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Use Green's Theorem to find the work done by the force \(\mathbf{F}(x, y)=x(x+y) \mathbf{i}+x y^{2} \mathbf{j}\) in moving a particle from the origin along the \(x\) -axis to \((1,0),\) then along the line segment to \((0,1),\) and then back to the origin along the \(y\) -axis.

Short Answer

Expert verified
The work done is \(-\frac{5}{36}\).

Step by step solution

01

Outline the Path and Define the Curve

The problem specifies a path composed of three segments: from the origin \((0,0)\) along the x-axis to \((1,0)\), then a straight line segment to \((0,1)\), and finally back to the origin along the y-axis. Define this path as a closed curve \(C\).
02

Express Force Vector Field in Component Form

The force field vector \( \mathbf{F} \) is given as \( \mathbf{F}(x,y) = x(x+y) \mathbf{i} + xy^2 \mathbf{j} \). This can be expressed in component form as \( \mathbf{F}(x, y) = (P(x, y), Q(x, y)) \) where \( P(x, y) = x(x+y) \) and \( Q(x, y) = xy^2 \).
03

Setup Green's Theorem

Green's Theorem states that \( \int_C (P \, dx + Q \, dy) = \int\int_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \), where \( R \) is the region enclosed by \( C \). This will be used to convert the line integral into a double integral over the region \( R \).
04

Differentiate to Find the Double Integral

Calculate the partial derivatives: \( \frac{\partial Q}{\partial x} = y^2 \) and \( \frac{\partial P}{\partial y} = x \). Plug these into Green's theorem to find the expression for the double integral: \( \int\int_R (y^2 - x) \, dA \).
05

Define Limits of Integration

The region \( R \) is a right triangle with vertices at \((0,0), (1,0), (0,1)\). Integrate over \( x \) from 0 to 1 and \( y \) from 0 to the line connecting \((1,0)\) to \((0,1)\), which is \(y = 1 - x\).
06

Compute the Double Integral

Perform the double integral over the region \( R \):\[\int_{0}^{1} \int_{0}^{1-x} (y^2 - x) \, dy \, dx\]Integrate with respect to \( y \) first, yielding:\[-\int_{0}^{1} \left[ \frac{y^3}{3} - xy \right]_{0}^{1-x} \, dx \]This results in \(-\int_{0}^{1} \left[ \frac{(1-x)^3}{3} - x(1-x) \right] \, dx \).Simplify and integrate with respect to \( x \) next. It would yield the final answer for the integral.
07

Integrate and Simplify

Simplify the expression and integrate:\[-\int_{0}^{1} \left( \frac{1}{3}(1 - 3x + 3x^2 - x^3) - x + x^2 \right) \, dx \]Rearrange the terms to find:\[-\int_{0}^{1} \left( \frac{1}{3} - x + \frac{2}{3}x^2 - \frac{1}{3}x^3 \right) \, dx \]Integrate term by term:\[-\left[ \frac{1}{3}x - \frac{x^2}{2} + \frac{2}{9}x^3 - \frac{1}{12}x^4 \right]_{0}^{1} \]Evaluate this from 0 to 1 to find the work done.
08

