Problem 178 - Calculus Volume 3

Question

David and Sandra are skating on a frictionless pond in the wind. David skates on the inside, going along a circle of radius 2 in a counterclockwise direction. Sandra skates once around a circle of radius \(3,\) also in the counterclockwise direction. Suppose the force of the wind at \(\quad\) point \(\quad(x, y) \quad(x, y) \quad(x, y)\) is \(\mathbf{F}(x, y)=\left(x^{2} y+10 y\right) \mathbf{i}+\left(x^{3}+2 x y^{2}\right) \mathbf{j} . \quad\) Use \(\quad\) Green's theorem to determine who does more work.

Step-by-Step Solution

Answer

Sandra does more work.

Step 1: Understanding the Problem

We need to compare the work done by David and Sandra while skating in circles of radius 2 and 3, respectively, under the influence of a wind force field given by \( \mathbf{F}(x, y)=(x^2 y + 10y) \mathbf{i} + (x^3 + 2x y^2) \mathbf{j} \). Green's theorem will be used to compute the work done along each path.

Step 2: Green's Theorem Setup

Green's theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \). It states: \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \int \int_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \,dA \), where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \).

Step 3: Differentiation in Green's Theorem

Identify \( M = x^2 y + 10y \) and \( N = x^3 + 2xy^2 \). Calculate the partial derivative \( \frac{\partial N}{\partial x} = 3x^2 + 2y^2 \) and \( \frac{\partial M}{\partial y} = x^2 + 10 \).

Step 4: Calculate the Integrals

Compute the area integral: \( \int \int_R \left( 3x^2 + 2y^2 - (x^2 + 10) \right) \,dA = \int \int_R \left( 2x^2 + 2y^2 - 10 \right) \,dA \). The area integrates over disks of radius 2 for David and 3 for Sandra.

Step 5: Compute David's Work

For David's circle (radius 2): \((2x^2 + 2y^2 - 10)\) in polar coordinates turns into \((2r^2 - 10)\). Integrate \( \int_{0}^{2\pi} \int_{0}^{2} (2r^2 - 10) r \,dr \,d\theta \).

Step 6: Calculate the Double Integral: David

Compute \( \int_{0}^{2\pi} \int_{0}^{2} (2r^3 - 10r) \,dr \,d\theta \). Yield \( -\frac{160\pi}{3} \) after evaluating the integral.

Step 7: Compute Sandra's Work

For Sandra's circle (radius 3): Convert \( (2x^2 + 2y^2 - 10) \) to polar coordinates, \((2r^2 - 10)\), and integrate \( \int_{0}^{2\pi} \int_{0}^{3} (2r^2 - 10) r \,dr \,d\theta \).

Step 8: Calculate the Double Integral: Sandra

Evaluate \( \int_{0}^{2\pi} \int_{0}^{3} (2r^3 - 10r) \,dr \,d\theta \) to get \( -\frac{567\pi}{3} \).

Step 9: Conclusion: Compare Results

David's work is \( -\frac{160\pi}{3} \) and Sandra's is \( -\frac{567\pi}{3} \). Sandra does more work, as her numerical value is larger.

Key Concepts

Line Integral

A line integral is a key concept in vector calculus, where you integrate a function along a curve. It's like summing up all tiny pieces of work done along a path in a vector field. In our context, this is used to calculate the work performed by David and Sandra as they skate different paths on the pond. The line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is denoted as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). It evaluates how much of the vector field aligns with the direction of the path.

• The curve \( C \) is the path traced out by David or Sandra as they skate on the pond.
• The vector field \( \mathbf{F} \) represents the wind force acting at any point \( (x, y) \) on the pond.

To simplify calculation, Green's theorem helps transition from this line integral to a more manageable double integral over the region enclosed by the path. This theorem converts the problem from integrating around a curve to integrating over an area, making it more straightforward when evaluating complex vector fields.

Force Field

A force field in mathematics is a vector field that represents a distribution of forces at each point in space. Here, the wind on the pond is described by the force field \( \mathbf{F}(x, y) = (x^2 y + 10y) \mathbf{i} + (x^3 + 2xy^2) \mathbf{j} \). A force field specifies how much force and in what direction it acts at any location.

• Component \( M = x^2 y + 10y \) represents the force along the x-direction.
• Component \( N = x^3 + 2xy^2 \) represents the force along the y-direction.

Using Green's theorem, we bypass calculating the work as line integrals and instead focus on partial derivatives of these components across the area bounded by each skater's path.
The differences in partial derivatives, \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \), indicate how these forces accumulate over the region, influencing the total work done.

Polar Coordinates

Polar coordinates offer a different way to express points in a plane, using a radius and angle rather than traditional Cartesian coordinates \((x, y)\). This system is particularly handy when dealing with circular paths, like those skated by David and Sandra.

• Radius \( r \) denotes the distance from the origin to the point.
• Angle \( \theta \) indicates the counterclockwise angle from a reference direction, typically the positive x-axis.

Using polar coordinates simplifies the conversion of vector field functions over circular paths. For instance, transforming \( 2x^2 + 2y^2 - 10 \) to polar form results in \( 2r^2 - 10 \).
This makes the double integral calculation over a circular region more straightforward, involving \( r \) and \( \theta \) instead of \( x \) and \( y \).
The integration bounds translate easily:

• \( r \) varies from 0 to the circle's radius - 2 for David and 3 for Sandra.
• \( \theta \), due to the full circle, ranges from 0 to \( 2\pi \).