Question

Find the value of the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}$$ (Hint: If \(\mathbf{F}\) is conservative, the integration may be easier on an alternative path.) \(\int_{C}\left(x^{2}+y^{2}\right) d x+2 x y d y\) (a) \(\mathbf{r}_{1}(t)=t^{3} \mathbf{i}+t^{2} \mathbf{j}, \quad 0 \leq t \leq 2\) (b) \(\mathbf{r}_{2}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}, \quad 0 \leq t \leq \frac{\pi}{2}\)

Short answer

The values of the line integrals for the given paths \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \) have to be computed individually. After the respective computation, the results for cases (a) and (b) can be obtained.

Step-by-step solution

- Identify whether the vector field is conservative

The given vector field \( \mathbf{F} = (x^{2}+y^{2})\mathbf{i} + 2xy\mathbf{j} \). A 2D vector field \( F = M(x,y)\mathbf{i} + N(x,y)\mathbf{j} \) is conservative if and only if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). For our vector field, \( \frac{\partial M}{\partial y} = 2y \) and \( \frac{\partial N}{\partial x} = 2y \), so the field is indeed conservative.

- Compute the line integral over the path prescribed by \( \mathbf{r}_1(t) \)

Make substitutions according to \( \mathbf{r_1}(t) \): \( x = t^3, y = t^2 \). Also, \( dx = 3t^2 dt, dy = 2t dt \). Substitute these values into \( \mathbf{F} \cdot d\mathbf{r} = (x^{2}+y^{2}) dx + 2xy dy \) which becomes \( (t^{6} + t^{4})(3t^2 dt) + 2(t^3)(t^2)(2t dt) \). Simplify and then evaluate the integral from t=0 to 2.

- Compute the line integral over the path prescribed by \( \mathbf{r}_2(t) \)

Make substitutions according to \( \mathbf{r}_2(t) \): \( x = 2\cos(t), y = 2\sin(t) \). Also, \( dx = -2\sin(t) dt, dy = 2\cos(t) dt \). Substitute these values into \( \mathbf{F} \cdot d\mathbf{r} = (x^{2}+y^{2}) dx + 2xy dy \) which becomes \( ((2\cos(t))^{2} + (2\sin(t))^{2})(-2\sin(t) dt) + 2(2\cos(t))(2\sin(t))(2\cos(t) dt) \). Simplify and then evaluate the integral from t=0 to \( \frac{\pi}{2} \).