# 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?

Oct 21, 2015

$12 \pi$

#### Explanation:

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

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, $A \left(x\right)$ 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 $f \left(x\right)$. In other words;

$A \left(x\right) = 2 \pi \left(r a \mathrm{di} u s\right) \left(h e i g h t\right) = 2 \pi x \left(1 + {x}^{2}\right)$

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

${\int}_{0}^{2} A \left(x\right) \mathrm{dx}$

Plug in the area function.

${\int}_{0}^{2} 2 \pi x \left(1 + {x}^{2}\right) \mathrm{dx}$

Move the $2 \pi$ out front and simplify inside the integral.

$2 \pi {\int}_{0}^{2} x + {x}^{3} \mathrm{dx}$

Solve the integral.

$2 \pi {\left({x}^{2} / 2 + {x}^{4} / 4\right)}_{0}^{2}$

$2 \pi \left({2}^{2} / 2 + {2}^{4} / 4\right)$

$12 \pi$