Volume of Solid: Answers & Clarification

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Thanks to the awesome help on this board, I am narrowing my problems down quickly. I have already answered problems #6, 7, 9, and 10 as b, f, c, and a respectively. I wouldn't mind if you checked them but it's not my biggest worry. Instead I have given many attempts at number 8 using the disc/washer and shell formula. The page is out of date because I have been working on it since that page has been scanned and I no longer have access to that scanner. Any clarification on the formulas and their uses in this problem would be great.

Cal1chapter6test_0004.jpg
 
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Problem 8 is "Find the volume of the solid obtained by rotating the region bounded by the curves y= x and y= x2 around the line y= 2.

First, of course, draw a picture. You should see that vertical line drawn from y= 2 will cross that region first at y= x and then at y= x2. That means that a "washer" formed by rotating the figure about y= 2 will have inner radius 2- x and outer radius 2- x2. The areas of the two circles with those radii will be \pi(2- x)^2 and \pi(2- x^2)^2. The area of the "washer" between them will have area equal to the difference of those. Integrate that with respect to dx from x= 0 to x= 1.

If you want to do it by "shells", each shell will be a thin cylinder of thickness dy, radius 2- y, and height \sqrt{y}- y. The surface area of that will be 2\pi(2- y)(\sqrt{y}- y) and the volume will be 2\pi(2- y)(\sqrt{y}- y)dy integrated from y= 0 to y= 1.
 
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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