To find the volume of the solid bounded by the surfaces 2z = x^2 + y^2 and z = x, we need to determine the limits of integration.

First, let's graph the surfaces to visualize the solid.

The surface 2z = x^2 + y^2 represents a paraboloid, while the surface z = x represents a plane.

By setting 2z = x^2 + y^2 equal to z = x, we can find the intersection points.

x^2 + y^2 = 2x
x^2 - 2x + y^2 = 0
(x - 1)^2 + y^2 = 1

This equation represents a circle centered at (1, 0) with a radius of 1.

To determine the limits of integration for x and y, we need to find the boundaries of the circle.

For x, the limits will be the x-values of the intersection points of the circle with the plane z = x.

(x - 1)^2 + y^2 = 1
(x - 1)^2 = 1 - y^2
x - 1 = ±√(1 - y^2)
x = 1 ± √(1 - y^2)

Since we want the volume of the solid bounded by the surfaces, we need to find the limits of integration for z as well.

From the equation z = x, we can see that z will vary from the minimum value of x to the maximum value of x.

The minimum value of x is 1 - √(1 - y^2) and the maximum value of x is 1 + √(1 - y^2).

Now, the volume V can be calculated using a triple integral:

V = ∫∫∫ dV
= ∫∫∫ dz dy dx

Since the region of integration is circular, we can use cylindrical coordinates.

V = ∫∫∫ r dz dy dx

The limits of integration for r, θ, and z will be as follows:

r: 0 to 1 (since we are integrating over a circle with radius 1)
θ: 0 to 2π (since we are integrating over the full circle)
z: x to x^2 + y^2

Now we can set up the triple integral:

V = ∫∫∫ r dz dy dx
= ∫[0 to 2π] ∫[0 to 1] ∫[x to x^2 + y^2] r dz dy dx

To evaluate this integral, we can integrate with respect to z first:

V = ∫[0 to 2π] ∫[0 to 1] [r(x^2 + y^2 - x)] dy dx

Next, we integrate with respect to y:

V = ∫[0 to 2π] [r(x^2y + (y^3/3) - xy)]|[0 to 1] dx
= ∫[0 to 2π] [r(x^2 + (1/3) - x)] dx

Finally, we integrate with respect to x:

V = [r(x^3/3 + (x/3) - (x^2/2))]|[0 to