Couple Calculus Problems (Mostly Solid Revolutions)

[myanmar]

1. Find the vol. of a solid generated by revolving the region bounded by [tex]y = 0[/tex] and [tex] = \frac {1}{\sqrt {x}}[/tex] for [tex]1 \le x \le 2[/tex] about the line [tex]y = - 1[/tex]
2. Find the volume of the solid generated by revolving the region bounded by the lines [tex]x=0[/tex], [tex]y = 0[/tex], [tex]x = 1[/tex], and the curve [tex]y = e^{x^{2}}[/tex]. about the [tex]y[/tex] - axis
3. Can you write the func. [tex]f(x) = x[/tex] as the product of two differentiable functions ([tex]g(x)[/tex] and [tex]h(x)[/tex]) if [tex]g(0) = h(0) = 0[/tex].
4. Show that if
[tex]\frac {a_{0}}{1} + \frac {a_{1}}{2} + ... + \frac {a_{n}}{n + 1}[/tex]
then
[tex]a_{0} + a_{1}x + ... + a_{n}x^{n} = 0[/tex]
for some x in [tex][0,1][/tex]
5. Find the area bounded by [tex]y = \sqrt {x}[/tex], [tex]\frac {1}{x}[/tex], [tex]\frac {1}{x^{2}}[/tex]
6. The region in the first quadrant bounded by [tex]x = 2[/tex] and [tex]y = 4[/tex] and [tex]x^{2} = 4y[/tex] is revolved around the [tex]y[/tex]-axis. Find the volume.


Attempts at solution (so far)
3. Yes. But I don't know how to show this. It's true for $x^{\frac {1}{2}}$ and $x^{\frac {1}{2}}$ or $x^{\frac {1}{3}}$ and $x^{\frac {2}{3}}$.

 

[lippyka]

4. Let f(x) = a_0 + a_1x + ... + a_nx^n. Then, by the Mean Value Theorem, there exists c between 0 and 1 such that f'(c) = \frac {f(1) - f(0)}{1-0} = \frac {a_0 + a_1 + ... + a_n}{n+1}. But since f'(c) = a_0 + 2a_1c +...+ na_nc^{n-1}, it follows that a_0 + a_1 + ... + a_n = (n+1)(a_0 + 2a_1c + ... + na_nc^{n-1}). Therefore, a_0 + a_1x + ... + a_nx^n = 0 where x=c.