Problem 39 - Calculus Early Transcendentals

Question

Evaluate the integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(C\) is the boundary of the region \(R\) and \(C\) is oriented so that the region is on the left when the boundary is traversed in the direction of its orientation. \(\mathbf{F}(x, y)=\left(x^{2}+y\right) \mathbf{i}+(4 x-\cos y) \mathbf{j} ; C\) is the boundary of the region \(R\) that is inside the square with vertices \((0,0)\) \((5,0),(5,5),(0,5)\) but is outside the rectangle with vertices \((1,1),(3,1),(3,2),(1,2)\)

Step-by-Step Solution

Answer

The value of the integral is 69.

Step 1: Verify Closed Curve Condition

The path of integration, \(C\), is composed of the outer square boundary minus the inner rectangle boundary. This creates a closed curve since both the outer square and inner rectangle are closed curves themselves.

Step 2: Verify That \(\mathbf{F}\) Is Continuously Differentiable

The vector field \(\mathbf{F}(x, y) = \left(x^2 + y\right) \mathbf{i} + (4x - \cos y) \mathbf{j}\) is continuously differentiable. This is evident because both component functions \(x^2 + y\) and \(4x - \cos y\) have continuous derivatives everywhere.

Step 3: Applying Green's Theorem

Green's Theorem states that \(\int_{C} \mathbf{F} \cdot d \mathbf{r} = \int\int_{R} \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \, dA\), where \(\mathbf{F} = M \mathbf{i} + N \mathbf{j}\). Substitute \(M = x^2 + y\) and \(N = 4x - \cos y\).

Step 4: Calculate Partial Derivatives

Compute \(\frac{\partial N}{\partial x} = 4\) and \(\frac{\partial M}{\partial y} = 1\). Substitute these into Green's theorem: \[\int\int_{R} (4 - 1) \, dA = \int\int_{R} 3 \, dA.\]

Step 5: Calculate the Area of Region \(R\) Using Subtraction

The area of the large square is \(5 \times 5 = 25\). The area of the inner rectangle is \(2 \times 1 = 2\). Thus, the area of the region \(R\) is \(25 - 2 = 23\).

Step 6: Evaluate the Double Integral

Substitute the area into the double integral: \[\int\int_{R} 3 \, dA = 3 \times 23 = 69.\] Therefore, the value of the line integral is 69.

Key Concepts

Vector Field

A vector field is essentially a function that assigns a vector to every point in a space. In the context of the problem, we have the vector field \( \mathbf{F}(x, y) = (x^2 + y) \mathbf{i} + (4x - \cos y) \mathbf{j} \). This means that at every point \((x, y)\) in the plane, \( \mathbf{F} \) assigns a vector with components based on \(x\) and \(y\).
These components, \(x^2 + y\) and \(4x - \cos y\), are vector-valued functions. They respectively describe how the vector field behaves in the horizontal and vertical directions throughout the plane.
These vectors can be visualized as arrows pointing in directions determined by their components, varying from point to point. A vector field can often be used to describe physical phenomena such as fluid flow or magnetic fields. Understanding how these vectors change across the space is vital for applying concepts like Green's Theorem in calculus.

Line Integral

A line integral is a method used to compute the cumulative effect of a field along a path or curve. We are evaluating the integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\), where \(C\) is a path or curve in the vector field.
The notation \(d \mathbf{r}\) represents a small segment along the path \(C\), indicating the curve is traversed infinitesimally.
In a line integral of a vector field, the dot product \( \mathbf{F} \cdot d \mathbf{r} \) sums up the components of \( \mathbf{F} \) projected onto the direction of \( d \mathbf{r} \), providing a measure of how much the vector field is doing work along the curve.
This cumulative measure is essential in physics and engineering to determine things like the work done by a force field when moving along a path.

Closed Curve

A closed curve is a path that starts and ends at the same point while enclosing a region. In the given problem, \(C\) represents the boundary of a region \(R\) formed by a square minus a rectangle, resulting in a complex closed curve.
This type of curve is vital when applying Green's Theorem, as the theorem requires the curve to be closed to relate a line integral around the curve to a double integral over the region it encloses.
The direction of traversal is crucial: the problem specifies that the region \(R\) should be on the left as you travel around \(C\), ensuring the integration follows the standard counter-clockwise direction.
Understanding closed curves helps in solving many types of calculus problems, particularly those involving the calculation of areas and fluxes via Green's Theorem.