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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?

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For the shell method we should assume

$$\int_0^2 2\pi R H dx\int_0^2 [2\pi \cdot x \cdot (5-(x^2+1))] dx=2\pi\int_0^2 (4x-x^3) dx=2\pi\left[2x^2-\frac{x^4}4\right]_0^2=8\pi$$

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  • $\begingroup$ Why use 5-x^2-1 instead of x^2 + 1? $\endgroup$ – Eric Wiener Mar 2 '18 at 12:06
  • $\begingroup$ the height is given by $5-f(x)$, make a sketch $\endgroup$ – user Mar 2 '18 at 12:07
  • $\begingroup$ It makes sense that it should be 5-f(x), but in the formula for the shell method, isn’t it 2pi * x * f(x)? $\endgroup$ – Eric Wiener Mar 2 '18 at 12:12
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    $\begingroup$ Yes if we assume H=f(x) but in this case $f(x)=5-(x^2+1)$, you should always make a sketch of the region to avoid this kind of confusion $\endgroup$ – user Mar 2 '18 at 12:14
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    $\begingroup$ Got it. Thank so much! $\endgroup$ – Eric Wiener Mar 2 '18 at 12:16

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