Problem 208 - Advanced Calculus

Question

Evaluate the following line integrals: a) \(\oint_{C} F^{~} \cdot d C^{-}\) where \(F^{\rightarrow}\) is the vector field \(F^{-}(x, y)=\left(x^{2}, x y\right)\) and \(\mathrm{C}\) consists of the segment of the parabola \(\mathrm{y}=\mathrm{x}^{2}\) between \((0,0)\) and \((1,1)\) and the line segment from \((1,1)\) to \((0,0) .\) (Figure 1) b) \(\int_{C} F^{-} \cdot d C^{\rightarrow}\) where \(F^{-}=\left(x\left[\left(1-y^{2}\right) /\left(y^{2}+z^{2}\right)\right]^{1 / 2}, 0,0\right)\) and \(\mathrm{C}\) is the portion of the curve (in the first octant) of the intersection of the plane \(\mathrm{x}=\mathrm{y}\) and the cylinder \(2 \mathrm{y}^{2}+\mathrm{z}^{2}=1\) from \((0,0,1)\) to \([(\sqrt{2} / 2),(\sqrt{2} / 2), 0]\)

Step-by-Step Solution

Answer

In summary: a) The line integral of the vector field $F^{-}(x, y) = (x^{2}, x y)$ along the curve C, which consists of the segment of the parabola $y = x^{2}$ between $(0,0)$ and $(1,1)$, and the line segment from $(1,1)$ to $(0,0)$, is equal to \(\frac{1}{3}\). b) The line integral of the vector field $F^{-} = \left(x\left[\left(1-y^{2}\right) /\left(y^{2}+z^{2}\right)\right]^{1/2}, 0, 0\right)$ along the curve C in the first octant of the intersection of the plane $x=y$ and the cylinder $2y^{2}+z^{2}=1$ from $(0,0,1)$ to $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}, 0\right)$ is equal to 0.

Step 1: Parameterize the Curve C

We have two segments for this curve, C1 and C2. For C1: it's the segment of the parabola y = x^2 between (0,0) and (1,1). We can parameterize it as: r1(t) = (t, t^2) where t ∈ [0, 1] For C2: it's the line segment from (1,1) to (0,0). We can parameterize it as: r2(t) = (1-t, 1-t) where t ∈ [0, 1]

Step 2: Compute the derivative of parameterization

For both parameterizations, we compute the derivatives with respect to t r1'(t) = (1, 2t) r2'(t) = (-1, -1)

Step 3: Evaluate the given vector field F along the parameterized curve

F(r1(t)) = (t^2, t^3) F(r2(t)) = ((1-t)^2, (1-t)^2)

Step 4: Compute the dot product

For both C1 and C2, we compute the dot product between the evaluated vector field and the derivative of the parameterization F(r1(t)).r1'(t) = t^2(1) + t^3(2t) = t^2 + 2t^4 F(r2(t)).r2'(t) = (1-t)^2(-1) + (1-t)^2(-1) = -2(1-t)^2

Step 5: Integrate over the parameter interval

We integrate the dot product over their respective intervals Integral for C1: \(\int_{0}^{1} (t^2 + 2t^4)dt\) Integral for C2: \(\int_{0}^{1} -2(1-t)^2 dt\) Summing these two integrals, we get \(\oint_{C} F^{~} \cdot d C^{-} = \int_{0}^{1} (t^2 + 2t^4)dt + \int_{0}^{1} -2(1-t)^2 dt = \frac{1}{3}\) b)

Step 1: Parameterize the Curve C

The curve C is the portion of the curve of intersection of the plane x=y and the cylinder 2y^2+z^2=1 from (0,0,1) to (sqrt(2)/2,sqrt(2)/2,0). We can parameterize it as follows: r(t) = (t, t, sqrt(1-2t^2)), where t ∈ [0, sqrt(2)/2]

Step 2: Compute the derivative of parameterization

The derivative of this parameterization with respect to t is: r'(t) = (1, 1, -2t / sqrt(1-2t^2))

Step 3: Evaluate the given vector field F along the parameterized curve

F(r(t)) = (t(1-1) / (t^2 + (1-2t^2)), 0, 0) = (0, 0, 0)

Step 4: Compute the dot product

The dot product between the evaluated vector field and the derivative of the parameterization is: F(r(t)).r'(t) = 0

Step 5: Integrate over the parameter interval

Since the dot product is 0, the integral of F along C is also 0: \(\int_{C} F^{-} \cdot d C^{\rightarrow} = \int_{0}^{\frac{\sqrt{2}}{2}} 0 dt = 0\)

Key Concepts

Vector Fields

A vector field is a construction in vector calculus that assigns a vector to every point in a subset of space. For instance, imagine a weather map that shows wind speed and direction at different locations - this is a simple example of a vector field. Vector fields are inherently multi-dimensional and can be used to represent various physical quantities, such as gravitational forces, electric fields, or fluid velocities.

In the context of the exercise, the vector field is given by a function of space coordinates. For example, the vector field could be defined as \( F(x, y) = (x^2, xy) \) in two dimensions, or more complex forms in three dimensions. When integrating over a curve in a vector field, one is effectively summing up the influence of the field along that path.

Parameterization of Curves

The parameterization of curves is a key concept in calculus that involves describing a curve using a parameter, typically denoted as \( t \). By parameterizing a curve, we convert it into a set of functions that can be easily manipulated mathematically. This is crucial for calculating line integrals over curves in vector fields.

As seen in the exercise, a segment of a parabola could be parameterized as \( r_1(t) = (t, t^2) \) where \( t \) ranges from 0 to 1. Parameterization not only helps in describing the curve but also in computing the integral by providing a way to express the differentials along the curve. Correct parameterization is essential for solving problems involving line integrals accurately and efficiently.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is key in calculating work done when a force is applied along a path, and in this context, it's instrumental in computing line integrals.

In the exercise, for each parameterized segment of the curve, the dot product of the vector field's value at each point and the derivative of the parameterized curve is calculated. This involves taking the derivative of the parameterized expression with respect to \( t \) and then taking the product of the corresponding components of the vector field and the derivative. For example, \( F(r_1(t)) \) is dot multiplied by \( r_1'(t) \) to yield an expression that is then integrated over the interval of \( t \) to find the total line integral along the curve.