Evaluate Final Expression

Substitute \( x = 1 \) and \( x = 0 \) to find the work:\[-\left( \frac{1}{3} - \frac{1}{2} + \frac{2}{9} - \frac{1}{12} \right) = -\left( \frac{4}{12} - \frac{6}{12} + \frac{8}{36} - \frac{3}{36} \right)= -\left( -\frac{1}{12} + \frac{5}{36} \right) = -\frac{5}{36} \]Thus, the work done is \(-\frac{5}{36}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Integral
In mathematics, a **line integral** is a type of integral where a function is integrated along a curve. In this context, it is relevant for calculating the work done by a vector field along a path.
The line integral of a vector field \( \mathbf{F} = (P(x, y), Q(x, y)) \) along a curve \( C \) is given by:
  • \( \int_C P \, dx + Q \, dy \)
Here, \( P \) and \( Q \) are the components of the vector field, and \( dx \) and \( dy \) are small changes along the path. In the problem, the curve \( C \) is the triangular route from \((0,0)\) to \((1,0)\), then to \((0,1)\), and back to \((0,0)\).
This concept is vital for understanding how the force field interacts with the path over which the particle is moved.
Double Integral
A **double integral** involves integrating over two dimensions. In this exercise, we use it to calculate the work done over a region \( R \) enclosed by the curve \( C \). We convert a line integral into a double integral using Green's Theorem.
The double integral of the expression \( y^2 - x \) over the region \( R \) is represented by:
  • \( \int\int_R (y^2 - x) \, dA \)
The bounds for this double integral span the limits of the triangular region described, from 0 to 1 along the x-axis, and within each section defined by the line \( y = 1 - x \).
Breaking it down into a nested integral, first integrate with respect to \( y \) and second with respect to \( x \), making sure calculations align with the region's boundaries. This approach is crucial to solve problems involving areas and forces over a region in two dimensions.
Vector Field
A **vector field** assigns a vector to every point in a subset of space. In this exercise, the vector field \( \mathbf{F}(x, y) = x(x+y) \mathbf{i} + xy^2 \mathbf{j} \) describes the force applied to a particle at any point \( (x, y) \) in the plane.
Understanding vector fields is crucial for tracking how a force influences an object moving through different paths.
For instance:
  • The component \( x(x + y) \mathbf{i} \) shows how the force acts in the x-direction.
  • The component \( xy^2 \mathbf{j} \) affects movement in the y-direction.
Visualizing the vector field can aid in understanding how these forces impact objects moving along the curve \( C \) and how they contribute to the total work done.
Partial Derivatives
**Partial derivatives** come into play when dealing with functions of multiple variables. They represent how a function changes as each variable changes while keeping others constant.
In Green's Theorem, partial derivatives help in converting a line integral to a double integral. Specifically,
  • \( \frac{\partial Q}{\partial x} = y^2 \)
  • \( \frac{\partial P}{\partial y} = x \)
These calculations isolate how each component of the vector field changes independently at a point.
When plugged back into Green's Theorem, joint variations inform the change over the region \( R \), contributing to the final integral and hence the work done by the force field. Partial derivatives thus simplify and resolve complex multidimensional integration problems, forming the backbone of many multivariable calculus applications.

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Most popular questions from this chapter

We have seen that all vector fields of the form \(\mathbf{F}=\nabla g\) satisfy the equation curl \(\mathbf{F}=\mathbf{0}\) and that all vector fields of the form \(\mathbf{F}=\) curl \(\mathbf{G}\) satisfy the equation div \(\mathbf{F}=0\) (assuming continuity of the appropriate partial derivatives). This suggests the question: Are there any equations that all functions of the form \(f=\) div \(\mathbf{G}\) must satisfy? Show that the answer to this question is "No" by proving that every continuous function \(f\) on \(\mathbb{R}^{3}\) is the divergence of some vector field. \([\)Hint: Let \(\mathbf{G}(x, y, z)=\langle g(x, y, z), 0,0\rangle,\) where \( g(x, y, z)=\int_{0}^{x} f(t, y, z) d t ]\)

(a) Show that the parametric equations \(x=a \cosh u \cos v\) \(y=b \cosh u \sin v, z=c \sinh u,\) represent a hyperboloid of one sheet. (b) Use the parametric equations in part (a) to graph the hyperboloid for the case \(a=1, b=2, c=3 .\) (c) Set up, but do not evaluate, a double integral for the sur- face area of the part of the hyperboloid in part (b) that lies between the planes \(z=-3\) and \(z=3 .\)

Prove each identity, assuming that \(S\) and \(E\) satisfy the conditions of the Divergence Theorem and the scalar functions and components of the vector fields have continuous second-order partial derivatives. $$ \iint_{S}(f \nabla g) \cdot \mathbf{n} d S=\iiint_{E}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V $$

\(19-30\) Evaluate the surface integral \(\iint_{S} \mathbf{F} \cdot d S\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S .\) In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. $$\mathbf{F}(x, y, z)=x y \mathbf{i}+y z \mathbf{j}+z x \mathbf{k}, \quad $$ S is the part of the paraboloid \(z=4-x^{2}-y^{2}\) that lies above the square \(0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1,\) and has upward orientation

\(33-36\) Find an equation of the tangent plane to the given para- metric surface at the specified point. If you have software that graphs parametric surfaces, use a computer to graph the surface and the tangent plane. $$x=u+v, \quad y=3 u^{2}, \quad z=u-v ; \quad(2,3,0)$$

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