Question

Evaluate the line integral \(\oint(x+2 y) d x-2 x d y\) along each of the following closed paths, taken counterclockwise: (a) the circle \(x^{2}+y^{2}=1 ;\) (b) the square with corners at \((1,1),(-1,1),(-1,-1),(1,-1) ;\) (c) the square with corners \((0,1),(-1,0),(0,-1),(1,0) .\)

Short answer

(a) -4\pi, (b) -16, (c) -8

Step-by-step solution

Define the Line Integral

The given line integral is \(\t\oint\limits_{C}{(x + 2y)dx - 2x dy}\). This integral will be evaluated along different closed paths specified in the problem.

Apply Green's Theorem

Green's Theorem can simplify the evaluation of a line integral over a closed curve by converting it to a double integral over the region it encloses. Green's Theorem states: \[\oint\limits_{C}{M dx + N dy} = \iint\limits_{R}{\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA}\] where M = x + 2y and N = -2x.

Calculate Partial Derivatives

Determine the partial derivatives required by Green's Theorem: \[\frac{\partial N}{\partial x} = \frac{\partial ( -2x )}{\partial x} = -2\] \[\frac{\partial M}{\partial y} = \frac{\partial (x + 2y)}{\partial y} = 2\] Then, find \[\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -2 - 2 = -4\]

Integrate the Region for Each Path

The double integral over region R becomes \[\iint\limits_{R}{(-4) dA} = -4 \iint\limits_{R}{dA}\]. Now we evaluate this for each given closed path separately.

Path (a): Circle \(x^2 + y^2 = 1\)

The area of the circle is \[A = \pi \times 1^2 = \pi\]. Applying Green's Theorem, the integral is \(-4 \times \pi = -4 \pi\).

Path (b): Square with Corners \((1,1), (-1,1), (-1,-1), (1,-1)\)

The area of the square is \[A = 2 \times 2 = 4\]. Applying Green's Theorem, the integral is \(-4 \times 4 = -16\).

Path (c): Square with Corners \((0,1), (-1,0), (0,-1), (1,0)\)

The area of the diamond-shaped square in this case is equivalent to that of a regular square, as it is simply rotated by 45 degrees. The area is still \[A = 2 \times 2/2 = 2\]. Applying Green's Theorem, the integral is \(-4 \times 2 = -8\).

Key concepts

Understanding Line Integrals and Their Evaluation

A line integral is a type of integral where the function to be integrated is evaluated along a curve or path. In our exercise, the line integral is given as \(\t\oint\limits_{C}{(x + 2y)dx - 2x dy}\). This integral takes into account both the vector field along the path and the path itself. The integration is performed over different closed paths defined in the problem like a circle or square.

To evaluate this integral, we use Green's Theorem. Green's Theorem relates the line integral around a simple closed curve to a double integral over the plane region bounded by the curve. This simplification can turn a potentially complex line integral into a more manageable double integral. Let's explore this further.

Green's Theorem and Partial Derivatives

Green's Theorem is an important tool in vector calculus. It converts a line integral around a closed curve into a double integral over the region it encloses. The theorem is stated as: \[\t\oint\limits_{C}{M dx + N dy} = \iint\limits_{R}{\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA}\] where M and N are functions of both x and y.

For our given line integral, we identify M as \(x + 2y\) and N as \(-2x\).

Next, we determine the partial derivatives which are: \[\frac{\partial N}{\partial x} = \frac{\partial ( -2x )}{\partial x} = -2\] \[\frac{\partial M}{\partial y} = \frac{\partial (x + 2y)}{\partial y} = 2\]

Subtracting these, we get: \[\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -2 - 2 = -4\]
Now, we can use this result in our double integral.

Double Integrals and Area Calculation

Using Green's Theorem, our original line integral is converted to the double integral: \[\t\iint\limits_{R}{(-4) dA} = -4 \iint\limits_{R}{dA}\] This simplifies our problem to finding the area of the region enclosed by the path.

Let's evaluate this for each path:

• Path (a): For the circle \(x^2 + y^2 = 1\), the area \( A= \pi \times 1^2 = \pi\). Applying Green's Theorem, the integral is \(-4 \times \pi = -4 \pi\).
• Path (b): For the square with corners \((1, 1), (-1, 1), (-1, -1), (1, -1)\), the area \(A = 2 \times 2 = 4\). The integral is \(-4 \times 4 = -16\).
• Path (c): For the square with corners \((0, 1), (-1, 0), (0, -1), (1, 0)\), this forms a diamond shape equivalent to the area of a square. The area is \(A = 2 \times 2/2 = 2\). The integral is \(-4 \times 2 = -8\).

Therefore, by using double integrals and areas, we can efficiently calculate the value of line integrals for different paths.