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April 16th, 2017, 09:37 AM   #1
Joined: Apr 2017
From: te

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shell method question

Use shell method to express the volume of the solid obtained by rotating the region bounded by the curves x = √cos 8y
-pi/8≦y≦pi/8 , and x = 0 about the line x =4?
no need to compute the integral


no idea to solve this when about the line x but not line y...
Thank you

Last edited by khen; April 16th, 2017 at 09:51 AM.
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April 16th, 2017, 09:59 AM   #2
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shells w/r to $x$ ...

$\displaystyle V = 2\pi \int_a^b r(x) \cdot f(x) \, dx$

$\displaystyle V = \dfrac{\pi}{2} \int_0^1 (1-x)\left[\arccos(x^2)\right] \, dx$

washers w/r to $y$ ...

$\displaystyle V = \pi \int_c^d [R(y)]^2 - [r(y)]^2 \, dy$

$\displaystyle V = 2\pi \int_0^{\pi/16} 1 - \left[1 - \sqrt{\cos(8y)}\right]^2 \, dy$

fyi, symmetry of the rotated region used in both integrals
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