∫_c ((x^2 +2xy^2 )dx+(x^2 y^2 −1)dy) where C is the boundary of region define by y^2 = 4x and y =1 ?


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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
	$note that C is a closed curve . observe that y^2 =4x intersects x=1 when y = ± 2 Green′s Theorem yields ∫_c ((x^2 +2xy^2 )dx+(x^2 y^2 −1)dy)= ∫_(−2) ^2 ∫_(y^2 /4) ^1 (2xy^2 −4xy) dxdy =∫_(−2) ^2 {(x^2 y^2 −2x^2 y)}∣_(y^2 /4) ^1 dy = 2∫_0 ^2 (y^2 −(y^6 /(16))) dy [ even function] = [(2/3)y^3 −(y^7 /(112)) ]_9 ^2 = ((16)/3) −((128)/(112)) =((16)/3)−(8/7) = ((88)/(21))$
	$n o t e t h a t C i s a c l o s e d c u r v e .$ $o b s e r v e t h a t y^{2} = 4 x i n t e r s e c t s$ $x = 1 w h e n y = \pm 2$ ${G r e e n}^{'} s T h e o r e m y i e l d s$ $\int_{c} ((x^{2} + 2 {x y}^{2}) d x + (x^{2} y^{2} - 1) d y) =$ $\underset{- 2}{\int^{2}} \underset{y^{2} / 4}{\int^{1}} (2 {x y}^{2} - 4 x y) d x d y$ $= \underset{- 2}{\int^{2}} {(x^{2} y^{2} - 2 x^{2} y)} ∣_{y^{2} / 4}^{1} d y$ $= 2 \underset{0}{\int^{2}} (y^{2} - \frac{y^{6}}{16}) d y [e v e n$ $f u n c t i o n]$ $= {[\frac{2}{3} y^{3} - \frac{y^{7}}{112}]}_{9}^{2} = \frac{16}{3} - \frac{128}{112}$ $= \frac{16}{3} - \frac{8}{7} = \frac{88}{21}$
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