Finding the volume of a solid when the solid is a region rotated around a line

jlt90
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Homework Statement



Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y = x^(6), y = 1 about y = 2

Homework Equations





The Attempt at a Solution


So i drew a picture and found the region to be fin shaped from 0 to 1 on both the x and y axis. I then rotated it around y=2 to get a cylinder with a missing inside with flat sides and an outside that is curved. I believe the thickness of these cylinders is delta y, so I'm putting everything in terms of y. I then find the radius to be 2-y, since the center of the cylinder is at y=2 and at y=0 the radius is 2, at y=1 it is 1. So that gives me a circumference of 2pi(2-y). Next I find the height, which my picture shows as the x values when x=y^(1/6), so my h=y^(1/6). I am now able to set up my integral so that 0int1 2pi(2-y)y^(1/6)dy. When I evaluate this, I come up 2pi(12/7-6/13) which is the wrong answer. Can someone please tell me where I went wrong?
 
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I found where I made my mistake :) So no need to help me anymore. Thanks anyways.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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