Weird Volume of Revolution

The intersection of the two curves is at $x = \sqrt{2} x^2$ , which is at $x=0$ and $x = \dfrac{1}{\sqrt{2}}$ . Given a point $(x , \sqrt{2} x^2)$ on the parabola, its perpendicular distance to the line $y = x$ is given by

$r = \dfrac{1}{\sqrt{2}} ( x - \sqrt{2} x^2 )$

The thickness of the disc is $\sqrt{2} dx$ , hence the volume is given by

$\displaystyle V = \int_{0}^{\frac{1}{\sqrt{2}}} \sqrt{2} \pi r^2 dx$

Plugging the expression for $r$ from above, we get

$\displaystyle V = \sqrt{2} \int_{0}^{\frac{1}{\sqrt{2}}} \pi \frac{1}{2} (x - \sqrt{2} x^2)^2 dx$

Expanding, and integrating, this becomes,

$\displaystyle V = \dfrac{\sqrt{2}}{2} \pi \int_0^{\frac{1}{\sqrt{2}}} (x^2 - 2 \sqrt{2} x^3 + 2 x^4 ) dx$

$\displaystyle V = \frac{\sqrt{2}}{2} \pi \left( \frac{1}{3} (\frac{1}{\sqrt{2}})^3 - \frac{\sqrt{2}}{2} (1/2) + \dfrac{2}{5} (\frac{1}{\sqrt{2}})^5 \right)$

$\displaystyle V = \frac{1}{2} \pi \left( \frac{1}{6} - \frac{1}{4} + \frac{1}{10} \right) = \frac{1}{2} \pi (\frac{1}{60} )(10 - 15 + 6) = \pi (\dfrac{1}{120})$

So that $a = 1 , b = 120$ , making the answer $1 + 120 = 121$ .

The parabola $y = \sqrt{2}x^2$ is equivalent to $\sqrt{2}x^2 - y = 0$ , an equation in standard form where $A = \sqrt{2}$ , $E = -1$ , and $B = C = D = 0$ . If we rotate it by $45°$ , then:

$A' = A \cos^2 45° + B \sin 45° \cos 45° + C \sin^2 45°= \frac{\sqrt{2}}{2}$

$B' = 2(C - A) \sin 45° \cos 45° + B(\cos^2 45° - \sin^2 45°) = -\sqrt{2}$

$C' = A \sin^2 45° - B \sin 45° \cos 45° + C \cos^2 45°= \frac{\sqrt{2}}{2}$

$D' = D \cos 45° + E \sin 45°= -\frac{\sqrt{2}}{2}$

$E' = -D \sin 45° + E \cos 45°= -\frac{\sqrt{2}}{2}$

for a new equation of $\frac{\sqrt{2}}{2}x^2 - \sqrt{2}xy + \frac{\sqrt{2}}{2}y^2 - \frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2}y = 0$ , which rearranges to $y = x + \frac{1}{2} - \frac{1}{2}\sqrt{8x + 1}$ .

The line $y = x$ rotates to $y = 0$ after a $45°$ rotation, which intersects $y = x + \frac{1}{2} - \frac{1}{2}\sqrt{8x + 1}$ at $(0, 0)$ and $(1, 0)$ .

The volume is then $V = \int_{0}^{1} \pi(x + \frac{1}{2} - \frac{1}{2}\sqrt{8x + 1})^2 dx$ .

Substituting $u = \sqrt{8x + 1}$ and rearranging gives $V = \int_{1}^{3} \cfrac{\pi}{256}(u^5 - 8u^4 + 22u^3 - 24u^2 + 9u) du = \cfrac{1}{120}\pi$ .

Therefore, $a = 1$ , $b = 120$ , and $a + b = \boxed{121}$ .