Problem 10 - Calculus: Early Transcendental Functions

Question

Use Green's Theorem to evaluate the indicated line integral. \(\int_{C}\left(\sqrt{x^{2}+1}-x^{2} y\right) d x+\left(x y^{2}-y^{5 / 3}\right) d y,\) where \(C\) is the circle \(x^{2}+y^{2}=4\) oriented clockwise

Step-by-Step Solution

Answer

The value of the line integral is -8π.

Step 1: Identify the Vector Field Components

The vector field F is \(F = = <\sqrt{x^{2}+1}-x^{2} y, x y^{2}-y^{5 / 3}>\) where \(P = \sqrt{x^{2}+1}-x^{2} y\), and \(Q = x y^{2}- y^{5 / 3}\).

Step 2: Use Green's Theorem

Green’s Theorem states \(\int_{C} F \cdot dr = \iint_D (Q_x - P_y)dA\). Here \(P_y\) is the partial derivative of \(P\) with respect to \(y\) and \(Q_x\) is the partial derivative of \(Q\) with respect to \(x\). Calculate \(P_y = -x^{2}\) and \(Q_x = y^{2}\). Then, \(Q_x - P_y = y^{2} + x^{2}\).

Step 3: Change to Polar Coordinates

Since the region D is a circle, it is simpler to convert to polar coordinates for the double integral. Thus \(Q_x - P_y = y^{2} + x^{2}\) becomes \(r^{2}\). The limits of integration are \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\).

Step 4: Solve the Double Integral

Now, solve the double integral: \(\iint_D (Q_x - P_y)dA = 2\pi \int_{0}^{2} r^{3} dr = 2\pi [ \frac{1}{4} r^4 ]_{0}^{2} = 2\pi [4 - 0] = 8\pi \)

Step 5: Reverse the Orientation

The original path C is oriented clockwise. But we have calculated the result above for counterclockwise orientation (which is standard for Green's Theorem). When C is reversed (clockwise), the line integral changes sign. Therefore \(\int_{C} F \cdot dr = -8\pi\)

Key Concepts

Line Integral

To build an understanding of the line integral, think of it like a path-based accumulation of a field's influence. Imagine walking along a path while pushing a cart, and the ground beneath you varies in hilliness (our field). Your effort depends on the path taken, the hilliness at each point, and the direction you’re pushing. In mathematics, a line integral is similar; it represents the integration of a function over a curve. In the context of vector fields, like in Green's Theorem, the line integral is concerned with the sum of the vector field's component along the path. Here, we're not only interested in 'how hilly' (the magnitude) but also 'which direction the hills are sloping' (the direction). So, if our cart was pushed along the circle represented by the path C, the line integral would give us a cumulative measure of our effort considering both magnitudes and direction, defined as \(\int_{C}\mathbf{F} \cdot d\mathbf{r}\).

Vector Field

A vector field is a fancy term for a map that assigns a vector (a direction and magnitude) to every point in space. In our cart analogy, if each location on your path displayed an arrow showing wind direction and strength, your map with all those arrows would be akin to a vector field. For a field defined by a function \(\mathbf{F}(x, y)\), each point \( (x, y) \) gets a vector. In the exercise, the vector field is defined by \( P \) and \( Q \) for each component, resulting into the vector field \(\mathbf{F} = \). This vector field guides the computation of the line integral by providing the 'wind' (or in mathematical terms, the field's direction and magnitude) at each point along the path C.

Polar Coordinates

Polar coordinates are a different way of describing positions in space, like using landmarks instead of street addresses. Regular coordinates (x and y) are switched for a tag team of a distance from a central point (called radius r) and an angle (theta θ) from a reference direction. It's like saying, ‘I am 5 steps from the lamp post, facing north-east.’ It's handy particularly for circular or spiral paths where speaking in terms of radius and angle is more natural. In our circle-tracing exercise, using polar coordinates (where \( x = r\cos(\theta) \), \( y = r\sin(\theta) \) ) simplifies the nasty business of integration over a circle using Cartesian coordinates.

Double Integral

For a single integral, imagine adding up slices to find an area; with a double integral, we're stacking up tiny patches to find a volume. In technical jargon, we're integrating a function of two variables over a two-dimensional region. The exercise at hand asks us to find the 'volume' under a surface over the circular region D; this is akin to finding out how much paint you'd need to cover a wavy ceiling that covers the entire circle. The double integral takes all those infinitesimally small patches within the region D and adds them up to compute the total sum, represented as \(\iint_D (Q_x - P_y)dA\).

Partial Derivative

When a recipe says, 'Adjust the sweetness to taste, but keep the spice constant,' you're only tweaking one ingredient while holding the others steady. The partial derivative is like that. It measures how much a function changes as you nudge one variable, ignoring the rest. In our vector field \(\mathbf{F} = \), when Green’s Theorem asks for \( P_y \) or \( Q_x \) it’s asking, 'if I travel North (increase y) or East (increase x), but not both at the same time, how does the wind (our function) change?' In the case of Green’s theorem, this helps pivot from summing up along a path (line integral) to summing up across an area (double integral), because it reveals how a tiny nudge in one direction affects the whole system.