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Question Number 47026 by ajfour last updated on 04/Nov/18

Commented by ajfour last updated on 04/Nov/18

$I f t h e c i r c l e h a s u n i t r a d i u s a n d$ $b o t h c o l o u r e d a r e a s a r e e q u a l,$ $f i n d t h e e q . o f p a r a b o l a .$
Commented by ajfour last updated on 04/Nov/18

$M r W 3 S i r, p l e a s e s o l v e t h i s, y o u$ $m a y l i k e i t t o o; p l u s m y a n s w e r$ $w i l l b e c h e c k e d . .$
Commented by Meritguide1234 last updated on 04/Nov/18

$h o w d o y o u d r a w t h i s d i a g r a m ?$
Commented by ajfour last updated on 04/Nov/18

$u s i n g L e k h D i a g r a m .$
	Answered by MrW3 last updated on 04/Nov/18

	$e q n . o f p a r a b o l a :$ $y = {a x}^{2}$ $\Rightarrow y^{'} = 2 a x$ $e q n . o f c i r c l e :$ $x^{2} + {(y - b)}^{2} = r^{2} w i t h r = 1$ $\Rightarrow x + (y - b) y^{'} = 0$ $l e t a r = λ$ $c o n t a c t p o i n t s (\pm h, k)$ $k = {a h}^{2} . . . (i)$ $m = 2 a h . . . (i i)$ $h^{2} + {(k - b)}^{2} = r^{2} . . . (i i i)$ $h + (k - b) m = 0 . . . (i v)$ $\Rightarrow k - b = - \frac{h}{m} = - \frac{1}{2 a}$ $\Rightarrow b = k + \frac{1}{2 a}$ $h^{2} + {(- \frac{1}{2 a})}^{2} = r^{2}$ $h^{2} = r^{2} - \frac{1}{4 a^{2}}$ $\Rightarrow h = \sqrt{r^{2} - \frac{1}{4 a^{2}}} = r \sqrt{1 - \frac{1}{4 λ^{2}}}$ $\Rightarrow k = {a h}^{2} = a (r^{2} - \frac{1}{4 a^{2}}) = r (a r - \frac{1}{4 a r}) = r (λ - \frac{1}{4 λ})$ $\Rightarrow b = k + \frac{1}{2 a} = {a r}^{2} + \frac{1}{4 a} = r (λ + \frac{1}{4 λ})$ $\Rightarrow \tan θ = m = 2 a h = 2 a r \sqrt{1 - \frac{1}{4 λ^{2}}} = \sqrt{4 λ^{2} - 1}$ $A_{y e l l o w / 2} = \frac{2 h k}{3} + \frac{1}{2} \times r \cos θ \times r \sin θ - \frac{θ r^{2}}{2}$ $A_{y e l l o w / 2} = \frac{2 r^{2}}{3} (λ - \frac{1}{4 λ}) \sqrt{1 - \frac{1}{4 λ^{2}}} + \frac{r^{2} \sin θ \cos θ}{2} - \frac{θ r^{2}}{2}$ $A_{y e l l o w / 2} = A_{b l u e / 2} = \frac{π r^{2}}{2}$ $\frac{2 r^{2}}{3} (λ - \frac{1}{4 λ}) \sqrt{1 - \frac{1}{4 λ^{2}}} + \frac{r^{2} \sin θ \cos θ}{2} - \frac{θ r^{2}}{2} = \frac{π r^{2}}{2}$ $\frac{4}{3} (λ - \frac{1}{4 λ}) \sqrt{1 - \frac{1}{4 λ^{2}}} + \sin θ \cos θ - θ = π$ $\frac{4}{3} (λ - \frac{1}{4 λ}) \sqrt{1 - \frac{1}{4 λ^{2}}} + \frac{\tan θ}{1 + \tan^{2} θ} - θ = π$ $(\frac{4 λ}{3} - \frac{1}{3 λ}) \sqrt{1 - \frac{1}{4 λ^{2}}} + \frac{1}{2 λ} \sqrt{1 - \frac{1}{4 λ^{2}}} - \tan^{- 1} \sqrt{4 λ^{2} - 1} = π$ $\frac{1}{3} (4 λ + \frac{1}{2 λ}) \sqrt{1 - \frac{1}{4 λ^{2}}} - \tan^{- 1} \sqrt{4 λ^{2} - 1} = π$ $\frac{1}{3} (2 + \frac{1}{4 λ^{2}}) \sqrt{4 λ^{2} - 1} - \tan^{- 1} \sqrt{4 λ^{2} - 1} = π$ $\Rightarrow λ = 3.425$ $\Rightarrow a = 3.425$ $\Rightarrow b = 3.498$
	Commented by ajfour last updated on 04/Nov/18

