Evaluate
\int_C \mathbf{F} \cdot \mathrm{d} \mathbf{s}where \mathbf{F}=5 y^2 \mathbf{i}+2 x y \mathbf{j} and C is the straight line joining the origin and the point (1, 1).
We compare the integrand with the standard form and recognize that in this case P(x, y)=5 y^2 \text { and } Q(x, y)=2 x y \text {. } The integral becomes
\int_C 5 y^2 \mathrm{~d} x+2 x y \mathrm{~d} yThe curve of interest, in this case a straight line, is shown in Figure 27.5. Along this curve it is clear that y = x at all points. We use this fact to simplify the integral by writing everything in terms of x. We could equally well choose to write everything in terms of. If y=x \text { then } \frac{\mathrm{d} y}{\mathrm{~d} x}=1 \text {, that is } \mathrm{d} y=\mathrm{d} x \text {. } As we move along the curve C, x ranges from 0 to 1, and the integral reduces to
\int_C 5 x^2 \mathrm{~d} x+2 x^2 \mathrm{~d} x=\int_C 7 x^2 \mathrm{~d} x=\int_{x=0}^{x=1} 7 x^2 \mathrm{~d} x=\left[\frac{7 x^3}{3}\right]_0^1=\frac{7}{3}