**Instruction:** Set up a double integral to find the volume of the solid region bounded by the graphs of the equations. Do not evaluate the integral.**Solution:**The volume \( V \) of the solid region is given by the double integral:\[ V = 2 \int_{0}^{4} \int_{0}^{\sqrt{4-(x-1)^2}} (4x-x^2-y^2) \, dy \, dx \]**Graphical Representation:**Below the integral setup, the image includes a 3D graph illustrating the solid region. The graph features bounded surfaces depicted in color, with axes labeled \(x\), \(y\), and \(z\).- The equations \( z = 16 - 4x \) and \( z = 16 - x^2 - y^2 \) represent surfaces that intersect to form the upper boundary of the solid.- The red surface represents the plane \(z = 16 - 4x\).- The blue and purple regions indicate the intersection of the surfaces, showing where the solid is cut off by the paraboloid \(z = 16 - x^2 - y^2\).- The base of the solid lies in the \(xy\)-plane, with a circular boundary.

The two provided equations are z=16-x2-y2 and z=16-4x. Find the intersection of these two surfaces by equating them as follows. 16-x2-y2 =16-4xx2+y2-4x=0x2+y2-4x+4=4x-22+y2=4⇒y=±4-x-22 This is an equation of a circle having center at 2,0. The volume of a solid region is provided by, Volume=Volume under paraboloid-Volume under plane=∫04∫-4-x-224-x-2216-x2-y2dydx-∫04∫-4-x-224-x-2216-4xdydx=∫04∫-4-x-224-x-2216-x2-y2-16+4x=2∫04∫04-x-224x-x2-y2dydx Thus, the required volume is 2∫04∫04-x-224x-x2-y2dydx.