Problem 14
Question
Use Green's Theorem to evaluate the indicated line integral. \(\oint_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}=\left\langle x e^{x y}+y, y e^{x y}+2 x\right\rangle\) and \(C\) is formed by \(y=x^{2}\) and \(y=4\)
Step-by-Step Solution
Verified Answer
The integration step can be difficult due to the complexity of the integrand, but with careful calculation and patience, it's feasible. Proceed by integrating step-by-step while recalling integral calculation rules.
1Step 1: Express the Field as P and Q
Here, the given field, \(\mathbf{F}\), is a two-dimensional vector field and can be written as \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j}\). So, identify P as \(P = xe^{xy}+y\) and Q as \(Q = ye^{xy}+2x\)
2Step 2: Apply Green's Theorem
Green's theorem says that \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA\). So, we first need to compute \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\). After doing this we find \(\frac{\partial Q}{\partial x} = ye^{xy}+2\) and \(\frac{\partial P}{\partial y} = xe^{xy} + e^{xy} + 1\). Putting these into Green's theorem gives us: \(\iint_D ((ye^{xy} + 2) - (xe^{xy} + e^{xy} + 1)) dA\)
3Step 3: Set up the double integral
Our region of integration \(\(D\)\) is on the x-y plane bounded by \(y=x^2\) and \(y=4\). We choose to solve in terms of y first because there are two x-values for each \(y\) in the region of consideration. By rearranging the curve \(y=x^2\) for \(x\), we know that \(x=+\sqrt{y}\) and \(x=-\sqrt{y}\). Therefore, we can set up the double integral as: \(\int_{0}^{4} \int_{-\sqrt{y}}^{+\sqrt{y}} ((ye^{xy} + 2) - (xe^{xy} + e^{xy} + 1)) dx dy\)
4Step 4: Compute the Integration
Solve this double integral by first integrating in \(dx\), then in \(dy\). This will provide us the answer for the given line integral.
Key Concepts
Double IntegralVector FieldPartial DerivativesTwo-Dimensional Flow
Double Integral
Understanding double integrals is essential for applying Green's Theorem to a two-dimensional region. A double integral allows us to compute the accumulated sum of a function over a two-dimensional area. Imagine a surface above a region in the x-y plane, and think of the double integral as a way to 'add up' all the values of the function over that surface to get the total sum.
In the context of Green's Theorem, double integrals help us turn line integrals around a closed curve into an area integral over the region inside the curve. With the function already given in terms of x and y, we set the limits of integration based on the geometric boundaries of the region. In this case, we're looking at the area bounded by the curves y=x^2 and y=4 to evaluate the accumulation of the vector field's divergence inside this region.
In the context of Green's Theorem, double integrals help us turn line integrals around a closed curve into an area integral over the region inside the curve. With the function already given in terms of x and y, we set the limits of integration based on the geometric boundaries of the region. In this case, we're looking at the area bounded by the curves y=x^2 and y=4 to evaluate the accumulation of the vector field's divergence inside this region.
Vector Field
A vector field is a mathematical construct that assigns a vector to every point in a region of space. In two dimensions, you can visualize this like arrows on a plane that show both a direction and a magnitude at every point. The given vector field, \( \mathbf{F} \), can tell us about movement or flow through a point, which is fundamental to many fields like physics and engineering.
In our exercise, \( \mathbf{F} \) is expressed as \( P\mathbf{i} + Q\mathbf{j} \), where P and Q are functions of both x and y. Understanding the nature of this vector field over the area of interest is key to applying Green's Theorem effectively.
In our exercise, \( \mathbf{F} \) is expressed as \( P\mathbf{i} + Q\mathbf{j} \), where P and Q are functions of both x and y. Understanding the nature of this vector field over the area of interest is key to applying Green's Theorem effectively.
Partial Derivatives
Partial derivatives are a cornerstone of multivariable calculus. They represent how a function changes as you alter just one of its input variables while keeping others constant. When we take the partial derivative of each component of our vector field with respect to the other variable, we're seeing how the flow changes as we move along one axis only.
In Green's Theorem, we subtract the partial derivative of P with respect to y from the partial derivative of Q with respect to x. This difference gives us what's known as the 'curl' of the vector field, which is related to the field's rotation around a point. The curl can reveal key properties of the field's behavior within the enclosed region.
In Green's Theorem, we subtract the partial derivative of P with respect to y from the partial derivative of Q with respect to x. This difference gives us what's known as the 'curl' of the vector field, which is related to the field's rotation around a point. The curl can reveal key properties of the field's behavior within the enclosed region.
Two-Dimensional Flow
A two-dimensional flow refers to the movement of a substance or force within a plane. In other words, it's how things move across the x-y axis, without accounting for any changes in the third dimension. In the exercise we're discussing, the vector field \( \mathbf{F} \) represents a two-dimensional flow.
Using Green's Theorem, we can analyze such flows within a closed curve without having to evaluate the line integral directly. Instead, we look at the properties of the flow over the area it encompasses. This makes the theorem incredibly useful for various applications in physics, such as electromagnetism and fluid dynamics, where understanding the circulation and divergence in a field is essential.
Using Green's Theorem, we can analyze such flows within a closed curve without having to evaluate the line integral directly. Instead, we look at the properties of the flow over the area it encompasses. This makes the theorem incredibly useful for various applications in physics, such as electromagnetism and fluid dynamics, where understanding the circulation and divergence in a field is essential.
Other exercises in this chapter
Problem 14
Use the Divergence Theorem to compute \(\iint_{\partial O} \mathbf{F} \cdot \mathbf{n} d S\). \(Q\) is bounded by \(x^{2}+y^{2}=4, z=1\) and \(z=8-y\) \(\mathbf
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Sketch a graph of the parametric surface. \(x=2 \cos v, y=2 \sin v, z=u\)
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Show that the line integral is independent of path and use a potential function to evaluate the integral. $$\int_{C} 3 x^{2} y^{2} d x+\left(2 x^{3} y-4\right)
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Evaluate the line integral. \(\int_{C} 3 y d s,\) where \(C\) is the portion of \(y=x^{2}\) from (0,0) to (2,4)
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