The value of $\int_{c} [(3 x - 8 y^{2}) d x + (4 y - 6 x y) d y]$ , where C is the boundary of the region bounded by x = 0, y = 0 and x + y = 1 is_____
1.67

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Solution

The correct option is A 1.67

$∵$ $\oint_{c} [(3 x - 8 y^{2}) . d x + (4 y - 6 x y) . d y]$
Applying Green theorem,
$\oint_{c} F_{1} d x + F_{2} d y = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) . d x . d y$
$H e r e F_{1} = 3 x - 8 y^{2}$
$F_{2} = 4 y - 6 x y$
$S o, \oint_{c} [(3 x - 8 y^{2}) . d x + (4 y - 6 x y) d y)]$
$= \iint_{R} [(\frac{\partial}{\partial x} (4 x - 6 x y)) - (\frac{\partial}{\partial y} (3 x - 8 y^{2}))] . d x d y$
$= \iint_{R} [- 6 y + 16 y] . d x . d y$
$= \iint_{R} (10 y) . d x . d y$
$= \int_{0}^{1} \int_{0}^{1 - x} (10 y) d y d x$
$= \int_{0}^{1} {[5 y^{2}]}_{0}^{2} . d x = 5 \int_{0}^{1} (1 - x)^{2} d x$
$= 5 {[\frac{(x - 1)^{3}}{3}]}_{0}^{1} = \frac{5}{3} [0 - (- 1)] = \frac{5}{3} = 1.67$

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