Find the volume of the wedgelike solid that lies beneath the surface z = 16 - x^2 - y^2 and above the region R bounded by the curve y = 2√x, the line y = 4x - 2, and the x-axis.

Steps to Solve

1. Identify the region R

   The curves that bound the region R are (y = 2\sqrt{x}), (y = 4x - 2), and the x-axis ((y = 0)). We first need to find the points of intersection of these curves to determine the limits of integration.

2. Find intersections of curves

   Set (2\sqrt{x} = 4x - 2):

   $$ 2\sqrt{x} = 4x - 2 $$ Squaring both sides gives:

   $$ 4x = (4x - 2)^2 $$ Simplifying this leads to:

   $$ 4x = 16x^2 - 16x + 4 $$ Rearranging provides:

   $$ 16x^2 - 20x + 4 = 0 $$ Using the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}):

   $$ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 16 \cdot 4}}{2 \cdot 16} $$ This results in the roots, which define the limits of integration.

3. Determine the volume using double integrals

   The volume (V) can be calculated as:

   $$ V = \int_{x_1}^{x_2} \int_0^{y(x)} (16 - x^2 - y^2) , dy , dx $$ where (y(x)) is the upper boundary of the region defined by the curves.

4. Calculate the integral

   First, integrate with respect to (y):

   $$ \int_0^{y(x)} (16 - x^2 - y^2) , dy = [16y - x^2y - \frac{y^3}{3}]_0^{y(x)} $$ Substitute (y(x)) to evaluate it.

5. Perform the outer integral

   Substitute the results from the inner integral into the outer integral:

   $$ V = \int_{x_1}^{x_2} V_y , dx $$