Volume of solids of revolution

The volume ${\bf V = \pi * \int_{0}^{\sqrt{3} * b} (x - a)^2 dy = }$

$$

${\bf \pi \frac{a^2}{b^2} * \int_{0}^{\sqrt{3} * b} 2b^2 + y^2 -2b\sqrt{b^2 + y^2} dy }$

$$

Let ${\bf I_{2} = 2b * \int_{0}^{\sqrt{3} * b} \sqrt{b^2 + y^2} dy' }$ $$

Using trig substitution, Let ${\bf y = b tan\theta \implies dy = b sec^2\theta \implies}$ $$

${\bf I_{2} = 2b^3 * \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta }$ $$

Using integration by parts let ${\bf u = sec\theta \implies du = sec\theta tan\theta}$ $$

${\bf dv = sec^2(\theta) \implies v = tan\theta \implies }$

${\bf \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta = sec\theta tan\theta|_{0}^{\frac{\pi}{3}} - \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta + \int_{0}^{\frac{\pi}{3}} sec\theta d\theta \implies }$ $$

${\bf 2 * \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta = (sec\theta tan\theta + ln|sec\theta + tan\theta|)|_{0}^{\frac{\pi}{3}} \implies }$ $$

${\bf \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta = \frac{1}{2} * (sec\theta tan\theta + ln|sec\theta + tan\theta|)|_{0}^{\frac{\pi}{3}} = }$

${\bf \frac{1}{2} * (2\sqrt{3} + ln(2 + \sqrt{3})) \implies }$ $$

${\bf I_{2} = b^3 * (2\sqrt{3} + ln(2 + \sqrt{3})) }$ $$

Let ${\bf I_{1} = \int_{0}^{\sqrt{3} * b} 2b^2 + y^2 dy \implies }$ $$

${\bf I_{1} = 2b^2 y + \frac{y^3}{3}|_{0}^{\sqrt{3} * b} = 3\sqrt{3} b^3 \implies}$ $$

${\bf V = a^2 b \pi * (\sqrt{3} - ln(2 + \sqrt{3})) = a^2b\pi * (\sqrt{m} - ln(n + \sqrt{m})) \implies }$ $$

${\bf m + n = 5. }$