Use the general slicing method to find the volume of the following solid. The solid whose base is the region bounded by the curves \( y=4 x^{2} \) and \( y=5-x^{2} \), and whose cross sections through the solid perpendicular to the \( x \)-axis are squares The volume of the solid is \( \square \) cubic units. (Type an exact answer.)

To find the volume of the solid with a square cross-section perpendicular to the \( x \)-axis, we first need to determine the area of the base region bounded by the curves \( y = 4x^2 \) and \( y = 5 - x^2 \). First, we find the points of intersection by setting the equations equal to each other: \( 4x^2 = 5 - x^2 \) Rearranging gives us: \( 5x^2 + 5 = 0 \) This can be solved as: \( 5x^2 = 5 \) \( x^2 = 1 \) \( x = \pm 1 \) The region of interest is bounded by \( x = -1 \) and \( x = 1 \). Next, the height of the square cross-section at a given \( x \) can be found by calculating the difference of the upper and lower curves: Height \( h(x) = (5 - x^2) - (4x^2) = 5 - 5x^2 \) Since the cross-section is a square, the area \( A \) of each square is: \( A(x) = (h(x))^2 = (5 - 5x^2)^2 \) Now, we can integrate this area from \( x = -1 \) to \( x = 1 \) to get the volume \( V \): \[ V = \int_{-1}^{1} (5 - 5x^2)^2 \, dx \] Expanding \( (5 - 5x^2)^2 \): \[ (5 - 5x^2)^2 = 25 - 50x^2 + 25x^4 \] Now, we integrate each term separately: \[ V = \int_{-1}^{1} (25 - 50x^2 + 25x^4) \, dx = \left[ 25x - \frac{50}{3}x^3 + \frac{25}{5}x^5 \right]_{-1}^{1} \] Calculating at the bounds: At \( x = 1 \): \[ 25(1) - \frac{50}{3}(1) + 5(1) = 25 - \frac{50}{3} + 5 = 30 - \frac{50}{3} = \frac{90 - 50}{3} = \frac{40}{3} \] At \( x = -1 \): \[ 25(-1) - \frac{50}{3}(-1) + 5(-1) = -25 + \frac{50}{3} - 5 = -30 + \frac{50}{3} = \frac{-90 + 50}{3} = \frac{-40}{3} \] Now, subtract the two results: \[ V = \frac{40}{3} - \left(\frac{-40}{3}\right) = \frac{40 + 40}{3} = \frac{80}{3} \] Thus, the volume of the solid is: \[ \boxed{\frac{80}{3}} \] cubic units.