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Question Number 67807 by TawaTawa last updated on 31/Aug/19

Commented by MJS last updated on 01/Sep/19

$perimeter or area ?$
Commented by TawaTawa last updated on 01/Sep/19

$Area and perimeter sir$
	Answered by MJS last updated on 02/Sep/19

	$parabola p : y = x^{2}$ $line A D l_{1} : y = - \sqrt{3} x + 2 \sqrt{3}$ $line B C l_{2} : y = - \sqrt{3} x + 4 \sqrt{3}$ $D = p \cap l_{1} = (\begin{matrix} \frac{- \sqrt{3} + \sqrt{3 + 8 \sqrt{3}}}{2} \\ \frac{3 + 4 \sqrt{3} - \sqrt{9 + 24 \sqrt{3}}}{2} \end{matrix})$ $C = p \cap l_{2} = (\begin{matrix} \frac{- \sqrt{3} + \sqrt{3 + 16 \sqrt{3}}}{2} \\ \frac{3 + 8 \sqrt{3} - \sqrt{9 + 48 \sqrt{3}}}{2} \end{matrix})$ $the length of the parabola$ $\underset{a}{\int^{b}} \sqrt{1 + {(f^{'} (x))}^{2}} d x$ $\underset{a}{\int^{b}} \sqrt{1 + 4 x^{2}} d x = {[\frac{1}{2} x \sqrt{1 + 4 x^{2}} + \frac{1}{4} \ln (2 x + \sqrt{1 + 4 x^{2}})]}_{a}^{b}$ $\approx 2.33536$ $∣ A B ∣= 2$ $∣ A D ∣\approx 1.62640$ $∣ B C ∣\approx 4.19014$ $perimeter \approx 10.1519$ $area = \underset{a}{\int^{b}} p d x + \underset{b}{\int^{4}} l_{2} d x - \underset{a}{\int^{2}} l_{1} d x \approx 4.97554$ $this example {doesn}^{'} t resolve to ‘ ‘ {nice}^{″} solutiond$
	Commented by TawaTawa last updated on 03/Sep/19

	$God bless you sir, i appreciate your time$
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