Solve Integration Problem: Volume of Rotor Ring in x-axis

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Homework Statement



A solid rotor ring is made by rotation round the x-axis the area enclosed between the lines y=x+5 and y=2x+7 between the x-values x=0 and x=1.
Find its volume.

Homework Equations


n/a


The Attempt at a Solution


ytobeintegrated = 2x+7-x-5 \\= x+2\\\int (x=2)dx \\=\left[\frac{x^2}{2}+2x\right](between 1 and 0) \\=3.5 units

What next? I think this is only giving me the area in one plane? Do i just multiply by 360 degrees?
 
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You cannot simply take away the two functions before integration or multiply by 360 degrees. Also, if you integrate any 2-dimensional, cartesian function without any manipulation you will only get the area under that function. So although your integration is correct (though your solution should be 2.5 cubic units not 3.5), the idea is not.

Try using:

Volume \ = \ \int_{a}^{b} \pi f(x)^{2} \ dx
where f(x) is one of the functions.

What do you think is the next step?

The Bob
 
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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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