Chapter 16: Problem 16
In Exercises 13–17, show that the differential forms in the integrals are exact. Then evaluate the integrals. $$ \int_{(0,0,0)}^{(3,3,1)} 2 x d x-y^{2} d y-\frac{4}{1+z^{2}} d z $$
Short Answer
Expert verified
The integral evaluates to \(-\pi\).
Step by step solution
01
Identify the vector field
The integral is over the differential form \( \omega = 2x \, dx - y^2 \, dy - \frac{4}{1+z^2} \, dz \). We can identify this as a vector field \( \mathbf{F} = (2x, -y^2, -\frac{4}{1+z^2}) \).
02
Check for exactness
The differential form is exact if there exists a function \( f(x,y,z) \) such that \( df = \mathbf{F} \cdot (dx, dy, dz) \). For this form to be exact, the mixed partial derivatives must satisfy: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, \frac{\partial M}{\partial z} = \frac{\partial P}{\partial x}, \text{and} \frac{\partial N}{\partial z} = \frac{\partial P}{\partial y} \). Calculating these gives: \( \frac{\partial (2x)}{\partial y} = 0, \frac{\partial (-y^2)}{\partial x} = 0, \frac{\partial (2x)}{\partial z} = 0, \frac{\partial (-\frac{4}{1+z^2})}{\partial x} = 0, \frac{\partial (-y^2)}{\partial z} = 0, \frac{\partial (-\frac{4}{1+z^2})}{\partial y} = 0 \). All conditions are satisfied, hence the form is exact.
03
Find potential function \( f(x,y,z) \)
We find \( f(x,y,z) \) such that \( abla f = \mathbf{F} \). Start with \( f_x = 2x \), integrate with respect to \( x \): \( f(x,y,z) = x^2 + h(y,z) \). Then, from \( f_y = -y^2 \), differentiate \( f(x,y,z) \) by \( y \) and equate: \( \frac{d}{dy} h(y,z) = -y^2 \). Integrate with respect to \( y \): \( h(y,z) = -\frac{y^3}{3} + g(z) \). From \( f_z = -\frac{4}{1+z^2} \), differentiate \( f(x,y,z) \) in terms of \( z \) and equate: \( \frac{d}{dz} g(z) = -\frac{4}{1+z^2} \). Integrate with respect to \( z \): \( g(z) = -4 \tan^{-1}(z) + C \). Thus, potential function \( f(x,y,z) = x^2 - \frac{y^3}{3} - 4 \tan^{-1}(z) + C \).
04
Evaluate the integral
To evaluate the integral of an exact form from \((0,0,0)\) to \((3,3,1)\), use the potential function found. Compute \( f(3,3,1) - f(0,0,0) \). Substitute into \( f(x,y,z) \): \[f(3,3,1) = 3^2 - \frac{3^3}{3} - 4 \tan^{-1}(1) = 9 - 9 - 4 \times \frac{\pi}{4} = -\pi.\] \[f(0,0,0) = 0^2 - \frac{0^3}{3} - 4 \tan^{-1}(0) = 0.\] The result is \( -\pi - 0 = -\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potential Function
In multivariable calculus, a potential function is a scalar function whose gradient matches a given vector field. If a vector field can be expressed as the gradient of a potential function, it is termed as a conservative or exact vector field.
A potential function simplifies the calculation of line integrals. For a given vector field \( \mathbf{F} = (P, Q, R) \), if there is a function \( f(x, y, z) \) such that the gradient \( abla f \) is equal to \( \mathbf{F} \), then:
A potential function simplifies the calculation of line integrals. For a given vector field \( \mathbf{F} = (P, Q, R) \), if there is a function \( f(x, y, z) \) such that the gradient \( abla f \) is equal to \( \mathbf{F} \), then:
- \( \frac{\partial f}{\partial x} = P \)
- \( \frac{\partial f}{\partial y} = Q \)
- \( \frac{\partial f}{\partial z} = R \)
Multivariable Calculus
Multivariable Calculus extends calculus to functions of several variables. It includes differentiation and integration in multiple dimensions and is vital for understanding real-world phenomena where conditions change over space and time.
One crucial part of multivariable calculus is evaluating whether a differential form or a function in several dimensions is exact. Exactness tells us about the conservation properties of a vector field, helping to find potentials and simplify integrals across paths.
In the given exercise, determining that the form \( 2x \, dx - y^2 \, dy - \frac{4}{1+z^2} \, dz \) is exact required checking whether the mixed partial derivatives meet the necessary conditions. This includes ensuring \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), alongside other symmetry requirements.By satisfying these, the vector field is conservative, which means:
One crucial part of multivariable calculus is evaluating whether a differential form or a function in several dimensions is exact. Exactness tells us about the conservation properties of a vector field, helping to find potentials and simplify integrals across paths.
In the given exercise, determining that the form \( 2x \, dx - y^2 \, dy - \frac{4}{1+z^2} \, dz \) is exact required checking whether the mixed partial derivatives meet the necessary conditions. This includes ensuring \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), alongside other symmetry requirements.By satisfying these, the vector field is conservative, which means:
- The integral will be path-independent.
- Potential functions can describe the changes along the path.
Vector Fields
Vector fields are mathematical constructions used to represent spatially varying quantities. They assign a vector to every point in a space.
For instance, a vector field in 3D space might represent the wind speed and direction at each point in the atmosphere. In this exercise, the vector field \( \mathbf{F} = (2x, -y^2, -\frac{4}{1+z^2}) \) is considered.
This field includes components that vary depending on their position in space, specifically along the \( x, y, \text{ and } z \) axes. These components:
In mathematical terms, vector fields are exploredthrough concepts of divergence, curl, and potential functions, helping to solve practical problems and study propertieslike fluid flow, electric fields, and gravitational fields.
For instance, a vector field in 3D space might represent the wind speed and direction at each point in the atmosphere. In this exercise, the vector field \( \mathbf{F} = (2x, -y^2, -\frac{4}{1+z^2}) \) is considered.
This field includes components that vary depending on their position in space, specifically along the \( x, y, \text{ and } z \) axes. These components:
- \( 2x \) related to the \( x \)-direction.
- \(-y^2 \) related to the \( y \)-direction.
- \(-\frac{4}{1+z^2} \) related to the \( z \)-direction.
In mathematical terms, vector fields are exploredthrough concepts of divergence, curl, and potential functions, helping to solve practical problems and study propertieslike fluid flow, electric fields, and gravitational fields.