Volume of Solids of Revolution

Econometricia
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Find the volume of the solid st,
1. y=cos x , y= 0 in [0,pi] ; Rotated around x=1




2. I am slightly confused, I see that the area will double around twice so I can just use the left half of the curve. I am just not sure how to do so.
 
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Are you sure that you copied this problem correctly. The line x = 1 does not divide the region between y = cosx and y = 0 into two equal halves, so rotating it around the line x = 1 makes a complicated volume of revolution. Could it be that you're supposed to rotate the region around the line y = 1?
 
Yes, not 2 equal halves but the rotation around the x=1 axis would cover the right hand side of the curve in rotation and double over so can't we ignore that piece?
 
No, I don't think so. The part of the region under y = cosx on [0, 1] would cover the part on [1, pi/2], so I suppose you could ignore that part. The part of the curve on [pi/2, pi] is below the x-axis. The part above [0, 1] is above the x-axis. This is a very unusual problem.
 
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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