Problem:

Find the volume of the region in the first octant (x>=0, y>=0, z>=0) bounded by the cylinder x^2 y^2=4, the plane z=2, and the plane y=4.

Solution:

To find the volume of the given region, we need to determine the limits of integration for each variable and set up the triple integral.

Limit of Integration:

To determine the limits of integration, we need to visualize the region in the first octant.

1. xy-plane (z=0):
- We are given that x>=0 and y>=0, so the region lies in the first quadrant of the xy-plane.
- The equation of the cylinder x^2 y^2=4 represents a hyperbola in the xy-plane.
- The hyperbola opens along the x-axis and y-axis and passes through the points (2, 0) and (-2, 0).
- The region bounded by the cylinder x^2 y^2=4 in the xy-plane is shown below:


2. xz-plane (y=0):
- Since y>=0, the plane y=0 is the xz-plane.
- The given plane z=2 intersects the xz-plane at z=2.
- The region bounded by the plane z=2 in the xz-plane is shown below:


3. yz-plane (x=0):
- Since x>=0, the plane x=0 is the yz-plane.
- The plane y=4 intersects the yz-plane at y=4.
- The region bounded by the plane y=4 in the yz-plane is shown below:


Triple Integral Setup:

We can set up the triple integral by integrating with respect to x, y, and z in the following order:

∫∫∫dV

1. The limits of integration for x are from 0 to x_max.
2. The limits of integration for y are from 0 to y_max(x).
3. The limits of integration for z are from 0 to 2.

Calculating x_max and y_max(x):

To calculate x_max, we need to find the x-coordinate where the hyperbola x^2 y^2=4 intersects the yz-plane (x=0).

Substituting x=0 into the equation, we get y^2=4, which gives y=±2.

Since we are in the first octant (y>=0), x_max=0.

To calculate y_max(x), we need to find the y-coordinate where the hyperbola x^2 y^2=4 intersects the xy-plane (z=0).

Substituting z=0 into the equation, we get x^2 y^2=4, which simplifies to y^2=4/x^2.