Use the general slicing method to find the volume of the following solid. The solid whose base is the region bounded by the curve \( y=8 \sqrt{\cos x} \) and the \( x \)-axis on \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), and whose cross sections through the solid perpendicular to the \( x \)-axis are isosceles riaht trianales with a horizontal lea in the \( x v \)-dlane and a vertical lea above the \( x \)-axis. Set up the integral that gives the volume of the solid. Use increasing limits of integration. Select the correct choice below and fill in the answer boxes to complete your choice.

To find the volume of the solid, we will use the general slicing method and set up the integral based on the characteristics of the isosceles right triangles defined by the function \( y = 8 \sqrt{\cos x} \). The base of each isosceles right triangle is represented by the function \( y = 8 \sqrt{\cos x} \). Since the triangles are isosceles right triangles, the height will also be equal to the length of a leg of the triangle. For a triangle with base \( b \) and height \( h \), the area \( A \) is given by: \[ A = \frac{1}{2} \cdot b \cdot h = \frac{1}{2} \cdot b \cdot b = \frac{1}{2} b^2 \] Here, the base \( b \) of our triangles is \( y = 8 \sqrt{\cos x} \), so we substitute that into the area formula: \[ A(x) = \frac{1}{2} (8 \sqrt{\cos x})^2 = \frac{1}{2} \cdot 64 \cos x = 32 \cos x \] To find the volume \( V \) of the solid, we integrate the area of the cross-sections from \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \): \[ V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} A(x) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 32 \cos x \, dx \] Thus, the final integral that gives the volume of the solid is: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 32 \cos x \, dx \]