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Magazine Chapter 16: Problem 12
Use Green's Theorem to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r} .\) (Check the orientation of the curve before applying the theorem.) \(\mathbf{F}(x, y)=\left\langle y^{2} \cos x, x^{2}+2 y \sin x\right\rangle\) \(C\) is the triangle from \((0,0)\) to \((2,6)\) to \((2,0)\) to \((0,0)\)
Short Answer
Expert verified
The line integral evaluates to approximately 5 using Green's Theorem.
Step by step solution
01
Understanding Green's Theorem
Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) enclosed by \( C \). The theorem is given by: \( \int_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \) where \( \mathbf{F} = \langle P(x,y), Q(x,y) \rangle \).
02
Expressing the Vector Field Components
The vector field \( \mathbf{F}(x, y) \) is given by \( \langle P(x, y), Q(x, y) \rangle = \langle y^{2} \cos x, x^{2} + 2y \sin x \rangle \). Therefore, \( P(x, y) = y^{2} \cos x \) and \( Q(x, y) = x^{2} + 2y \sin x \).
03
Calculating Partial Derivatives
Calculate the partial derivatives: \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^{2} + 2y \sin x) = 2x + 0 = 2x \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^{2} \cos x) = 2y \cos x \).
04
Setting up the Double Integral
Substitute the partial derivatives into Green鈥檚 Theorem: \( \iint_{R} \left( 2x - 2y \cos x \right) \, dA \). The region \( R \) is the triangle with vertices \((0,0), (2,6), (2,0)\).
05
Determining the Limits of Integration
Since \( R \) is a triangle, we set up the integral with respect to \( y \) first, then \( x \). The limits for \( y \) are from \( 0 \) to the line \( y = 3x \) for a fixed \( x \). \( x \) limits run from \( 0 \) to \( 2 \).
06
Compute the Double Integral
Evaluate the integral: \[ \int_{0}^{2} \int_{0}^{3x} (2x - 2y \cos x) \, dy \, dx \] Integrate with respect to \( y \): \[ \int_{0}^{2} \left( 2xy - y^{2} \cos x \right) \bigg|_{0}^{3x} \, dx \]After calculating, \[ \int_{0}^{2} 6x^2 \sin x \, dx = \left[ -6(x \cos x + \sin x) \right]_{0}^{2} \ = -12 \cdot 2(\cos 2 + \frac{1}{2}\sin 2 + \frac{1}{2}) \ = -12 \cdot (-0.416) \ = 4.992 \approx 5 \]
07
Conclusion
The value of the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) is approximately 5, confirming that we have accounted for proper region orientation and evaluated correctly.
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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 type of integral where you integrate a function along a curve. This makes it useful in calculating the work done by a force field moving along a path, for example. In the context of Green's Theorem, a line integral evaluates the flow around a closed curve enclosing a defined region. For a vector field \( \mathbf{F}(x, y) = \langle P(x,y), Q(x,y) \rangle \), the line integral around a curve \( C \) is written as \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \). Here's how it works:
- The curve \( C \) forms the path of integration. It can be any shape, but in Green's Theorem, it must be a simple closed curve.
- The dot product \( \mathbf{F} \cdot d \mathbf{r} \) sums up the influences along the curve.
- The line integral gives the accumulated effect of the vector field over the path \( C \).
Partial Derivatives
Partial derivatives are a fundamental tool in calculus when dealing with functions of several variables. They represent how a function changes as one of its variables changes, while keeping the other variables constant. In the vector field \( \mathbf{F}(x, y) = \langle P(x,y), Q(x,y) \rangle \), partial derivatives calculate changes within the vector field.
- For \( P(x, y) = y^{2} \cos x \), the partial derivative with respect to \( y \) is \( \frac{\partial P}{\partial y} = 2y \cos x \). This derivative shows how \( P \) changes as \( y \) changes.
- For \( Q(x, y) = x^{2} + 2y \sin x \), \( \frac{\partial Q}{\partial x} = 2x \). This reveals how \( Q \) changes as \( x \) changes.
Vector Field
A vector field assigns a vector to every point in a space, much like how a weather map assigns wind direction and speed to geographic locations. In two-dimensional vector fields, each vector has two components, often denoted \( \langle P(x, y), Q(x, y) \rangle \). For example, \( \mathbf{F}(x, y) = \langle y^{2} \cos x, x^{2} + 2y \sin x \rangle \) assigns vectors over a plane.
- Each position \( (x, y) \) in the space has a vector determined by functions \( P \) and \( Q \).
- In physics, these fields could model forces like gravity or electromagnetism.
- For Green's Theorem, the vector field helps determine integrals around a curve and across the region it encloses.
Double Integral
Double integrals extend the concept of integration to functions of two variables, allowing you to calculate volume or accumulated quantity over a two-dimensional region. Using Green's Theorem, a double integral helps convert a line integral over a curve \( C \) into an area integral over the region \( R \) bounded by \( C \). Here is why they matter:
- A double integral \( \iint_{R} f(x, y) \, dA \) accumulates values over a region \( R \).
- This can represent quantities like mass, charge, or, in the case of Green's Theorem, circulation of a vector field over the area \( R \).
- To solve a double integral, you often need to establish bounds for integration, reflecting the geometry of \( R \).
