Problem 38 - Calculus Early Transcendentals

Question

(a) Let \(C\) be the line segment from a point \((a, b)\) to a point \((c, d) .\) Show that $$ \int_{C}-y d x+x d y=a d-b c $$ (b) Use the result in part (a) to show that the area \(A\) of a triangle with successive vertices \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right)\) going counterclockwise is $$ \begin{aligned} A=\frac{1}{2}\left[\left(x_{1} y_{2}\right.\right.&\left.-x_{2} y_{1}\right) \\ &\left.+\left(x_{2} y_{3}-x_{3} y_{2}\right)+\left(x_{3} y_{1}-x_{1} y_{3}\right)\right] \end{aligned} $$ (c) Find a formula for the area of a polygon with successive vertices \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) going counterclockwise. (d) Use the result in part (c) to find the area of a quadrilateral with vertices \((0,0),(3,4),(-2,2),(-1,0)\)

Step-by-Step Solution

Answer

The area of the quadrilateral is 8.

Step 1: Parametrize the Line Segment

To parametrize the line segment from \((a, b)\) to \((c, d)\), set \((x(t), y(t)) = ((1-t)a + tc, (1-t)b + td),\) where \(t \in [0, 1].\)

Step 2: Compute Differential Elements

Calculate the differentials: \[ dx = (c-a)dt, \quad dy = (d-b)dt. \]

Step 3: Substitute and Integrate Line Integral

Substitute \(y(t)\), \(dx\), and \(dy\) into the integral and integrate:\[ \int_{C} -y\,dx + x\,dy = \int_{0}^{1} \left[ -((1-t)b + td)(c-a) + ((1-t)a + tc)(d-b) \right] dt. \]

Step 4: Simplify the Integrand

Simplify the integrand:\[ = \int_{0}^{1} [ -bc + bct - tdc + tad - tad + tcb - tq(a-c)(b-d) ] dt \ = t(a(d-b) - b(c-a)) \mid_{0}^{1}. \]

Step 5: Evaluate the Integral

Evaluating the definite integral from 0 to 1 gives:\[ = a(d-b) - b(c-a) = ad - bc. \] This completes part (a).

Step 6: Apply Result to Triangle Area

Use the result from part (a) for each side of the triangle:\[ A = \frac{1}{2} \left[ x_1y_2 - y_1x_2 + x_2y_3 - y_2x_3 + x_3y_1 - y_3x_1 \right]. \]

Step 7: Generalize to Polygon Area

Apply the result iteratively for all sides of a polygon with vertices \((x_i, y_i)\):\[ A = \frac{1}{2} \left[ \sum_{i=1}^{n-1} (x_iy_{i+1} - y_ix_{i+1}) + (x_ny_1 - y_nx_1) \right]. \]

Step 8: Compute Area for Given Quadrilateral

Substitute the given vertices \((0,0),(3,4),(-2,2),(-1,0))\) into the polygon area formula:- Compute: \[ A = \frac{1}{2} \left[ 0\cdot4 - 0\cdot3 + 3\cdot2 - 4\cdot(-2) + (-2)\cdot0 - 2\cdot(-1) + (-1)\cdot0 - 0\cdot0 \right] \]- Evaluate: \[ A = \frac{1}{2} \left[ 0 + 6 + 8 + 0 + 2 + 0 \right] = \frac{1}{2} \times 16 = 8.\]

Key Concepts

Triangle Area Formula

Finding the area of a triangle using its vertices is a crucial concept in geometry, which can be easily calculated using the triangle area formula derived from linear algebra concepts. When a triangle is given by vertices \( \left(x_1, y_1\right), \left(x_2, y_2\right), \left(x_3, y_3\right) \), the area can be computed without directly measuring base and height, but using the determinant method.
The formula is expressed as \[A = \frac{1}{2} \left[ x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3 \right].\]This formula is derived from analyzing the geometrical properties and using the integrals of line segments making up the polygon.
Why does this work? It exploits the geometric property that the determinant of a matrix formed by these points encodes information about the area in its scalar value. This formula ensures that you account for the entire area by summing the clockwise areas and subtracting the counterclockwise ones.

Polygon Area Formula

The polygon area formula is a generalization of the triangle area formula and allows us to find the area of a polygon (a shape with multiple sides). To compute the area of a polygon with vertices labeled in sequence, \( \left(x_1, y_1\right), \left(x_2, y_2\right), \, \ldots, \left(x_n, y_n\right) \), the formula becomes:
\[A = \frac{1}{2} \left[\sum_{i=1}^{n-1} (x_iy_{i+1} - y_ix_{i+1}) + (x_ny_1 - y_nx_1)\right].\]
This method is sometimes known as the "Shoelace formula" or the "Surveyor's formula" due to its pattern when written vertically. It's particularly useful for computing the area of any polygon, regardless of how many sides it might have.
Identifying each vertex correctly and listing them in the proper sequence is crucial because the approach depends strongly on the vertices being ordered either clockwise or counterclockwise to maintain consistent sign conventions.

Definite Integral Evaluation

Definite integral evaluation is a core concept in calculus, often used to compute the accumulated quantity, like areas under a curve or between curves, in a given interval. In the context of line integrals, it's about integrating over a curve rather than between two points on the x-axis.
Line integrals, like in the exercise, are useful for calculating work done by a force along a path or in physics contexts when field lines are involved. In this exercise, parametrization of the curve is crucial, which involves expressing the endpoints \( \left(a, b\right) \) and \( \left(c, d\right) \) as functions of a parameter \( t \).
The evaluation process involves:

• Substituting these parameterized expressions into the integral.
• Carrying out the integration over the specific interval, typically from 0 to 1, to find the result.

By carefully performing each step, we calculate \( a(d-b) - bc \), the result linked to the orientation and length of the segment, iteratively leading to finding polygon areas as demonstrated in the exercise.