	$T h a n k s e n o r m o u s l y!$
	Answered by ajfour last updated on 04/Nov/18

	Commented by ajfour last updated on 05/Nov/18

	$y = {A x}^{2}, h = R \sin θ$ $\frac{d y}{d x} ∣_{(h, k)} = 2 A h = \tan θ$ $\Rightarrow 2 A R \sin θ = \tan θ . . . . (i)$ $R^{2} θ + π R^{2} (y e l l o w) = R^{2} \sin θ \cos θ$ $+ \frac{4 {A h}^{3}}{3}$ $\Rightarrow θ + π = \sin θ \cos θ + \frac{2 \sin^{3} θ}{3 \cos θ}$ $θ + π = \frac{\tan θ}{3} (2 + \cos^{2} θ)$ $\Rightarrow θ = 1.42428 r a d$ $A = \frac{1}{2 \cos θ} = 3.425 .$
	Commented by MrW3 last updated on 05/Nov/18

	$n i c e a p p r o a c h, s h o r t a n d d i r e c t!$ $w h a t i s A i n 3 D c a s e, i . e . b l u e a n d$ $y e l l o w v o l u m n s a r e t h e s a m e ?$
	Commented by ajfour last updated on 09/Nov/18
	$z_P = Ar_P ^2 ; r_P = Rsin α (dz/dr) = tan α = 2ARsin α ⇒ 2ARcos α = 1 .....(I) z_P = AR^2 sin^2 α = (R/2)sin αtan α ..(II) yellow volume = blue volume 2π∫_0 ^( Rsin α) r{z_P +Rcos α−(√(R^2 −r^2 ))−Ar^2 )}dr = (4/3)πR^3 ⇒ [(z_P +Rcos α)(r^2 /2) + (((R^2 −r^2 )^(3/2) )/3) −((Ar^4 )/4)]_0 ^(Rsin α) = ((2R^3 )/3) (((sin^2 α)/(2cos α))+cos α)((sin^2 α)/2)−(((1−cos^3 α))/3) −((sin^4 α)/(8cos α)) = (2/3) dividing by cos^3 α , & with m= tan α (((m^2 +2)m^2 )/4)−(((1+m^2 )^(3/2) )/3)+(1/3) −(m^4 /8) = ((2(1+m^2 )^(3/2) )/3) ⇒ 3m^4 −24(1+m^2 )^(3/2) +12m^2 +8= 0 ⇒ m = 7.67754 A = (1/(2cos (tan^(−1) 7.67754))) ⇒ A= 3.8712 .$
	$z_{P} = {A r}_{P}^{2}; r_{P} = R \sin α$ $\frac{d z}{d r} = \tan α = 2 A R \sin α$ $\Rightarrow 2 A R \cos α = 1 . . . . . (I)$ $z_{P} = {A R}^{2} \sin^{2} α = \frac{R}{2} \sin α \tan α . . (I I)$ $y e l l o w v o l u m e = b l u e v o l u m e$ $2 π \int_{0}^{R \sin α} r {z_{P} + R \cos α - \sqrt{R^{2} - r^{2}} - {A r}^{2})} d r$ $= \frac{4}{3} π R^{3}$ $\Rightarrow [(z_{P} + R \cos α) \frac{r^{2}}{2} + \frac{{(R^{2} - r^{2})}^{3 / 2}}{3}$ ${- \frac{{A r}^{4}}{4}]}_{0}^{R \sin α} = \frac{2 R^{3}}{3}$ $(\frac{\sin^{2} α}{2 \cos α} + \cos α) \frac{\sin^{2} α}{2} - \frac{(1 - \cos^{3} α)}{3}$ $- \frac{\sin^{4} α}{8 \cos α} = \frac{2}{3}$ $d i v i d i n g b y \cos^{3} α, & w i t h$ $m = \tan α$ $\frac{(m^{2} + 2) m^{2}}{4} - \frac{{(1 + m^{2})}^{3 / 2}}{3} + \frac{1}{3}$ $- \frac{m^{4}}{8} = \frac{2 {(1 + m^{2})}^{3 / 2}}{3}$ $\Rightarrow 3 m^{4} - 24 {(1 + m^{2})}^{3 / 2} + 12 m^{2} + 8 = 0$ $\Rightarrow m = 7.67754$ $A = \frac{1}{2 \cos (\tan^{- 1} 7.67754)}$ $\Rightarrow A = 3.8712 .$
	Commented by ajfour last updated on 09/Nov/18

