I was attempting to solve this problem using the shell method:
The region R in the first quadrant is enclosed by the lines $x=0$ and $y=5$ and the graph of $y=x^2+1$. The volume of the solid generated when R is revolved around the y axis is
The correct answer is $8\pi$.
I attempted to solve this using the shell method as follows: $\int_0^2 (2\pi*x*(x^2+1)) dx$. However, I get $12\pi$ as a result.
Solving with the disk method, I get the correct answer: $\int_1^5 (\pi*(y-1))dy = 8\pi.$
What mistake did I make using the shell method?
For the shell method we should assume
$$\int_0^2 2\pi R H dx\int_0^2 [2\pi \cdot x \cdot (5-(x^2+1))] dx=2\pi\int_0^2 (4x-x^3) dx=2\pi\left[2x^2-\frac{x^4}4\right]_0^2=8\pi$$