The question doesn't state whether it's x or y that goes from -4 to 4, so we'll assume it's y because these are the y values for x=0. We'll also assume initially that the curve is rotated around the y-axis. The axis of rotation determines that x is going to be the radius variable. When y=0 (the x-axis), x=16 (x-intercept) so we know that x goes from 0 to 16. If we consider the volume of a cylinder radius x and height h we get πx2h. If the radius is x+dx, the volume of the cylinder is π(x+dx)2h, so the volume of the cylindrical shell between these two cylinders is the difference in the volumes=π(x+dx)2h-πx2h=π(x2+2xdx+dx2)h-πx2h=2πxhdx if dx is very small so that dx2 can be ignored, especially as dx→0 (becomes infinitesimal). The height h is given by y, where y=√(16-x) because x+y2=16.
So we have the volume of the cylindrical shell=2πx√(16-x)dx, and the volume of the whole shape when rotated is the integral of this between x=0 and x=16.
x+y2=16 so 1+2ydy/dx=0, or dx=-2ydy and the integral becomes:
∫2π(16-y2)y(-2ydy)=-4π∫y2(16-y2)dy which is an easier integral. The parabola is symmetrical, with half the shape above the x-axis and half below, so the y limits can be y=4 to y=0. Note that y is decreasing. We can double this integral to get the volume of the whole shape=-8π4∫0(16y2-y4)dy=8π[16y3/3-y5/5]40=8π(1024/3-1024/5)=8192π(2/15)=16384π/15=3431.457 cubic units approx.
Now let's assume that the curve is rotated around the x-axis. In this case y is the radius and x is the height. The integral is 2π∫xydy, and, again, we can work with limits y=0 and y=4, but increasing from 0 to 4, and we don't have to double this time.
2π∫(16-y2)ydy=2π0∫4(16y-y3)dy=2π[8y2-y3/3]04=2π(128-64/3)=640π/3=670.206 cubic units approx.