$$ \int_0^2\!\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\!\int_0^{2x+y}\mathrm{dz}\,\mathrm{d}x\,\mathrm{dy} $$

To evaluate the triple integral 

\[
\int_0^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_0^{2x+y} \, dz \, dy \, dx,
\]

let's proceed step by step.

### Step 1: Evaluate the innermost integral
The innermost integral is with respect to $ z $:

\[
\int_0^{2x+y} dz.
\]

The integral of $ 1 $ with respect to $ z $ is simply $ z $, so:

\[
\int_0^{2x+y} dz = \left[ z \right]_0^{2x+y} = (2x + y) - 0 = 2x + y.
\]

### Step 2: Update the integral
Substituting this result, the integral becomes:

\[
\int_0^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (2x + y) \, dy \, dx.
\]

### Step 3: Evaluate the second integral
The second integral is with respect to $ y $:

\[
\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (2x + y) \, dy.
\]

Split this integral into two parts:

\[
\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 2x \, dy + \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} y \, dy.
\]

#### First term:
The integral of $ 2x $ with respect to $ y $ is:

\[
\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 2x \, dy = 2x \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 1 \, dy = 2x \left[ y \right]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}.
\]

Substitute the limits:

\[
2x \left( \sqrt{4-x^2} - (-\sqrt{4-x^2}) \right) = 2x \cdot 2\sqrt{4-x^2} = 4x\sqrt{4-x^2}.
\]

#### Second term:
The integral of $ y $ with respect to $ y $ is:

\[
\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} y \, dy = \left[ \frac{y^2}{2} \right]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}.
\]

Evaluate this:

\[
\frac{\left(\sqrt{4-x^2}\right)^2}{2} - \frac{\left(-\sqrt{4-x^2}\right)^2}{2}.
\]

Since $ \left(\sqrt{4-x^2}\right)^2 = \left(-\sqrt{4-x^2}\right)^2 = 4-x^2 $, the result is:

\[
\frac{4-x^2}{2} - \frac{4-x^2}{2} = 0.
\]

Thus, the second term vanishes.

### Combine the results
The second integral simplifies to:

\[
\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (2x + y) \, dy = 4x\sqrt{4-x^2}.
\]

### Step 4: Evaluate the outermost integral
The outermost integral is:

\[
\int_0^2 4x\sqrt{4-x^2} \, dx.
\]

Let $ u = 4 - x^2 $, so $ du = -2x \, dx $ or $ x \, dx = -\frac{1}{2} \, du $. Change the limits:
- When $ x = 0 $, $ u = 4 $.
- When $ x = 2 $, $ u = 0 $.

Rewriting the integral:

\[
\int_0^2 4x\sqrt{4-x^2} \, dx = \int_4^0 4\sqrt{u} \left(-\frac{1}{2}\right) \, du = \int_0^4 2\sqrt{u} \, du.
\]

The integral of $ 2\sqrt{u} $ is:

\[
\int_0^4 2u^{1/2} \, du = 2 \cdot \frac{u^{3/2}}{3/2} \Big|_0^4 = \frac{4}{3} \cdot \left[ u^{3/2} \right]_0^4.
\]

Evaluate this:

\[
\frac{4}{3} \cdot \left[ (4)^{3/2} - (0)^{3/2} \right] = \frac{4}{3} \cdot \left( 8 - 0 \right) = \frac{32}{3}.
\]

### Final Answer:
The value of the integral is:

\[
\boxed{\frac{32}{3}}
\]

Would you like more details or clarifications? 

### Follow-up Questions:
1. What are some examples of problems where polar coordinates might simplify integration?
2. How does the choice of variable substitution affect integral evaluation?
3. What is the geometric interpretation of this triple integral?
4. Could this problem be approached using symmetry? Why or why not?
5. How does the Jacobian factor into changing coordinate systems in triple integrals?

### Tip:
Always visualize the region of integration to determine if switching coordinate systems (e.g., polar or spherical) could simplify the computation.

The query submitted by the user: $$ \int_0^2\!\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\!\int_0^{2x+y}\mathrm{dz}\,\mathrm{d}x\,\mathrm{dy} $$

A detailed solution to the triple integral $$ \int_0^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_0^{2x+y} dz\,dy\,dx $$ using step-by-step integration techniques, including substitution and solving the geometry of circular regions. Suitable for advanced students in calculus.

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