Calculating Volume of Revolution: Bounded Figure Rotated about y = -1

ChaoticLlama · 2006-12-12T23:55:19+00:00

What is the volume of the figure bounded by y = 2x - x^2, y = 2x, x = 2, and rotated about the line y = -1. Is this the correct integral? \[ V = \pi \int_0^2 {((2x + 1)^2 - (2x - x^2 + 1)^2 } )dx \] Thank you for your time.

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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Thread 'Prove that the integral is equal to ##\pi^2/8##'

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