Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x^2 + y^2 = 4, and the plane z + y = 3.

Steps to Solve

1. Identify the boundaries of the volume

The solid is in the first octant, so the boundaries are:

• The coordinate planes: $x=0$, $y=0$, $z=0$.
• The cylinder described by $x^2 + y^2 = 4$.
• The plane described by the equation $z + y = 3$, which can be rewritten as $z = 3 - y$.

2. Set up the ranges for the triple integral

In the first octant:

• For the cylinder $x^2 + y^2 = 4$, $x$ and $y$ are both non-negative.
• The maximum for $y$ is determined by the intersection with the cylinder, so:

$$ y = 0 \text{ to } y = \sqrt{4 - x^2} $$

• For $z$, the upper limit is given by the plane, so:

$$ z = 0 \text{ to } z = 3 - y $$

The limits for $x$ will be $0$ to $2$ since $x^2 \leq 4$.

3. Set up the triple integral

The volume $V$ can be found using the triple integral:

$$ V = \int_0^2 \int_0^{\sqrt{4 - x^2}} \int_0^{3 - y} dz , dy , dx $$

4. Calculate the innermost integral with respect to $z$

The innermost integral is:

$$ \int_0^{3 - y} dz = (3 - y) $$

Thus, the volume integral becomes:

$$ V = \int_0^2 \int_0^{\sqrt{4 - x^2}} (3 - y) , dy , dx $$

5. Calculate the middle integral with respect to $y$

Next, evaluate the integral with respect to $y$:

$$ \int_0^{\sqrt{4 - x^2}} (3 - y) , dy $$

This can be computed as follows:

$$ = \left[ 3y - \frac{y^2}{2} \right]_0^{\sqrt{4 - x^2}} $$

Substituting the limits:

$$ = 3\sqrt{4 - x^2} - \frac{(\sqrt{4 - x^2})^2}{2} $$

$$ = 3\sqrt{4 - x^2} - \frac{4 - x^2}{2} $$

6. Combine and simplify to compute the outer integral

The outer integral now is:

$$ V = \int_0^2 \left( 3\sqrt{4 - x^2} - \frac{4 - x^2}{2} \right) dx $$

Simplify:

$$ = \int_0^2 3\sqrt{4 - x^2} , dx - \frac{1}{2} \int_0^2 (4 - x^2) , dx $$

The first part can be computed using a trigonometric substitution, while the second can be computed directly.