	$M r W S i r, p l e a s e c h e c k t h i s .$
	Commented by MrW3 last updated on 09/Nov/18

	$t h a n k y o u s i r!$ $p l e a s e c h e c k t h e e q n .$
	Answered by MJS last updated on 04/Nov/18

	$I turn it upside down and put$ $parabola : y = - {a x}^{2} + b; y^{'} = - 2 a x$ $circle : x^{2} + {(y + c)}^{2} = 1$ $y = - c + \sqrt{1 - x^{2}}$ $y^{'} = - \frac{x}{\sqrt{1 - x^{2}}}$ $zeros$ $x = \pm \sqrt{1 - c^{2}}$ $slope of tangents in zeros$ $s = \mp \frac{\sqrt{1 - c^{2}}}{c}$ $parabola : zeros at x = \pm \sqrt{1 - c^{2}}$ $- a (1 - c^{2}) + b = 0 \Rightarrow b = a (1 - c^{2})$ $slope = \mp \frac{\sqrt{1 - c^{2}}}{c} at x = \pm \sqrt{1 - c^{2}}$ $\frac{\sqrt{1 - c^{2}}}{c} = 2 a \sqrt{1 - c^{2}} \Rightarrow a = \frac{1}{2 c}; b = \frac{1 - c^{2}}{2 c}$ $parabola : y = - \frac{x^{2}}{2 c} + \frac{1 - c^{2}}{2 c}$ $- \frac{1}{2 c} \underset{- \sqrt{1 - c^{2}}}{\int^{\sqrt{1 - c^{2}}}} (x^{2} - 1 + c^{2}) d x + \underset{- \sqrt{1 - c^{2}}}{\int^{\sqrt{1 - c^{2}}}} (c - \sqrt{1 - x^{2}}) d x =$ $= \frac{(2 + c^{2}) \sqrt{1 - c^{2}}}{3 c} - \arcsin \sqrt{1 - c^{2}}$ $\frac{(2 + c^{2}) \sqrt{1 - c^{2}}}{3 c} - \arcsin \sqrt{1 - c^{2}} = π$ $\Rightarrow c = . 145986$ $\Rightarrow a = 3.42500$ $\Rightarrow b = 3.35200$
	Commented by ajfour last updated on 05/Nov/18

	$T h a n k y o u S i r, m a s t e r m e t h o d!$
	Commented by MJS last updated on 09/Nov/18

