Area of a Region

${\bf Begin }$ $$

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: $$

${\bf (16 cos^2\theta - 12 sin(2\theta) + 9 sin^2\theta)x'^2 - (7 sin(2\theta) + 24 cos(2\theta))x'y' +}$

${\bf (16 sin^2\theta + 12 sin(2\theta) + 9 cos^2\theta)y'^2 - 5(3 cos\theta + 4 sin\theta)x' + 5 (3 sin\theta - 4 cos\theta)y' = 0. }$

To eliminate the ${\bf x'y' }$ term set ${\bf 7 sin(2\theta) + 24 cos(2\theta) = 0 \implies tan(2\theta) = -\frac{24}{7} }$

${\bf \implies tan2\theta = \frac{2 tan\theta}{1 - tan^2\theta} = -\frac{24}{7} \implies }$

${\bf 12 tan^2\theta - 7 tan\theta - 12 = 0 \implies tan\theta = \frac{7 \:\pm \:25}{24} }$

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Since the sign of xy term is negative we choose ${\bf tan\theta = -\frac{3}{4} \implies cos\theta = \frac{4}{\sqrt{5}} }$ and ${\bf sin\theta = -\frac{3}{\sqrt{5}} \implies }$

${\bf x = \frac{4}{5} x' + \frac{3}{5} y'}$

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${\bf y = -\frac{3}{5} x' + \frac{4}{5} y' }$

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

${\bf y' = x'^2 }$

$$

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

Two points on the line ${\bf x - 7 y + 10 = 0 \: are }$

${\bf (-10, 0) }$ and ${\bf (0 , \frac{10}{7}) }$ $$

For ${\bf (-10, 0) }$ we have:

$$ ${\bf -10 = \frac{4}{5} x' + \frac{3}{5} y' }$ $$

${\bf 0 = -\frac{3}{5} x' + \frac{4}{5} y' }$

${\bf \implies (x' = -8, y' = -6) }$ $$

Similarly, ${\bf (0, \frac{10}{7}) \rightarrow (x' = -\frac{6}{7}, y' = \frac{8}{7}) }$

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Using the points in the ${\bf x'y' \: system \: (x' = -8, y' = -6) \:, \: (x' = -\frac{6}{7} , y' = \frac{8}{7}) \implies }$ $$

${\bf slope \: m = 1 \implies y' = x' + 2 }$ $$

Two points on the line ${\bf 7 x + y - 10 = 0 \: are }$ $$

${\bf (0 , 10) \: and (\frac{10}{7}, 0) }$

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Similarly, ${\bf (0, 10) \rightarrow (x' = -6 , y' = 8) }$ and ${\bf (\frac{10}{7} , 0) \rightarrow (x' = \frac{8}{7}, y' = \frac{6}{7}) }$

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Using the points in the ${\bf x'y' \: system \: (x' = -6 , y' = 8) , (x' = \frac{8}{7} , y' = \frac{6}{7}) \implies }$

${\bf \implies slope \: m = -1 \implies y' = -x' + 2 }$

${\bf y' = -x' + 2 \: and \: y' = x'^2 \implies x'^2 + x' - 2 = 0 \implies x' = 1 \:, x' = -2\: \implies }$

The points of intersection of ${\bf y' = -x' + 2 \: and \: y' = x'^2 }$ are ${\bf (1 \:, 1) \:, (-2 \:, 4) }$

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Simiarily, ${\bf y' = x'^2 \: and \: y' = x' + 2 }$ intersect at ${\bf (-1 , 1) \: and \: (2 , 4). }$

Using the symmetry about the ${\bf y' \: axis }$ the total area ${\bf A = 2 * \int_{0}^{2} x' + 2 - x'^2 dx' = 2 * \int_{0}^{2} -x'^2 + x' + 2 dx' = }$

${\bf 2 * (-\frac{x'^3}{3} + \frac{x'^2}{2} + 2 x') \: |_{0}^{2} = \frac{20}{3} = \frac{a}{b} \implies a + b = 23. }$ $$

${\bf End. }$

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Another method to find the desired area, although it's not practical, is as follows:

In the ${\bf x'y' }$ system: $$

Let ${\ A:(2,4) \: B:(1,1) \: C:(-1,1) \: D:(-2,4) \: O:(0,2) }$

Since the lines are perpendicular ${\bf \angle AOB, \angle BOC \angle DOC }$ are all right angles. ${\bf \therefore \triangle AOB, \triangle BOC, \triangle DOC }$ are all right triangles.

${\bf OA = 2\sqrt{2} = OD \: and \: OB = \sqrt{2} \implies \: Area_{\triangle AOB} = Area_{\triangle DOC} = 2. }$

${\bf OC = \sqrt{2} \implies \: Area_{\triangle BOC} = 1 \implies A_{1} = }$ Area of all triangles ${\bf = 5 }$

${\bf slope \: m_{AB} = 3 \implies y' = 3 x' - 2 \: on \: (1 <= x' <= 2) }$

${\bf y' = 1 \: on \: (-1 <= x' <= 1) }$

${\bf \: slope \: m_{CD} = -3 \implies y' = -3 x' - 2 \: on \: (-2 <= x' <= -1) }$

${\bf A_{2} = 2 * \int_{1}^{2} (3 x' - 2 - x'^2) dx' }$ ${\bf + \int_{-1}^{1} (1 - x'^2) dx' }$ = ${\bf \frac{5}{3} }$

${\bf \therefore }$ Total Area ${\bf A = A_{1} + A_{2} = 5 + \frac{5}{3} = \frac{20}{3}. }$