The Surface of Revolution

If you rotate every point of the curve and the line in the xy-plane clockwise through $\phi=\arctan\left(\frac{4}{3}\right)$ about the origin of the coordinate system, the line is on the x-axis and the curve is described by $x\to\left(x'(x),y'(x)\right)~~~,~0\leq x\leq 1$ , where ( $u=\frac 43$ )

$\begin{aligned} x'&=x\cos{\phi}+y\sin{\phi}=\frac{1}{\sqrt{1+u^2}}(x+yu)~~~~~~~~~~~~~~~~~~~~~~,~{\sin\left(\arctan{u}\right)=\frac{u}{\sqrt{1+u^2}}~~,~~\cos\left(\arctan{u}\right)=\frac{1}{\sqrt{1+u^2}}}\\ &=\frac 35\left(\frac{11}{3}x+\frac 49x^3\right)=\frac{11}{5}x+\frac{4}{15}x^3\end{aligned}$

and

$\begin{aligned} y'&=-x\sin{\phi}+y\cos{\phi}=\frac 35\left(-\frac 43x+y\right)\\ &=\frac 35\left(\frac 23x+\frac 13x^3\right)=\frac 25x+\frac 15x^3 \end{aligned}$

$~\\ ~\\ ~$

This curve is now rotating about the x-axis and the surface element is

$\begin{aligned} \mathrm dA&=2\pi\sqrt{x_x'^2+y_x'^2}\cdot y'\mathrm dx=2\pi\sqrt{5+4x^2+x^4}\left(\frac 25x+\frac 15x^3\right)\mathrm dx\\ &=\frac{2\pi}{20}\sqrt{5+4x^2+x^4}\left(8x+4x^3\right)\mathrm dx\\ &=\frac{\pi}{10}\sqrt{5+4x^2+x^4}\left[\frac{\mathrm d}{\mathrm dx}\left(5+4x^2+x^4\right)\right]\mathrm dx \end{aligned}$

.

$\begin{aligned} A&=\int\mathrm dA=\int\limits_0^1\frac{\pi}{10}\sqrt{5+4x^2+x^4}\left[\frac{\mathrm d}{\mathrm dx}\left(5+4x^2+x^4\right)\right]\mathrm dx\\\ &=\left[\frac 23\frac{\pi}{10}\sqrt{5+4x^2+x^4}^{\frac 32}\right]^1_0=\frac{\pi}{3\cdot 5}\left(10\sqrt{10}-5\sqrt{5}\right)\\ &=\frac{\sqrt{40}-\sqrt{5}}{3}\pi \end{aligned}$

So $A+B=\boxed{45}$