How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y = 1 + x^2#, #y = 0#, #x = 0#, #x = 2# rotated about the y-axis?

1 Answer
Oct 21, 2015

#12pi #

Explanation:

With a problem like this, it is always a good idea to start with a graph.

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We are going to rotate this shape around the #y#-axis, which will result in a cylinder with a parabolic indentation in the top. In order to evaluate this using shells, we will be integrating a bunch of ring surfaces between the #x#-axis and the function, thus we will get an area function, #A(x)# for surfaces at each #x# and add them all together to get the volume.

We can find our area function by stacking a bunch of circles of radius #x# until they reach a height of #f(x)#. In other words;

#A(x)=2pi (radius) (height)=2pi x(1+x^2)#

Now that we have an area function, we just need to integrate it from the #y#-axis to #x=2#.

#int_0^2 A(x) dx#

Plug in the area function.

#int_0^2 2pi x(1+x^2) dx#

Move the #2pi# out front and simplify inside the integral.

#2pi int_0^2 x+x^3 dx#

Solve the integral.

#2pi (x^2/2 +x^4/4)_0^2#

#2pi (2^2/2+2^4/4)#

#12pi #