[Math] Volume of a Parabola Rotated Around a Line

calculusvolume

The question is as follows: Find the volume of the solid generated by revolving the region bounded by the parabola $y = x^2$ and the line $y = 1$ about the line $y = -1$.

My attempt: Using the washer method, I set the outer radius to $1 + x^2$ and the inner radius to be $1$. This gives me the integral:
$V = \pi\int_{-1}^1(1+x^2)^2-1^2dx = \pi\int_{-1}^12x^2+x^4dx$, which when evaluated gives the answer of $26\pi/15$. This however, is wrong. The correct answer is $64\pi/15$.

I put it through Wolfram Alpha and it gave me the same answer: https://www.wolframalpha.com/input/?i=rotate+the+region+between+0+and+x%5E2+with+-1%3Cx%3C1+around+the+line+y+%3D+-1

Any help would be appreciated.

EDIT: I tried using the shell method and got the correct answer, using $y+1$ as the radius and $\sqrt y$ as the height. So now the question is how would you do this using the washer method.

Best Answer

  1. Your wolfram alpha link has the wrong bounds, using $y=0$ instead of $y=1$ as one of the bounds. Even with the correct bounds, wolfram|alpha seems to be screwing up the integral and missing the central "missing" piece.
  2. Using the washer method, the outer radius is $R_{outer} = 1 - (-1) = 2$ and the inner radius is $R_{inner} = x^2 - (-1) = x^2 + 1$. Thus \begin{align*} V &= \int_{-1}^1 \pi( R_{outer}^2 - R_{inner}^2) ~\mathrm{d} x \\ &= \pi \int_{-1}^1 2^2 - (x^2+1)^2 ~\mathrm{d}x \end{align*} which evaluates to the correct result.
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