Problem:
Find the volume of the solid which is in the first octant bounded by the cylinder x^2 y^2=a^2 and y^2 z^2=a^2

Solution:
To find the volume of the solid bounded by the given surfaces, we need to set up a triple integral in Cartesian coordinates.

Step 1: Identifying the Boundaries
In the first octant, the boundaries for x, y, and z are all positive. We can determine these boundaries by analyzing the given surfaces:
- The cylinder x^2 y^2 = a^2 intersects the x-y plane at y = 0 and x = a. Thus, the boundaries for x and y are 0 ≤ x ≤ a and 0 ≤ y ≤ a.
- The surface y^2 z^2 = a^2 intersects the y-z plane at y = 0 and z = a. Therefore, the boundaries for y and z are 0 ≤ y ≤ a and 0 ≤ z ≤ a.

Step 2: Setting Up the Triple Integral
Since we are working in Cartesian coordinates, the volume element is dx dy dz. Therefore, the volume of the solid can be expressed as the triple integral:

V = ∫∫∫ dx dy dz

We need to determine the limits of integration for each variable based on the identified boundaries in Step 1.

Step 3: Evaluating the Integral
Now, we can evaluate the triple integral:

V = ∫[0,a] ∫[0,a] ∫[0,a] dx dy dz

The innermost integral with respect to x is simply the variable x evaluated between 0 and a:

V = ∫[0,a] ∫[0,a] (x=a - x=0) dy dz

Simplifying further, we get:

V = ∫[0,a] ∫[0,a] a dy dz

The next integral is with respect to y, and the limits of integration are from 0 to a:

V = ∫[0,a] (a - 0) dz

Finally, we integrate with respect to z, from 0 to a:

V = (a - 0)(z=a - z=0)

V = a^2

Therefore, the volume of the solid bounded by the given surfaces in the first octant is a^2.