Calculus of a Single Variable, Chapter 7, 7.3, Section 7.3, Problem 20

We can use a shell method when a bounded region represented by a rectangular strip is parallel to the axis of revolution. It forms an infinite number of thin hollow pipes or “representative cylinders”.
In this method, we follow the formula: V = int_a^b (length * height * thickness)
or V = int_a^b 2pi* radius*height*thickness
For the bounded region, as shown on the attached image, the rectangular strip is parallel to x-axis (axis of rotation). We can let:
r=y
h =f(x) or h=x_2 - x_1
The x_2 will be based on the equation y =4x^2 rearranged into x= sqrt(y/4) or x =sqrt(y)/2
h =sqrt(y)/2-0
h=sqrt(y)/2
For boundary values, we have y_1=0 to y_2=4 (based from the boundary line).
Plug-in the values:
V = int_a^b 2pi*radius*height*thickness, we get:
V = int_0^4 2pi*y*sqrt(y)/2*dy
V = int_0^4 2pi*y*y^(1/2)/2*dy
V = int_0^4 piy^(3/2)dy
Apply basic integration property: intc*f(x) dx = c int f(x) dx.
V = pi int_0^4 y^(3/2)dy
Apply power rule for integration: int y^n dy= y^(n+1)/(n+1).
V = pi *(y^(3/2+1))/((3/2+1))|_0^4
V = pi *(y^(5/2))/((5/2))|_0^4
V = pi *y^(5/2)*2/5|_0^4
V = (2pi y^(5/2))/5|_0^4
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = (2pi (4)^(5/2))/5-(2pi (0)^(5/2))/5
V = (64pi)/5 -0 V = (64pi)/5 or 40.21 (approximated value)