To solve the problem using a triple integral, let’s break it down step by step.

Problem Breakdown:

• Cylinder equation: $x^2 + z^2 = 4$ (a cylinder symmetric about the $y$-axis with a radius of 2).
• Planes: $y = -1$ and $y + z = 4$, which can also be rewritten as $z = 4 - y$.

We need to compute the volume of the region bounded by these surfaces.

Step 1: Setting up the bounds

The bounds are determined as follows:

1. Cylinder bounds: The cylinder is defined by $x^2 + z^2 = 4$. This bounds $x$ and $z$ in the $xz$-plane.

   • For $x$, we have: $-2 \leq x \leq 2$.
   • For $z$, given $x$, $z$ satisfies $-\sqrt{4 - x^2} \leq z \leq \sqrt{4 - x^2}$.

2. Plane bounds: The planes $y = -1$ and $y + z = 4$ bound $y$. Substituting $z = 4 - y$ into the bounds of $z$:

   • Rearrange for $y$: $-1 \leq y \leq 4 - z$.

Step 2: The Triple Integral

The volume is given by:

$V = \int_{x=-2}^{2} \int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{y=-1}^{4-z} 1 \, dy \, dz \, dx$

Step 3: Evaluate the Integral

We compute the integral step by step:

1. Inner integral (with respect to $y$): $\int_{y=-1}^{4-z} 1 \, dy = \Big[y\Big]_{-1}^{4-z} = (4-z) - (-1) = 5 - z$

2. Middle integral (with respect to $z$): Substitute $5 - z$ into the $z$-bounds: $\int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5 - z) \, dz$ Split the integral into two parts: $\int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 5 \, dz - \int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} z \, dz$

   For the first term: $\int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 5 \, dz = 5 \int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 1 \, dz = 5 \cdot (2\sqrt{4-x^2}) = 10\sqrt{4-x^2}$

   For the second term: Since $z$ is an odd function integrated over symmetric limits: $\int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} z \, dz = 0$

   Thus: $\int_{z=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5 - z) \, dz = 10\sqrt{4-x^2}$

3. Outer integral (with respect to $x$): Substitute $10\sqrt{4-x^2}$ into the $x$-bounds: $\int_{x=-2}^{2} 10\sqrt{4-x^2} \, dx$

   Factor out 10: $10 \int_{x=-2}^{2} \sqrt{4-x^2} \, dx$

   This integral represents the area of a semicircle of radius 2: $\int_{x=-2}^{2} \sqrt{4-x^2} \, dx = \frac{\pi (2^2)}{2} = 2\pi$

   Thus: $10 \int_{x=-2}^{2} \sqrt{4-x^2} \, dx = 10 \cdot 2\pi = 20\pi$

Final Volume:

$V = 20\pi$

Let me know if you'd like further explanation or details! Here are related questions you might explore:

1. How is the symmetry of the cylinder exploited in setting the bounds?
2. What does the integral $\sqrt{4-x^2}$ geometrically represent?
3. How would the solution change if the cylinder radius were different?
4. What is the effect of changing the plane $y + z = 4$?
5. Can this problem be solved using cylindrical coordinates? How?

Tip: Always consider symmetry in geometry problems to simplify integration!