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consisting of straight lines from (0,0) to (2,0) and then to (3,2)

To evaluate the line integral of the vector field \( \mathbf{A} = (2x+y)\mathbf{i} + (3y-x)\mathbf{j} \) along the curve \( C \), which consists of two straight segments from \( (0,0) \) to \( (2,0) \) and then to \( (3,2) \), we will break the problem down into manageable parts. This involves calculating the integral over each segment of the curve and then summing the results.

Defining the Curve Segments

The curve \( C \) can be divided into two segments:

• Segment 1: From \( (0,0) \) to \( (2,0) \)
• Segment 2: From \( (2,0) \) to \( (3,2) \)

Segment 1: From (0,0) to (2,0)

For the first segment, we can parameterize the line as follows:

\( \mathbf{r}_1(t) = (t, 0) \) where \( t \) ranges from \( 0 \) to \( 2 \).

Next, we compute \( d\mathbf{r} \):

\( d\mathbf{r} = (dx, dy) = (dt, 0) \).

Now, substituting \( x = t \) and \( y = 0 \) into the vector field \( \mathbf{A} \):

\( \mathbf{A} = (2t + 0)\mathbf{i} + (3(0) - t)\mathbf{j} = 2t\mathbf{i} - t\mathbf{j} \).

Now, we can evaluate the integral over this segment:

\( \int_{C_1} \mathbf{A} \cdot d\mathbf{r} = \int_0^2 (2t\mathbf{i} - t\mathbf{j}) \cdot (dt, 0) = \int_0^2 2t \, dt \).

Calculating this integral:

\( \int_0^2 2t \, dt = [t^2]_0^2 = 2^2 - 0^2 = 4 \).

Segment 2: From (2,0) to (3,2)

For the second segment, we can parameterize the line as:

\( \mathbf{r}_2(t) = (2 + t, 2t) \) where \( t \) ranges from \( 0 \) to \( 1 \).

Here, \( d\mathbf{r} = (dx, dy) = (dt, 2dt) \).

Substituting \( x = 2 + t \) and \( y = 2t \) into the vector field \( \mathbf{A} \):

\( \mathbf{A} = (2(2+t) + 2t)\mathbf{i} + (3(2t) - (2+t))\mathbf{j} = (4 + 2t + 2t)\mathbf{i} + (6t - 2 - t)\mathbf{j} = (4 + 4t)\mathbf{i} + (5t - 2)\mathbf{j} \).

Now we evaluate the integral over this segment:

\( \int_{C_2} \mathbf{A} \cdot d\mathbf{r} = \int_0^1 ((4 + 4t)\mathbf{i} + (5t - 2)\mathbf{j}) \cdot (dt, 2dt) \).

This simplifies to:

\( \int_0^1 ((4 + 4t)dt + 2(5t - 2)dt) = \int_0^1 (4 + 4t + 10t - 4)dt = \int_0^1 (14t)dt \).

Calculating this integral:

\( \int_0^1 14t \, dt = [7t^2]_0^1 = 7(1^2) - 7(0^2) = 7 \).

Combining the Results

Now, we can sum the results from both segments:

\( \int_C \mathbf{A} \cdot d\mathbf{r} = \int_{C_1} \mathbf{A} \cdot d\mathbf{r} + \int_{C_2} \mathbf{A} \cdot d\mathbf{r} = 4 + 7 = 11 \).

Final Answer

The value of the line integral \( \int_C \mathbf{A} \cdot d\mathbf{r} \) is \( 11 \).