	$R^{3}$ $parabola : x^{2} = - 2 c y - c^{2} + 1$ $V_{p} = π \underset{0}{\int^{\frac{1 - c^{2}}{2 c}}} (- 2 c y - c^{2} + 1) d y = π \frac{c^{4} - 2 c^{2} + 1}{4 c}$ $circle : x^{2} = 1 - {(y + c)}^{2}$ $V_{c} = π \underset{0}{\int^{1 - c}} (1 - {(y + c)}^{2}) d y = π \frac{c^{3} - 3 c + 2}{3}$ $V = V_{p} - V_{c} = \frac{4 π}{3}$ $\frac{c^{4} - 2 c^{2} + 1}{4 c} - \frac{c^{3} - 3 c + 2}{3} - \frac{4}{3} = 0$ $c^{4} - 6 c^{2} + 24 c - 3 = 0$ $this can be solved exactly using$ $(c - α - i \sqrt{β}) (c - α + i \sqrt{β}) (c - γ - \sqrt{δ}) (c - γ + \sqrt{δ}) = 0$ $\Rightarrow α = - \sqrt{3}; β = - 2 \sqrt{3}; γ = \sqrt{3}; δ = - 2 \sqrt{3}$ $c_{1} = - \sqrt{3} - \sqrt{2 \sqrt{3}} < 0$ $c_{2} = - \sqrt{3} + \sqrt{2 \sqrt{3}} > 0 \Rightarrow c = c_{2} \approx . 129159$ $c_{3} = \sqrt{3} - i \sqrt{2 \sqrt{3}}$ $c_{4} = \sqrt{3} + i \sqrt{2 \sqrt{3}}$ $a = 1 + \frac{\sqrt{3}}{2} + (\frac{\sqrt{3}}{3} + \frac{1}{2}) \sqrt{2 \sqrt{3}} \approx 3.87120$ $b = 1 + (1 + \frac{\sqrt{2 \sqrt{3}}}{3}) \sqrt{3} \approx 3.80662$ $c = \sqrt{2 \sqrt{3}} - \sqrt{3} \approx . 129159$ $please check my method$
	Commented by MrW3 last updated on 09/Nov/18

	$g r e a t s i r!$ $w e g e t t h e s a m e e q n . a n d r e s u l t .$
	Answered by MrW3 last updated on 09/Nov/18

	Commented by MrW3 last updated on 09/Nov/18

	$y = {a x}^{2}$ $y^{'} = 2 a x$ $m = \tan θ = 2 a h$ $\Rightarrow h = \frac{\tan θ}{2 a} = R \sin θ$ $\Rightarrow a = \frac{1}{2 R \cos θ}$ $k = {a h}^{2} = \frac{R^{2} \sin^{2} θ}{2 R \cos θ} = \frac{R \sin θ \tan θ}{2}$ $V_{y e l l o w} = π h^{2} k - π R^{2} {(1 - \cos θ)}^{2} [R - \frac{1 - \cos θ}{3} R] - 2 π \int_{0}^{h} {a x}^{3} d x$ $V_{y e l l o w} = π h^{2} k - π R^{3} {(1 - \cos θ)}^{2} (\frac{2 + \cos θ}{3}) - \frac{π {a h}^{4}}{2}$ $V_{y e l l o w} = \frac{π R^{3} \sin^{3} θ \tan θ}{2} - π R^{3} {(1 - \cos θ)}^{2} (\frac{2 + \cos θ}{3}) - \frac{π R^{3} \sin^{4} θ}{4 \cos θ} = \frac{4 π R^{3}}{3}$ $\frac{\sin^{3} θ \tan θ}{2} - {(1 - \cos θ)}^{2} (\frac{2 + \cos θ}{3}) - \frac{\sin^{4} θ}{4 \cos θ} = \frac{4}{3}$ $3 \sin^{4} θ - 4 {(1 - \cos θ)}^{2} (2 + \cos θ) \cos θ = 16 \cos θ$ $3 \sin^{4} θ - 24 \cos θ + 12 \cos^{2} θ - {4 \cos}^{4} θ = 0$ $\Rightarrow \cos^{4} θ - 6 \cos^{2} θ + 24 \cos θ - 3 = 0$ $\Rightarrow \cos θ = 0.1292 \Rightarrow θ = 82.576 °$ $\Rightarrow a = \frac{1}{2 \times 0.1292} = 3.87$
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