Question 771777
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*[tex \LARGE y\ =\ x^2\ -\ 1] is a parabola with vertex at *[tex \LARGE (0,\,-1)].  *[tex \LARGE y\ =\ 3] is the horizontal line that passes through *[tex \LARGE (0,\,3)] and that intersects the parabola, in the desired interval, at *[tex \LARGE (2,\,3)].


It will be convenient to write *[tex \LARGE x] as a function of *[tex \LARGE y]:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ -\ 1]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x\ =\ \sqrt{y\ +\ 1}]


Rotate the parabola, then consider a thin disk with radius *[tex \LARGE \sqrt{y\ +\ 1}].  The volume of this disk is *[tex \LARGE \pi\left(\sqrt{y\ +\ 1}\right)^2\,dy\ =\ \pi(y\ +\ 1)\,dy]


Integrate from -1 to 3


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi\,\int_{-1}^3\,y\ +\ 1\,dy]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi\left|\frac{y^2}{2}\ +\ y\right|^3_{-1}\  =\ \pi\left(\left(\frac{9}{2}\ +\ 3\right)\ -\ \left(\frac{-1}{2}\ +\ (-1)\right)\right)\ =\ 9\pi]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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