Evaluate-r-0-x-r-x-dx-

Question Number 27400 by Tinkutara last updated on 06/Jan/18

E v a l u a t e \underset{r}{\int^{0}} \sqrt{\frac{x}{r - x}} d x

Commented by Tinkutara last updated on 07/Jan/18

B u t t h i s i n t e g r a l w a s u s e d i n a P h y s i c s

q u e s t i o n w h e r e i t s v a l u e {c a n}^{'} t b e

n e g a t i v e . I t s h o u l d b e o n l y \frac{π r}{2} . I t i s

u s e d i n q u e s t i o n 27445 .

Commented by abdo imad last updated on 06/Jan/18

l e t d o t h e c h . \sqrt{\frac{x}{r - x}} = t \Leftrightarrow \frac{x}{r - x} = t^{2} \Leftrightarrow x = (r - x) t^{2}

\Leftrightarrow x + {x t}^{2} = {r t}^{2} \Leftrightarrow (1 + t^{2}) x = {r t}^{2} \Leftrightarrow x = \frac{{r t}^{2}}{1 + t^{2}}

x = \frac{r (1 + t^{2} - 1)}{1 + t^{2}} = r - \frac{r}{1 + t^{2}} \Leftrightarrow \frac{d x}{d t} = \frac{2 r t}{{(1 + t^{2})}^{2}}

\int_{r}^{0} \sqrt{\frac{x}{r - x}} d x = \int_{\propto}^{0} t \frac{2 r t}{{(1 + t^{2})}^{2}} d t = - 2 r \int_{0}^{\infty} \frac{t^{2}}{{(1 + t^{2})}^{2}} d t

b u t \int_{0}^{\infty} \frac{t^{2}}{{(1 + t^{2})}^{2}} d t = \int_{0}^{\infty} \frac{d t}{1 + t^{2}} - \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{2}}

= \frac{π}{2} - \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{2}} a n d b y r e s i d u s t h e o r e m

\int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{2}} = \frac{1}{2} \int_{R} \frac{d t}{{(1 + t^{2})}^{2}} = \frac{1}{2} 2 i π R e s (f, i) w i t h

f (z) = \frac{1}{{(1 + z^{2})}^{2}} = \frac{1}{{(z - i)}^{2} {(z + i)}^{2}} a n d i i s a d o u b l e p o l e o f f

R e s (f, i) = {l i m}_{z - > i} \frac{1}{(2 - 1)!} {({(z - i)}^{2} f (z))}^{,}

= {l i m}_{z - > i} {({(z + i)}^{- 2})}^{,} = {l i m}_{z - > i} - 2 {(z + i)}^{- 3}

= - 2 {(2 i)}^{- 3} = - 2 \frac{1}{{(2 i)}^{3}} = \frac{1}{4 i}

\int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{2}} = i π \frac{1}{4 i} = \frac{π}{4}

\int_{r}^{0} \sqrt{\frac{x}{r - x}} d x = - 2 r (\frac{π}{2} - \frac{π}{4}) = - 2 r \frac{π}{4} = \frac{- r π}{2} .

Commented by mrW1 last updated on 07/Jan/18

t h e r e s u l t a b o v e i s c o r r e c t .

\int_{r}^{0} \sqrt{\dots} d x i s n e g a t i v e

\int_{0}^{r} \sqrt{\dots} d x i s p o s i t i v e

Commented by abdo imad last updated on 07/Jan/18

w e m u s t h a v e 0 < x < r b e c a u s e o f \sqrt{\dots} \Rightarrow \int_{r}^{o} (\dots) d x = - \int_{0}^{r} (\dots) ⩽ 0

Answered by prakash jain last updated on 07/Jan/18

x = r \cos^{2} θ

d x = - 2 r \cos θ \sin θ d θ

\int - 2 r \cos θ \sin θ . \frac{\cos θ}{\sin θ} d θ

= - r \int {2 \cos}^{2} θ d θ

= - r \int (1 + \cos 2 θ) d θ

= - r [θ + \frac{\sin 2 θ}{2}] + C

= - r [\cos^{- 1} \sqrt{\frac{x}{r}} + \sqrt{\frac{x}{r} \times \frac{r - x}{r}}] + C

= - r [\cos^{- 1} \sqrt{\frac{x}{r}} + \frac{1}{r} \sqrt{x (r - x)}] + C

t a k i n g l i m i t s

= - r [\frac{π}{2} - 0] = - \frac{π r}{2}

Commented by mrW1 last updated on 07/Jan/18

v e r y n i c e w a y!

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