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To find the volume of the solid bounded above by the cone \( z = \sqrt{x^2 + y^2} \) and below by the disk \( x^2 + y^2 \leq 4 \) in the \( xy \)-plane, we use cylindrical coordinates for simplicity. Here’s the step-by-step solution:

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### Step 1: Representation in Cylindrical Coordinates
In cylindrical coordinates:
- \( x = r \cos\theta \)
- \( y = r \sin\theta \)
- \( z = z \)
- The volume element \( dV = r \, dr \, d\theta \, dz \).

The cone equation \( z = \sqrt{x^2 + y^2} \) becomes \( z = r \), and the disk \( x^2 + y^2 \leq 4 \) implies \( r \leq 2 \).

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### Step 2: Volume Integral
The volume of the solid is given by:
\[
V = \int \int \int r \, dz \, dr \, d\theta
\]
with:
- \( \theta \) ranging from \( 0 \) to \( 2\pi \),
- \( r \) ranging from \( 0 \) to \( 2 \),
- \( z \) ranging from \( 0 \) to \( r \) (the cone surface).

Thus, the integral becomes:
\[
V = \int_0^{2\pi} \int_0^2 \int_0^r r \, dz \, dr \, d\theta
\]

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### Step 3: Evaluate the Integral
1. **Integrate with respect to \( z \):**
\[
\int_0^r r \, dz = r \cdot \left[z\right]_0^r = r \cdot (r - 0) = r^2
\]

2. **Update the integral:**
\[
V = \int_0^{2\pi} \int_0^2 r^2 \, dr \, d\theta
\]

3. **Integrate with respect to \( r \):**
\[
\int_0^2 r^2 \, dr = \left[\frac{r^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}
\]

4. **Update the integral:**
\[
V = \int_0^{2\pi} \frac{8}{3} \, d\theta
\]

5. **Integrate with respect to \( \theta \):**
\[
\int_0^{2\pi} \frac{8}{3} \, d\theta = \frac{8}{3} \cdot \left[\theta\right]_0^{2\pi} = \frac{8}{3} \cdot (2\pi - 0) = \frac{16\pi}{3}
\]

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### Final Answer:
The volume of the solid is:
\[
V = \frac{16\pi}{3}
\]

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Do you have any questions, or would you like additional explanations? Here are 5 related questions to explore further:

1. How would the volume change if the cone had a different height?
2. What is the significance of using cylindrical coordinates for this problem?
3. Can this problem be solved using spherical coordinates? Why or why not?
4. How do we visualize the bounds of integration geometrically?
5. What happens to the volume if the cone is bounded by a smaller disk (e.g., \( x^2 + y^2 \leq 1 \))?

**Tip:** Always consider symmetry when setting up integrals in cylindrical or spherical coordinates—it simplifies computation significantly!

Find the volume of the solid bounded above by the cone z = sqrt(x^2 + y^2) and below by the disk x^2 + y^2 <= 4 in the xy-plane.

Learn how to calculate the volume of a solid bounded above by the cone z = sqrt(x^2 + y^2) and below by the disk x^2 + y^2 <= 4 in the xy-plane. This detailed step-by-step solution involves triple integration in cylindrical coordinates and is ideal for students studying advanced calculus.

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