Answer
(a) $\int_C \textbf F \cdot \textbf{dr}_1=\frac{236}{5}$
(b) $\int_C \textbf F \cdot \textbf{dr}_2=-\frac{236}{5}$
The answer in part (b) is the negative of that in part (a) because the paths in the part (a) and (b) represent opposite orientations of the same curve segment.
Work Step by Step
$\textbf F(x,x) = x^2x\textbf i + xy^{3/2} \textbf j$
(a) $\textbf r_1(t) = (1+t)\textbf i +t^2 \textbf j, 0\le t \le 2\\
\therefore x(t) = (1+t), y(t) =t^2\\
\textbf r_1' = \textbf i + 2t \textbf j\\
\textbf F(x(t),y(t)) = (1+t)^2t^2 \textbf i + (1+t)t^3\textbf j\\
\therefore \int_C \textbf F \cdot d\textbf r_1 = \int_{t=0}^2 \textbf F \cdot \textbf r_1' dt\\
= \int_0^1 [(1+t)^2t^2 \textbf i + (1+t)t^3\textbf j]\cdot (\textbf i + 2t \textbf j)dt\\
=\int_0^2(t^2+2t^3+3t^4+2t^5)dt \\
= [\frac{t^3}{3}+\frac{2t^4}{4}+\frac{3t^5}{5}+\frac{2t^6}{6}]_0^2 = \frac{236}{5}$
(b) $\textbf r_2(t) = (1+2\cos t)\textbf i +(4\cos^2t)\textbf j, 0\le t \le \pi/2\\
\therefore x(t) = (1+2\cos t), y(t) = (4\cos^2t)\\
\textbf r_2' = -2\sin t \textbf i -8\cos t. \sin t \textbf j\\
\textbf F(x(t),y(t)) = (1+2\cos t )^2.4\cos^2t \textbf i +(1+2\cos t).8\cos^3t \textbf j\\
\therefore \int_C \textbf F \cdot d\textbf r_2 = \int_{t=0}^{\pi/2} \textbf F \cdot \textbf r_2' dt\\
= \int_{0}^{\pi/2} [(1+2\cos t )^2.4\cos^2t \textbf i +(1+2\cos t).8\cos^3t \textbf j]\cdot [-2\sin t \textbf i -8\cos t. \sin t \textbf j] dt\\
= \int_{0}^{\pi/2} (-128\cos^5t.\sin t-96\cos^4t.\sin t - 32\cos^3t\sin t - 8\cos^2t.\sin t) dt\\
= [\frac{128}{6}\cos^6t + \frac{96}{5}\cos^5 t + \frac{32}{4}\cos^4t+\frac{8}{3}\cos^3t]_0^{\pi/2} = -\frac{236}{5}$
The answer in part (b) is the negative of that in part (a) because the paths in the part (a) and (b) represent opposite orientations of the same curve segment.
