Find the area of the region described. The region bounded by \( y=4 x^{3} \) and \( y=36 x \) The area of the region is \( \square \). (Type an exact answer, using radicals as needed.)

To find the area of the region bounded by \( y = 4x^3 \) and \( y = 36x \), we first need to determine the points of intersection of the two curves. Setting them equal to each other, we have: \[ 4x^3 = 36x \] Rearranging, we get: \[ 4x^3 - 36x = 0 \] Factoring out \( 4x \): \[ 4x(x^2 - 9) = 0 \] This gives us the solutions: \[ 4x = 0 \quad \Rightarrow \quad x = 0 \] \[ x^2 - 9 = 0 \quad \Rightarrow \quad x = 3 \quad \text{or} \quad x = -3 \] So, the points of intersection are \( x = -3, 0, 3 \). Next, to find the area between the curves from \( x = -3 \) to \( x = 3 \), we calculate the integral of the upper function minus the lower function. The upper function in this interval is \( y = 36x \) and the lower function is \( y = 4x^3 \). Thus, we set up the integral: \[ \text{Area} = \int_{-3}^{3} (36x - 4x^3) \, dx \] Calculating the integral: \[ = \int_{-3}^{3} 36x \, dx - \int_{-3}^{3} 4x^3 \, dx \] Calculating the first integral: \[ \int 36x \, dx = 18x^2 \Big|_{-3}^{3} = 18(3^2 - (-3)^2) = 18(9 - 9) = 0 \] Calculating the second integral: \[ \int 4x^3 \, dx = x^4 \Big|_{-3}^{3} = (3^4 - (-3)^4) = (81 - 81) = 0 \] Since both integrals evaluate to zero, it suggests that we should break down the integral into parts to calculate the area properly because the functions intersect at points that divide the intervals. Realizing our mistake in finding the absolute area, we decide to find area in one quadrant, as both areas will be mirrored due to symmetry. Thus, we can compute the area from \( x = 0 \) to \( x = 3 \): \[ \text{Area}_{0 \text{ to } 3} = \int_{0}^{3} (36x - 4x^3) \, dx \] Calculating this integral: \[ = \left[ 18x^2 - x^4 \right]_{0}^{3} = \left[ 18(3^2) - (3^4) \right] - \left[ 18(0) - (0) \right] = 18(9) - (81) \] \[ = 162 - 81 = 81 \] Since there is symmetry, we multiply by 2 for the area between \( x = -3 \) and \( x = 3\): \[ \text{Total Area} = 2 \cdot 81 = 162 \] Thus, the area of the region is \( \boxed{162} \).