Problem 25 - Thomas Calculus

Question

Find the work done by \(\mathbf{F}\) in moving a particle once counterclockwise around the given curve. \(\mathbf{F}=2 x y^{3} \mathbf{i}+4 x^{2} y^{2} \mathbf{j}\) C: The boundary of the "triangular" region in the first quadrant enclosed by the \(x\) -axis, the line \(x=1,\) and the curve \(y=x^{3}\)

Step-by-Step Solution

Answer

The work done is \( \frac{1}{12} \).

Step 1: Understand the problem

We want to find the work done by the vector field \( \mathbf{F} = 2xy^3 \mathbf{i} + 4x^2y^2 \mathbf{j} \) as a particle moves once counterclockwise around the boundary of the region enclosed by the \(x\)-axis, the line \(x=1\), and the curve \(y=x^3\) in the first quadrant.

Step 2: Verify independence

Check if the vector field \( \mathbf{F} \) is conservative. Compute \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \), where \( M = 2xy^3 \) and \( N = 4x^2y^2 \). If these partial derivatives are equal, the field is conservative, and the integral over a closed curve would be zero.

Step 3: Compute partial derivatives

Calculate \( \frac{\partial N}{\partial x} = \frac{\partial }{\partial x}(4x^2y^2) = 8xy^2 \) and \( \frac{\partial M}{\partial y} = \frac{\partial }{\partial y}(2xy^3) = 6xy^2 \). Since \( 8xy^2 eq 6xy^2 \), the field is not conservative.

Step 4: Setup Work Integral using Green's Theorem

According to Green's Theorem, the work done \( W \) around a closed curve \( C \) is given by the integral \( W = \oint_C M \, dx + N \, dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \). We now calculate the double integral over the region \( R \).

Step 5: Evaluate the double integral

The region \( R \) is bounded by the \( x \)-axis, the line \( x=1 \), and the curve \( y=x^3 \). The limits of integration are from \( x=0 \) to \( x=1 \) and from \( y=0 \) to \( y=x^3 \). Therefore, the integral becomes \( \int_0^1 \int_0^{x^3} (8xy^2 - 6xy^2) \, dy \, dx = \int_0^1 \int_0^{x^3} 2xy^2 \, dy \, dx \).

Step 6: Simplify and compute the inner integral

Compute the inner integral: \( \int_0^{x^3} 2xy^2 \, dy = 2x \left[ \frac{y^3}{3} \right]_0^{x^3} = 2x \left( \frac{(x^3)^3}{3} \right) = \frac{2x^7}{3} \).

Step 7: Compute the outer integral

Now compute the outer integral: \( \int_0^1 \frac{2x^7}{3} \, dx = \frac{2}{3} \left[ \frac{x^8}{8} \right]_0^1 = \frac{2}{3} \times \frac{1}{8} = \frac{1}{12} \).

Step 8: Conclude the result

The work done by the force field \( \mathbf{F} \) around the curve \( C \) is \( \frac{1}{12} \).

Key Concepts

Work Done

In the context of vector calculus, 'Work Done' often involves evaluating how a force described by a vector field affects a particle as it moves through a path. Specifically, work is calculated as a line integral of a vector field along a given path. Consider a particle moving in a force field represented by \( \mathbf{F} = 2xy^3 \mathbf{i} + 4x^2y^2 \mathbf{j} \). The particle follows a path that forms a closed curve, moving counterclockwise around a specific boundary in the first quadrant. The work done by the vector field on the particle along this path can be calculated using line integrals.

• Work is calculated over closed curves using line integrals.
• The work done by a vector field moving a particle is vital in understanding energy and force interactions.

In this case, computing work using line integrals involves breaking down the work into components along coordinate paths using vector calculus concepts. Once calculated, this work gives you a scalar value representing energy change along the path.

Green's Theorem

Green’s Theorem is a crucial tool in vector calculus that connects the concept of a line integral around a closed curve to a double integral over the area the curve encloses. For a vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), Green's Theorem relates the work done around a closed path to the area integral, allowing us to evaluate more complex integrals with ease. The theorem is expressed as: \[ \oint_C M \, dx + N \, dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]This powerful tool provides a means to convert a line integral into a double integral, simplifying the problem.

• Relates the circulation around a path to the total curl within it.
• Helps in simplifying the calculation of work done by connecting it to an area integral.

In this exercise, the work done by the field \( \mathbf{F} \) around the boundary can be effectively computed using the double integral as stated by Green's Theorem, significantly simplifying the calculation process.

Conservative Vector Field

A vector field is called conservative if the work done in moving a particle around any closed path is zero, indicating that the field has potential energy stored. A key test to verify if a field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \) is conservative involves checking if the equality \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \) holds. If true, the field is conservative, and the work done over any closed loop would be zero.

• Conservative fields indicate path-independent work, valuable for calculating potential energy.
• In this exercise, checking conservativeness simplifies the computational process of work done if confirmed true.

In this problem, however, the computations showed \( 8xy^2 eq 6xy^2 \), proving \( \mathbf{F} \) is not conservative. This distinction is important as it dictates the subsequent computation approach using Green's Theorem rather than relying on the simpler properties of conservative vector fields.

Double Integral

The double integral is a way to calculate the volume under a surface defined by a function of two variables, integrating over a specified region. In the context of this exercise, it calculates the area under the 'curl' of the vector field to find the work done along a path using Green's Theorem.For the region bounded by the \( x \)-axis, the line \( x=1 \), and the curve \( y=x^3 \), our task here is to evaluate the double integral:\[ \iint_R (8xy^2 - 6xy^2) \, dA = \int_0^1 \int_0^{x^3} 2xy^2 \, dy \, dx \]

• The choice of limits reflects the region of interest.
• Handles integration sequentially: first \( y \) then \( x \).

Using the integration boundaries and simplifying the integral step by step reveals the total work done over the region by the applied vector field. Mastery of double integrals is vital in multivariable calculus, offering a tool to solve area and volume problems across regions.