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Question Number 103773 by bemath last updated on 17/Jul/20

\int_{c} ((x^{2} + 2 {x y}^{2}) d x + (x^{2} y^{2} - 1) d y)

w h e r e C i s t h e b o u n d a r y o f

r e g i o n d e f i n e b y y^{2} = 4 x a n d y

= 1 ?

Answered by bramlex last updated on 17/Jul/20

n o t e t h a t C i s a c l o s e d c u r v e .

o b s e r v e t h a t y^{2} = 4 x i n t e r s e c t s

x = 1 w h e n y = \pm 2

{G r e e n}^{'} s T h e o r e m y i e l d s

\int_{c} ((x^{2} + 2 {x y}^{2}) d x + (x^{2} y^{2} - 1) d y) =

\underset{- 2}{\int^{2}} \underset{y^{2} / 4}{\int^{1}} (2 {x y}^{2} - 4 x y) d x d y

= \underset{- 2}{\int^{2}} {(x^{2} y^{2} - 2 x^{2} y)} ∣_{y^{2} / 4}^{1} d y

= 2 \underset{0}{\int^{2}} (y^{2} - \frac{y^{6}}{16}) d y [e v e n

f u n c t i o n]

= {[\frac{2}{3} y^{3} - \frac{y^{7}}{112}]}_{9}^{2} = \frac{16}{3} - \frac{128}{112}

= \frac{16}{3} - \frac{8}{7} = \frac{88}{21}

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