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Magazine Chapter 6: Problem 177
A particle starts at point \((-2,0),\) moves along the \(x\) -axis to \((2,0),\) and then travels along semicircle \(y=\sqrt{4-x^{2}}\) to the starting point. Use Green's theorem to find the work done on this particle by force field \(\mathbf{F}(x, y)=x \mathbf{i}+\left(x^{3}+3 x y^{2}\right) \mathbf{j}\).
Short Answer
Expert verified
The work done by the force field is \( 12\pi \).
Step by step solution
01
Understanding Green's Theorem
Green's theorem relates a line integral around a closed curve to a double integral over the plane region bounded by the curve. The theorem 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 \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \). For our force field \( \mathbf{F}(x, y) = x \mathbf{i} + (x^3 + 3xy^2) \mathbf{j} \), we need \( M(x, y) = x \) and \( N(x, y) = x^3 + 3xy^2 \).
02
Compute the Partial Derivatives
Calculate the partial derivatives of \( N \) with respect to \( x \) and \( M \) with respect to \( y \). This gives us: \( \frac{\partial N}{\partial x} = 3x^2 + 3y^2 \) and \( \frac{\partial M}{\partial y} = 0 \).
03
Set Up the Double Integral
According to Green's theorem, the work done \( W \) is \( \int \int_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA = \int \int_R (3x^2 + 3y^2) \, dA \). The region \( R \) is the semicircle described by \( y=\sqrt{4-x^2} \) for \( -2 \leq x \leq 2 \).
04
Convert to Polar Coordinates
For a semicircle, it's convenient to switch to polar coordinates. Substitute \( x = r\cos(\theta) \), \( y = r\sin(\theta) \), and \( dA = r \, dr \, d\theta \). The limits for \( r \) are from 0 to 2 and for \( \theta \) are from 0 to \( \pi \).
05
Evaluate the Double Integral
Substitute the polar coordinates into the integral: \( \int_0^{\pi} \int_0^2 3r^2(\cos^2(\theta) + \sin^2(\theta))r \, dr \, d\theta \). Simplifying gives \( \int_0^{\pi} \int_0^2 3r^3 \, dr \, d\theta = \int_0^{\pi} \left. \frac{3}{4}r^4 \right|_0^2 \, d\theta = \int_0^{\pi} 12 \, d\theta \).
06
Calculate the Final Answer
Solve the integral \( \int_0^{\pi} 12 \, d\theta = [12\theta]_0^{\pi} = 12\pi \). Therefore, the work done on the particle by the force field is \( 12\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integral
A line integral is a special type of integral where a function is integrated along a curve. Picture a particle moving along a path, and imagine calculating the total amount of work done by a force acting on the particle. This calculation is precisely what a line integral determines. In mathematical terms, a line integral of a vector field \( \mathbf{F} \) over a curve \( C \) is expressed as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \).
In the context of Green's Theorem, which relates the line integral around a closed curve to a double integral over the region it encloses, this method allows us to turn a challenging path integral problem into a typically simpler area integral problem.
When solving the initial problem, we computed line integrals to grasp the entire curve's work done, utilizing Green's Theorem to facilitate the process.
In the context of Green's Theorem, which relates the line integral around a closed curve to a double integral over the region it encloses, this method allows us to turn a challenging path integral problem into a typically simpler area integral problem.
When solving the initial problem, we computed line integrals to grasp the entire curve's work done, utilizing Green's Theorem to facilitate the process.
Force Field
Force fields are vector fields that describe the influence a force has at various points in space. Imagine an invisible wind pushing a leaf along its path; a force field functions similarly for particles moving in space. In our problem, the force field \( \mathbf{F}(x, y) = x \mathbf{i} + (x^3 + 3xy^2) \mathbf{j} \) applies a different amount of force at each point depending on the coordinates \( x \) and \( y \).
To use Green's Theorem, the force field is broken down into its components \( M \) and \( N \) where \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \). Understanding how the force changes at every point on the path is key to using Green's Theorem to evaluate the work done around a closed curve.
To use Green's Theorem, the force field is broken down into its components \( M \) and \( N \) where \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \). Understanding how the force changes at every point on the path is key to using Green's Theorem to evaluate the work done around a closed curve.
Polar Coordinates
Polar coordinates are an alternate system for defining points based on a distance from a fixed point and an angle from a fixed direction. While Cartesian coordinates use \( (x, y) \), polar coordinates use \( (r, \theta) \), representing the radius and angle.
In our exercise, the semicircle is better suited for polar coordinates because of its symmetry around the origin. By converting from \( (x, y) \) to \( (r, \theta) \), where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), the integration limits become more straightforward to handle.
In our exercise, the semicircle is better suited for polar coordinates because of its symmetry around the origin. By converting from \( (x, y) \) to \( (r, \theta) \), where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), the integration limits become more straightforward to handle.
- \( r \) varies from 0 to radius, which here is 2.
- \( \theta \) varies from 0 to \( \pi \).
Double Integral
A double integral extends the concept of integrals to functions of two variables, allowing us to calculate the accumulation of quantities over a region in the plane. In this problem, using a double integral helps compute the work done by the force field over a semicircular area.
After converting to polar coordinates, our double integral \( \int \int_R (3x^2 + 3y^2) \, dA \) becomes \( \int_0^{\pi} \int_0^2 3r^3 \, dr \, d\theta \) because \( x^2 + y^2 \) simplifies to \( r^2 \) and \( dA \) switches to \( r \, dr \, d\theta \).
When evaluated, this integral measures the total effect the force field has on the entire region, resulting in the work done being \( 12\pi \).
This application demonstrates how double integrals, combined with polar coordinates, become potent tools in solving real-world physical problems.
After converting to polar coordinates, our double integral \( \int \int_R (3x^2 + 3y^2) \, dA \) becomes \( \int_0^{\pi} \int_0^2 3r^3 \, dr \, d\theta \) because \( x^2 + y^2 \) simplifies to \( r^2 \) and \( dA \) switches to \( r \, dr \, d\theta \).
When evaluated, this integral measures the total effect the force field has on the entire region, resulting in the work done being \( 12\pi \).
This application demonstrates how double integrals, combined with polar coordinates, become potent tools in solving real-world physical problems.
