Use the shell method to find the volume of the region below the graph of f(x)=cos(x^2) [0,1/2sqrt(pi)] is rotated about the y-axis

Question

find the volume generated by revolving the region between them about the x-axis: f(x)=2 and g(x)=cos(x) [0,Pi/2]

Answer 1

In the shell method, the dV = 2(pi)xy dx where x is the radius, y is the height and dx is the thickness of a shell (laminate).

 

The x values vary from 0 to ½•√pi on the definite integral which has integrand 2(pi)x cos (x²), to get V from integration (accumulation) of infinitely many laminates (shells) of thickness dx.

 

Since 2pi can be factored out in front of the integral symbol, the basic integrand is x cos x², and the antiderivative is simply   ½ sin x².  Therefore the answer is pi [sin (pi/4) - sin 0] = pi/√2.

 

I trust you will check this out carefully, step by step, on paper, and find that you agree.  If there are any questions, don't hesitate to comment.