Question

for the vector field ( mathbf{f}=(y e^{x y}+3 cos (3 x+y)) mathbf{i}+(x e^{x y}+cos (3 x+y)) mathbf{j} ), evaluate ( int_{c} mathbf{f} cdot d mathbf{r} ), where ( c ) is the path from the origin along the ( x )-axis to the point ( (2,0) ) and then counterclockwise around the perimeter of the circle ( x^{2}+y^{2}=4 ) to the point ( (2 / sqrt{2}, 2 / sqrt{2}) ).
( int_{c} mathbf{f} cdot d mathbf{r}= )

Explanation:

Step1: Check if F is conservative

Let \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} = (ye^{xy} + 3\cos(3x+y))\mathbf{i} + (xe^{xy} + \cos(3x+y))\mathbf{j} \). Compute \( \frac{\partial P}{\partial y} = e^{xy} + xye^{xy} - 3\sin(3x+y) \), \( \frac{\partial Q}{\partial x} = e^{xy} + xye^{xy} - 3\sin(3x+y) \). Since \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), F is conservative.

Step2: Find potential function f

Integrate \( P \) with respect to x: \( f(x,y) = \int (ye^{xy} + 3\cos(3x+y))dx = e^{xy} + \sin(3x+y) + g(y) \). Differentiate f with respect to y: \( f_y = xe^{xy} + \cos(3x+y) + g'(y) = Q \), so \( g'(y)=0 \Rightarrow g(y)=C \). Take \( C=0 \), so \( f(x,y)=e^{xy} + \sin(3x+y) \).

Step3: Apply Fundamental Theorem of Line Integrals

The path C goes from origin (0,0) to \( (2/\sqrt{2}, 2/\sqrt{2})=(√2,√2) \). So \( \int_C \mathbf{F}·d\mathbf{r} = f(√2,√2) - f(0,0) = [e^{(√2)(√2)} + \sin(3√2 + √2)] - [e^{0} + \sin(0)] = e^2 + \sin(4√2) - 1 - 0 = e^2 + \sin(4√2) - 1 \).

Answer:

\( e^2 + \sin(4\sqrt{2}) - 1 \)