Math Problem Statement

Evaluate, ∫F.dr where F = x²i + xyj and C is the boundary of square in the plane z=0 and bounded by the lines x=0, y=0, x=a & y=a

Solution

To evaluate the line integral CFdr\int_C \mathbf{F} \cdot d\mathbf{r}, where F=x2i^+xyj^\mathbf{F} = x^2 \hat{i} + xy \hat{j} and CC is the boundary of a square in the plane z=0z = 0 bounded by x=0x = 0, y=0y = 0, x=ax = a, and y=ay = a, we proceed as follows:

Step 1: Parameterize the boundary curve CC

The boundary of the square consists of four straight line segments. We will break the boundary curve into four parts and parameterize each one.

  1. Segment 1 (from (0,0)(0, 0) to (a,0)(a, 0)): x=t,y=0,wheret[0,a]x = t, \quad y = 0, \quad \text{where} \quad t \in [0, a] The vector dr=(dx,dy)=(dt,0)d\mathbf{r} = (dx, dy) = (dt, 0).

  2. Segment 2 (from (a,0)(a, 0) to (a,a)(a, a)): x=a,y=t,wheret[0,a]x = a, \quad y = t, \quad \text{where} \quad t \in [0, a] The vector dr=(dx,dy)=(0,dt)d\mathbf{r} = (dx, dy) = (0, dt).

  3. Segment 3 (from (a,a)(a, a) to (0,a)(0, a)): x=t,y=a,wheret[a,0]x = t, \quad y = a, \quad \text{where} \quad t \in [a, 0] The vector dr=(dx,dy)=(dt,0)d\mathbf{r} = (dx, dy) = (-dt, 0).

  4. Segment 4 (from (0,a)(0, a) to (0,0)(0, 0)): x=0,y=t,wheret[a,0]x = 0, \quad y = t, \quad \text{where} \quad t \in [a, 0] The vector dr=(dx,dy)=(0,dt)d\mathbf{r} = (dx, dy) = (0, -dt).

Step 2: Compute the line integral for each segment

Segment 1: x=tx = t, y=0y = 0

F=(x2i^+xyj^)=(t2i^+0j^)=t2i^\mathbf{F} = (x^2 \hat{i} + xy \hat{j}) = (t^2 \hat{i} + 0 \hat{j}) = t^2 \hat{i} The line integral over this segment is: C1Fdr=0a(t2i^)(dti^)=0at2dt=t330a=a33\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^a (t^2 \hat{i}) \cdot (dt \hat{i}) = \int_0^a t^2 \, dt = \frac{t^3}{3} \Big|_0^a = \frac{a^3}{3}

Segment 2: x=ax = a, y=ty = t

F=(x2i^+xyj^)=(a2i^+atj^)\mathbf{F} = (x^2 \hat{i} + xy \hat{j}) = (a^2 \hat{i} + at \hat{j}) The line integral over this segment is: C2Fdr=0a(a2i^+atj^)(0i^+dtj^)=0aat2dt=at330a=a43\int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_0^a (a^2 \hat{i} + at \hat{j}) \cdot (0 \hat{i} + dt \hat{j}) = \int_0^a at^2 \, dt = \frac{at^3}{3} \Big|_0^a = \frac{a^4}{3}

Segment 3: x=tx = t, y=ay = a

F=(x2i^+xyj^)=(t2i^+atj^)\mathbf{F} = (x^2 \hat{i} + xy \hat{j}) = (t^2 \hat{i} + at \hat{j}) The line integral over this segment is: C3Fdr=a0(t2i^+atj^)(dti^+0j^)=a0t2dt=t33a0=a33\int_{C_3} \mathbf{F} \cdot d\mathbf{r} = \int_a^0 (t^2 \hat{i} + at \hat{j}) \cdot (-dt \hat{i} + 0 \hat{j}) = \int_a^0 -t^2 \, dt = \frac{t^3}{3} \Big|_a^0 = -\frac{a^3}{3}

Segment 4: x=0x = 0, y=ty = t

F=(x2i^+xyj^)=(0i^+0j^)=0\mathbf{F} = (x^2 \hat{i} + xy \hat{j}) = (0 \hat{i} + 0 \hat{j}) = 0 The line integral over this segment is: C4Fdr=a00dt=0\int_{C_4} \mathbf{F} \cdot d\mathbf{r} = \int_a^0 0 \, dt = 0

Step 3: Add the contributions from all segments

Now we add up the integrals from all four segments: CFdr=a33+a43a33+0=a43\int_C \mathbf{F} \cdot d\mathbf{r} = \frac{a^3}{3} + \frac{a^4}{3} - \frac{a^3}{3} + 0 = \frac{a^4}{3}

Final Answer:

CFdr=a43\int_C \mathbf{F} \cdot d\mathbf{r} = \frac{a^4}{3}


Would you like any more details or clarifications on the steps?

Here are some related questions to explore further:

  1. What is the line integral when the vector field is conservative?
  2. How do you apply Green's Theorem to this problem?
  3. Can you find the work done by a force field along a different path?
  4. What happens if the boundary is a circle instead of a square?
  5. How does the integral change if the square is rotated?

Tip: When solving line integrals, always pay attention to the direction of traversal along the boundary—incorrect parametrization can lead to sign errors.

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Math Problem Analysis

Mathematical Concepts

Line Integrals
Vector Fields
Parameterization of Curves
Boundary Conditions

Formulas

∫F.dr = ∫ (F·dr)
Line integral formula for vector fields
Parameterization of curve segments

Theorems

Green's Theorem (as a related concept for line integrals)
Fundamental Theorem of Line Integrals

Suitable Grade Level

Grades 11-12 (Advanced Calculus/Vector Calculus)