Volume with Shell Method for Revolving Curves and Lines

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Homework Statement



Use the shell method to find the volumes of the solid generated by revolving the regions bounded by the curves and lines.

x=2y-y2, x=0

Homework Equations



The shell method is of the format: V = 2 \pi\int x * height dx


The Attempt at a Solution



I cannot picture the problem (not sure exactly how to graph it)
 
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x=2y-y^2 is a parabola and x=0 is a line. What's the problem with graphing them? Where do they intersect? And what what axis are you going to rotate the region around?
 
rotating around the x-axis


how can I tell its a parabola?
 
A usual parabola is given by y = ax^2 + bx + c. In this case, we have x = ay^2 + by + c. The two are much the same really, except that in the second case, the parabola is 'on it's side'.

It is actually two square roots, pasted together though. You can see this by solving the equation for y, resulting in your usual "y = f(x)" graph. For general a, b, c:
y = \frac{-b \pm \sqrt{b^2 - 4ac + 4ax} }{2a}
(Two equations, one for + and one for -!)

If you graph these, we get (using a = 1, b = 2, c = -1 for example):
sqrz8o.jpg

(The two don't meet in the middle exactly because Maple has trouble graphic them there.)If you really can't figure out the shape of a curve, why not simply try to draw a few easy points on paper? Like (0,0) or (0,1), (1,0) etc... You will most likely recognize a familiar shape from that, and you can then go on and analyze the curve equation further.
 
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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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