Problem 17 - Calculus: Early Transcendental Functions

Question

Use Green's Theorem to evaluate the indicated line integral. \(\oint_{C} x^{2} d x+2 x d y+(z-2) d z,\) where \(C\) is the triangle from (0,0,2) to (2,0,2) to (2,2,2) to (0,0,2)

Step-by-Step Solution

Answer

The value of the line integral is \(20/3\).

Step 1: Parameterization of the curve

The triangle \(C\) can be parameterized by a piecewise function for each segment. Since we have three points, we define three vectors: For (0,0,2) to (2,0,2): \(r_1(t) = (2t, 0, 2)\), for \(0 \leq t \leq 1\). For (2,0,2) to (2,2,2): \(r_2(t) = (2, 2t, 2)\), for \(0 \leq t \leq 1\). For (2,2,2) to (0,0,2): \(r_3(t) = (2-2t, 2-2t, 2)\), for \(0 \leq t \leq 1\).

Step 2: Compute the differential forms

Firstly, differentiate \(r_i(t)\) to find \(dr_i\). Then compute each differential form for \(i\) from 1 to 3. \(d r_1 = 2 dt, dr_2 = 2 dt, dr_3 = -2\sqrt{2} dt\). Therefore, the integral becomes: \[\oint_{C} = \int_0^1 x^2 dx + 2x dy + (z-2)dz = \int_0^1 r_1(t) dr_1 + r_2(t) dr_2 + r_3(t) dr_3\].

Step 3: Evaluate the Integrals

In this step, just substitute \(r_i(t)\) and \(dr_i\) into the integral and evaluate them: \(\int_0^1 (2t)^2 (2 dt) + 2 * 2t (2 dt) + [(2) -2] * 2 dt = 8 \int_0^1 t^2 dt + 8 \int_0^1 t dt + 0 = 8 [\frac{t^3}{3}]_0^1 + 8 [\frac{t^2}{2}]_0^1 = 8 [\frac{1}{3} + \frac{1}{2}] = 8\frac{5}{6} = 20/3\). For \(i\) equals 2 and 3, because \(dy\) and \(dz\) are 0 and they have no \(x^2 dx\) term in the integral, they will be 0. Therefore, \(\oint_{C} = 20/3 + 0 + 0 = 20/3\).

Key Concepts

Line Integral

A line integral involves integrating a function over a curve or path. In this exercise, we're dealing with a line integral around a closed curve which is denoted by \( \oint \). This is different from a regular integral which typically involves an interval on the real number line.
Line integrals are used to find the work done by a force field in moving an object along a curve, among other applications. They consider both the function values and the path taken on the curve.

• In this context, we integrate the function \( x^2 \, dx + 2x \, dy + (z-2) \, dz \) around the triangular path \( C \).
• The curve \( C \) is given by the triangle with specific points: \((0,0,2)\), \((2,0,2)\), and \((2,2,2)\).

The calculation of the line integral depends on this path and considers how the function \( x^2 \, dx + 2x \, dy + (z-2) \, dz \) changes along it.

Parameterization of the Curve

Parameterization is crucial when working with line integrals because it provides a way to express the curve explicitly in terms of a single parameter, often \( t \). For a curve split into segments, a piecewise parameterization handles each segment separately.
In this exercise:

• The path \( C \) is broken into three segments corresponding to the triangle's sides.
• Each segment is parameterized using a function of \( t \):
  - From \((0,0,2)\) to \((2,0,2)\): \( r_1(t) = (2t, 0, 2) \) for \( 0 \leq t \leq 1 \).
  - From \((2,0,2)\) to \((2,2,2)\): \( r_2(t) = (2, 2t, 2) \) for \( 0 \leq t \leq 1 \).
  - From \((2,2,2)\) to \((0,0,2)\): \( r_3(t) = (2-2t, 2-2t, 2) \) for \( 0 \leq t \leq 1 \).

This step is fundamental because it reformulates the original path into a form that's easier to integrate.

Differential Forms

Differential forms allow us to express the integrals in terms of small changes along the path. They are extremely useful in converting multi-variable integrals to a form that is easier to handle with calculus.
For our parameterized curve:

• We start by differentiating the parameterization functions to find \( dr_i(t) \).
• This yields:
  - \( d r_1(t) = 2 \, dt \).
  - \( d r_2(t) = 2 \, dt \).
  - \( d r_3(t) = -2\sqrt{2} \, dt \).

By understanding the differential forms \( dx, dy, dz \) along the curve, we can substitute them into the integral to make calculations possible. The integral becomes an expression involving these differentials and the path segments.

Evaluation of Integrals

Once we parameterize the curve and calculate differential forms, the final step is evaluating the integrals. Substituting the parametrized functions and differential forms into the integral expression allows us to compute its value.
The steps involved:

• For the segment from \((0,0,2)\) to \((2,0,2)\), the integral is:
  \( \int_0^1 (2t)^2 (2 \, dt) + 2 \cdot 2t (2 \, dt) + 0 \, dt = 8 \int_0^1 t^2 \, dt + 8 \int_0^1 t \, dt = \frac{20}{3} \).
• The other segments contribute \( 0 \) since the terms not present in these segments simplify to zero.

Through these calculations, the entire line integral around the curve \( C \) is found to be \( \frac{20}{3} \). This verifies the application of Green's Theorem in the context of evaluating line integrals.