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Question Number 35294 by ajfour last updated on 17/May/18

Commented by ajfour last updated on 17/May/18

$F i n d m o m e n t o f i n e r t i a o f a$ $s q u a r e p l a t e i f i t s d e n s i t y a t a$ $p o i n t (s a y P) i s p r o p o r t i o n a l t o t h e$ $d i s t a n c e o f t h a t p o i n t f r o m$ $v e r t e x A .$
	Answered by ajfour last updated on 18/May/18

	$l e t s t a k e t h e r o t a t i o n a l a x i s t h e$ $y a x i s . A t h e o r i g i n .$ $d I = ρ x^{2} (d y) (d x)$ $l e t ρ = k r$ $\Rightarrow I = k \int_{0}^{a} [\int_{0}^{a} x^{2} \sqrt{x^{2} + y^{2}} d y] d x$ $o r$ $l e t x = r \cos θ, y = r \sin θ$ $I = \int_{0}^{π / 4} [\int_{0}^{a \sec θ} k r (r^{2} \cos^{2} θ) (r d r)] d θ$ $+ \int_{π / 4}^{π / 2} [\int_{0}^{a cosec θ} k r (r^{2} \cos^{2} θ) r d r] d θ$ $= \frac{{k a}^{5}}{5} [\int_{0}^{π / 4} \sec^{3} θ d θ + \int_{π / 4}^{π / 2} \cot^{2} θ cosec^{3} θ d θ]$ $= \frac{{k a}^{5}}{5} (I_{1} + I_{2})$ $I_{1} = \int_{0}^{π / 4} \sec θ \sec^{2} θ d θ$ $= \sec θ \tan θ ∣_{0}^{π / 4} - \int_{0}^{π / 4} (\sec^{2} θ - 1) \sec θ d θ$ $\Rightarrow 2 I_{1} = \sqrt{2} + \ln ∣ \sec θ + \tan θ ∣_{0}^{π / 4}$ $\Rightarrow I_{1} = \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (1 + \sqrt{2})$ $I_{2} = \int_{π / 4}^{π / 2} \cot^{2} θ cosec^{3} θ d θ$ $l e t ϕ = \frac{π}{2} - θ \Rightarrow d ϕ = - d θ$ $I_{2} = \int_{0}^{π / 4} \tan^{2} θ \sec^{3} θ d θ$ $l e t \tan θ = t \Rightarrow d t = \sec^{2} θ d θ$ $= \int_{0}^{1} t^{2} \sqrt{1 + t^{2}} d t$ $= \frac{t^{3}}{3} \sqrt{1 + t^{2}} ∣_{0}^{1} - \int_{0}^{1} \frac{t^{3}}{3} \times \frac{t}{\sqrt{1 + t^{2}}} d t$ $3 I_{2} = \sqrt{2} - \int_{0}^{1} \frac{t^{4} - 1}{\sqrt{1 + t^{2}}} d t - \int_{0}^{1} \frac{d t}{\sqrt{1 + t^{2}}}$ $3 I_{2} = \sqrt{2} - \int_{0}^{1} (t^{2} - 1) \sqrt{1 + t^{2}} d t - \int \frac{d t}{\sqrt{1 + t^{2}}}$ $4 I_{2} = \sqrt{2} + \int_{0}^{1} \sqrt{1 + t^{2}} - \int_{0}^{1} \frac{d t}{\sqrt{1 + t^{2}}}$ $= \sqrt{2} + (\frac{t}{2} \sqrt{1 + t^{2}}) ∣_{0}^{1} + \frac{1}{2} \ln ∣ t + \sqrt{1 + t^{2}} ∣_{0}^{1}$ $- \ln ∣ t + \sqrt{1 + t^{2}} ∣_{0}^{1}$ $\Rightarrow I_{2} = \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{8} - \frac{1}{8} \ln (1 + \sqrt{2})$ $A s I = \frac{{k a}^{5}}{5} (I_{1} + I_{2})$ $I = \frac{{k a}^{5}}{5} [\frac{\sqrt{2}}{2} + \frac{1}{2} \ln (1 + \sqrt{2}) +$ $+ \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{8} - \frac{1}{8} \ln (1 + \sqrt{2})]$ $\Rightarrow I = \frac{{k a}^{5}}{40} [7 \sqrt{2} + 3 \ln (1 + \sqrt{2}] .$
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