A calculus problem by Rocco Dalto

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

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

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

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

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

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To eliminate the ${\bf x' y' }$ term set ${\bf sin^2\theta - cos^2\theta = 0 \implies tan^2\theta = 0 \implies tan\theta = +-1 }$

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For ${\bf 0 < tan\theta < 90 \implies \theta = 45 \implies }$

${\bf 4 x'^2 + 4 y'^2 + y'^2 - x'^2 = 4 \implies 3x'^2 + 5y'^2 = 4 }$ and we have an ellipse. $$

The area ${\bf I = \frac{2}{\sqrt{5}} \int_{-2/\sqrt{3}}^{2/\sqrt{3}} \sqrt{4 - 3 x'^2} dx' }$

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Let ${\bf \sqrt{3} x' = 2 sin\theta \implies dx' = \frac{2}{\sqrt{3}} cos\theta d\theta \implies }$

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${\bf I = \frac{4}{\sqrt{15}} \int_{-\pi/2}^{\pi/2} (1 + cos(2\theta)) d\theta = }$

${\bf \frac{4}{\sqrt{15}} * (\theta + \frac{1}{2} sin(2\theta))|_{-\pi/2}^{\pi/2} = }$

${\bf \frac{4\pi}{\sqrt{15}} = \frac{a\pi}{\sqrt{b}} \implies a + b = 19 }$

Arrange as quadratic:

$2y^2-xy+2(x^2-1)=0$

So $\displaystyle y=\frac{x \pm \sqrt{x^2-16(x^2-1)}}{2} \\ \displaystyle y=\frac{x \pm \sqrt{16-15 x^2}}{2}$

As $y \in R$ discriminant should be greater than zero.So $16-15 x^2 >0 i.e. x<|\frac{4}{\sqrt{15}}|$

The area will be $A=\displaystyle \int_{\frac{-4}{\sqrt{15}}}^{\frac{4}{\sqrt{15}}} y dx$

integrating the above expression using basic integration technique the area will be = $\frac{4 \pi}{\sqrt{15}}$