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For \( \bar{F}=x^{2} \hat{\imath}+x y \hat{\jmath} \), the value of \( \int_{C} \bar{F} \cdot d \bar{r} \) for the curve \( y=x \) joining the points \( (0,0) \) and \( (1,1) \) is |
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Answer» \(\int_c \vec F. d \vec r =\int_c (x^2 \hat i + xy\hat j) (dx\,\hat i + dy \hat j)\) = \(\int _c (x^2dx+xydy)\) = \(\int_cx^2dx+\int_cy^2dy\) (∵ y = x) = \(\int\limits_0^1 x^2dx + \int\limits_0^1y^2dy\) = \((\frac {x^3}{3})_0^1 + (\frac {y^3}{3})_0^1\) = \(\frac {1}{3} + \frac {1}{3}= \frac 23\) |
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