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Question Number 103607 by bemath last updated on 16/Jul/20

w h a t i s t h e v a l u e o f \int_{c} (x + 2 y) d x + (4 - 2 x) d y

a r o u n d t h e e l l i p s e C : \frac{x^{2}}{16} + \frac{y^{2}}{8} = 1

i n t h e c o u n t e r c l o c k w i s e

d i r e c t i o n ?

Answered by bobhans last updated on 16/Jul/20

s i n c e t h e e l l i p s e i s a c l o s e d c u r v e, w i t h

{G r e e n}^{'} s T h e o r e m .

\int_{C} ((x + 2 y) d x + (4 - 2 x) d y) =

\int \int_{R} (\frac{\partial}{\partial x} (4 - 2 x) - \frac{\partial}{\partial y} (x + 2 y)) d A

= \int \int_{R} (- 2 - 2) d A = - 4 \int \int_{R} d A

b e c a u s e t h e a r e a o f a n e l l i p s e w i t h

e q u a t i o n \frac{x^{2}}{16} + \frac{y^{2}}{8} = 1 e q u a l t o π . 4 . \sqrt{8} = 8 π \sqrt{2}

w e g e t - 4 \int \int_{R} = - 4 \times 8 π \sqrt{2} = - 32 π \sqrt{2}

w e c o n c l u d e

\int_{C} ((x + 2 y) d x + (4 - 2 x) d y) = - 32 π \sqrt{2}

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