Discussion
The value of the line integral \(I = \int \vec{A}.\vec{dr}\) where \(\vec{r} = x\hat{i} + y\hat{j}\) and \(\vec{A} = (x+y)\hat{i} + (y-x)\hat{j}\) along the path \(y^{2} = x\) from the point (1, 1) to the point (4, 2) is :
Answer: B
The line integral is 11.33.
On the path \(x = y^{2}\), so \(dx = 2y\,dy\), and \(y\) runs from 1 to 2. The integrand \(\vec{A}\cdot d\vec{r} = (x+y)\,dx + (y-x)\,dy\) becomes a function of \(y\) alone:
\[ \begin{aligned} I &= \int_{1}^{2} [(y^{2}+y)2y + (y - y^{2})]\,dy \\ &= \int_{1}^{2} (2y^{3} + y^{2} + y)\,dy \\ &= \left[\tfrac{y^{4}}{2} + \tfrac{y^{3}}{3} + \tfrac{y^{2}}{2}\right]_{1}^{2} \\ &= \tfrac{38}{3} - \tfrac{4}{3} = \tfrac{34}{3} \approx 11.33 \end{aligned} \]
The field is not conservative (\(\partial A_y/\partial x = -1\) while \(\partial A_x/\partial y = 1\)), so the value depends on the path and must be computed along the parabola. The other options differ from 34/3 by whole units and do not come out of this integral.
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