left[ \frac{z^3}{3} \right]_{0}^{\sqrt{16 - x^2}} = \frac{10}{3} (16 - x^2)^{3/2}$

Step 3: Combine the Results

Substituting back, we have:

$\int_{-4}^{4} \left( 160\sqrt{16 - x^2} - 10x^2\sqrt{16 - x^2} - \frac{10}{3}(16 - x^2)^{3/2} \right) \, dx$

This is now a single integral in terms of $x$:

$\int_{-4}^{4} \left( 160\sqrt{16 - x^2} - 10x^2\sqrt{16 - x^2} - \frac{10}{3}(16 - x^2)^{3/2} \right) \, dx$

Would you like further details on evaluating this integral, or any particular part of the simplification? Here are some related questions:

1. How do we evaluate integrals involving square roots?
2. Can we simplify expressions further by using trigonometric substitutions?
3. What techniques are best for evaluating definite integrals like this?
4. How does symmetry affect the integration limits here?
5. How do we interpret the physical meaning of each part of this integral?

Tip: Symmetry can often simplify integrals, especially when integrating over symmetric bounds.