Area of a Region

Using the equations of rotation to rotate the x and y axis we have:

$$

${\bf x = x' cos\theta - y' sin\theta }$

${\bf y = x' sin\theta + y' cos\theta }$

Replacing the equations of rotation in the original equation and simplifying we obtain: $$

${\bf (2 cos^2\theta + 12 cos\theta sin\theta - 3 sin^2\theta)x'^2 + (12 cos(2\theta) - 5 sin(2\theta))x'y' +}$

${\bf (2 sin^2\theta - 12 cos\theta sin\theta - 3 cos^2\theta)y'^2 - 42 = 0. }$

To eliminate the ${\bf x'y' }$ term set ${\bf 12 cos(2\theta) - 5 cos(2\theta) = 0 \implies tan(2\theta) = \frac{12}{5} }$

${\bf \implies tan2\theta = \frac{2 tan\theta}{1 - tan^2\theta} = \frac{12}{5} \implies }$

${\bf 6 tan^2\theta + 5 tan\theta - 6 = 0 \implies tan\theta = \frac{-5 \:\pm \:13}{12} }$

$$

Choosing the positive value of ${\bf tan\theta \implies tan\theta = \frac{2}{3} \implies cos\theta = \frac{3}{\sqrt{13}} }$ and ${\bf sin\theta = \frac{2}{\sqrt{13}} \implies }$

${\bf x = \frac{3}{\sqrt{13}} x' - \frac{2}{\sqrt{13}} y'}$

$$

${\bf y = \frac{2}{\sqrt{13}} x' + \frac{3}{\sqrt{13}} y' }$

Replacing the equations of rotation in the original equation and simplifying we obtain: $$

${\bf 78 x'^2 - 91 y'^2 = 546 }$

${\bf \implies \frac{x'^2}{7} - \frac{y'2}{6} = 1 \: or \: 6 x'^2 - 7 y'^2 = 42 }$ $$

So we have a hyperbola in the ${\bf x'y' }$ system.

${\bf -3\sqrt{13}\:x - 2\sqrt{13}\:y + 39 = 0 \implies y = -\frac{3}{\sqrt{2}}\:x + \frac{39}{2\sqrt{13}} }$ $$

Two point on this line are ${\bf (0, \frac{39}{2\sqrt{13}}) }$ and ${\bf (\sqrt{13}, 0) }$ $$

Rotating the point ${\bf (\sqrt{13}, 0) }$ we have: $$

${\bf \sqrt{13} = \frac{3}{\sqrt{13}} x' - \frac{2}{\sqrt{13}} y' }$ $$

${\bf 0 = \frac{2}{\sqrt{13}} x' + \frac{3}{\sqrt{13}} y' \implies (x' = 3,y' = -2) }$

Similarily, rotating the point ${\bf (0, \frac{39}{2\sqrt{13}}) ) \implies (x' = 3,y' = \frac{9}{2}) }$

$$

${\bf \therefore }$ In the ${\bf x'y'}$ system we have the line ${\bf x' = 3, \:}$ and the curve ${\bf 6 x'^2 - 7 y'^2 = 42 }$ and the line ${\bf x' = 3 }$ intersect at ${\bf (3,2\sqrt{\frac{3}{7}}) }$ and ${\bf (3,-2\sqrt{\frac{3}{7}}) \implies }$ $$

${\bf Area A_{1} = \int_{-2\sqrt{\frac{3}{7}}}^{2\sqrt{\frac{3}{7}}} 3 dy' }$ ${\bf - \sqrt{\frac{7}{6}} \int_{-2\sqrt{\frac{3}{7}}}^{2\sqrt{\frac{3}{7}}} \sqrt{y'^2 + 6} dy' = 12\sqrt{\frac{3}{7}} - \sqrt{\frac{7}{6}} \int_{-2\sqrt{\frac{3}{7}}}^{2\sqrt{\frac{3}{7}}} \sqrt{y'^2 + 6} dy' }$

$$

For ${\bf I(y') = \sqrt{\frac{7}{6}} \int \sqrt{y'^2 + 6} \: dy' }$ $$

Let ${\bf y' = \sqrt{6} tan\theta \implies dy' = \sqrt{6} sec^2\theta \implies }$

${\bf I(\theta) = \sqrt{42} \int sec^3\theta d\theta }$

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

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

${\bf \int sec^3\theta d\theta = \frac{1}{2} (sec\theta tan\theta + ln|sec\theta + tan\theta)|) }$

For ${\bf -\frac{\pi}{2} < \theta < \frac{\pi}{2} \: Let \: 0 < \theta < \frac{\pi}{2} \implies }$

${\bf I_{\theta} = \frac{\sqrt{42}}{2} \int_{-\theta}^{\theta} sec^3\theta d\theta = }$

${\bf \sqrt{42} (sec\theta tan\theta + ln|\frac{1 + sin\theta}{cos\theta}|) }$

$$

${\bf y' = \sqrt{6} tan\theta \implies tan\theta = \frac{2}{\sqrt{14}}\:, sec\theta = \frac{3}{\sqrt{7}} }$

${\bf sin\theta = \frac{\sqrt{2}}{3}, \: and \: cos\theta = \frac{\sqrt{7}}{3} \implies }$

${\bf I = \frac{6\sqrt{21}}{7} + \sqrt{42} ln(\frac{3 + \sqrt{2}}{\sqrt{7}}) \implies }$

Area ${\bf A_{1} = 12\sqrt{\frac{3}{7}} - 6\sqrt{\frac{3}{7}} + \sqrt{42} ln(\frac{3 + \sqrt{2}}{\sqrt{7}}) = }$

${\bf 6\sqrt{\frac{3}{7}} + \sqrt{42} ln(\frac{3 + \sqrt{2}}{\sqrt{7}}) = }$

${\bf \sqrt{6} (3\sqrt{\frac{2}{7}} + \sqrt{7} ln(\frac{3 + \sqrt{2}}{\sqrt{7}})) }$

$$

By symmetry ${\bf Area \: A_{1} = A_{2} \implies }$

${\bf \:Total \: Area \: A = 2 A_{1} = 2\sqrt{6} (3\sqrt{\frac{2}{7}} + \sqrt{7} ln(\frac{3 + \sqrt{2}}{\sqrt{7}})) = }$

${\bf 2\sqrt{3 * 2} (3\sqrt{\frac{2}{7}} + \sqrt{7} ln(\frac{3 + \sqrt{2}}{\sqrt{7}})) = }$

${\bf b\sqrt{a * b} (a\sqrt{\frac{b}{c}} + \sqrt{c} ln(\frac{a + \sqrt{b}}{\sqrt{c}})) \implies }$

${\bf a = 3,\: b = 2,\: and \: c = 7 \implies a + b + c = 12. }$