Find the volume enclosed by the cone (x^{2} + y^{2} = z^{2}) and the plane 2z-y-3=0

We are tasked with finding the volume enclosed by the cone \( x^2 + y^2 = z^2 \) and the plane \( 2z - y - 3 = 0 \).

### Step 1: Express the plane equation in a simpler form.
We begin by solving for \( y \) in terms of \( z \) from the plane equation:
\[
2z - y - 3 = 0 \quad \Rightarrow \quad y = 2z - 3
\]
Thus, the equation of the plane is \( y = 2z - 3 \).

### Step 2: Find the limits of \( z \).
To determine the limits of \( z \), we find the intersection of the cone \( x^2 + y^2 = z^2 \) with the plane \( y = 2z - 3 \). Substituting \( y = 2z - 3 \) into the cone equation, we have:
\[
x^2 + (2z - 3)^2 = z^2
\]
Expanding the square:
\[
x^2 + (4z^2 - 12z + 9) = z^2 \quad \Rightarrow \quad x^2 + 4z^2 - 12z + 9 = z^2
\]
Simplifying:
\[
x^2 + 3z^2 - 12z + 9 = 0
\]
Now, since \( x^2 \geq 0 \), the quadratic equation \( 3z^2 - 12z + 9 \) must be non-negative for real values of \( z \). To solve for \( z \), we factor the quadratic:
\[
3(z^2 - 4z + 3) = 0 \quad \Rightarrow \quad 3(z - 3)(z - 1) = 0
\]
Thus, \( z = 1 \) and \( z = 3 \).

### Step 3: Set up the volume integral.
We use cylindrical coordinates to set up the volume integral. In cylindrical coordinates, the cone equation becomes \( r = z \), where \( r = \sqrt{x^2 + y^2} \), and the limits for \( z \) are from 1 to 3. The plane equation in cylindrical coordinates is \( y = 2z - 3 \), which translates to the corresponding limits on \( r \) and \( \theta \).

The volume element in cylindrical coordinates is given by \( dV = r \, dr \, d\theta \, dz \).

Thus, the volume integral becomes:
\[
V = \int_{\theta=0}^{2\pi} \int_{z=1}^{3} \int_{r=0}^{z} r \, dr \, dz \, d\theta
\]

### Step 4: Solve the integral.
First, integrate with respect to \( r \):
\[
\int_{0}^{z} r \, dr = \left[\frac{r^2}{2}\right]_{0}^{z} = \frac{z^2}{2}
\]
Now, the volume integral becomes:
\[
V = \int_{0}^{2\pi} \int_{1}^{3} \frac{z^2}{2} \, dz \, d\theta
\]
Next, integrate with respect to \( z \):
\[
\int_{1}^{3} \frac{z^2}{2} \, dz = \frac{1}{2} \left[\frac{z^3}{3}\right]_{1}^{3} = \frac{1}{2} \left(\frac{27}{3} - \frac{1}{3}\right) = \frac{1}{2} \times \frac{26}{3} = \frac{13}{3}
\]
Finally, integrate with respect to \( \theta \):
\[
V = \frac{13}{3} \int_{0}^{2\pi} d\theta = \frac{13}{3} \times 2\pi = \frac{26\pi}{3}
\]

### Final Answer:
The volume enclosed by the cone and the plane is:
\[
V = \frac{26\pi}{3}
\]

Would you like further explanation on any part of the solution?

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Here are 5 related questions you might want to explore:
1. How do you compute the volume of a sphere using triple integrals?
2. What is the volume of the region between two paraboloids?
3. How do you find the volume of a cone using cylindrical coordinates?
4. What are the differences between using cylindrical and spherical coordinates for volume integrals?
5. How can you find the volume of a solid using divergence theorem?

**Tip:** Always carefully choose the coordinate system that best simplifies the problem when calculating volumes using integrals!

Find the volume enclosed by the cone (x^{2} + y^{2} = z^{2}) and the plane 2z - y - 3 = 0.

Learn how to compute the volume enclosed by the cone x^2 + y^2 = z^2 and the plane 2z - y - 3 = 0 using cylindrical coordinates and volume integrals. A detailed step-by-step solution is provided, ideal for college-level multivariable calculus students.

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