Question

Use Green's Theorem to calculate the work done by the force \(F\) on a particle that is moving counterclockwise around the closed path \(C\). $$ \mathbf{F}(x, y)=\left(3 x^{2}+y\right) \mathbf{i}+4 x y^{2} \mathbf{j} $$ \(C:\) boundary of the region lying between the graphs of \(y=\sqrt{x}, y=0,\) and \(x=9\)

Short answer

The numerical solution to the double integral represents the work done by the force on a particle moving counterclockwise around the closed path \(C\). Use appropriate integral calculation techniques to find the exact value.

Step-by-step solution

Understanding the Given Information

The force field is given as \(F(x, y) = (3x^2+y)i + 4xy^2j\). The path \(C\) corresponds to the boundary of the region between the graphs of \(y=\sqrt{x}, y=0,\) and \(x=9\). This boundary is a counterclockwise path.

Express the Force Field in Divergence Form

Express the force field \(F(x, y)\) in the form \(Pdx + Qdy\). Here, \(P = 3x^2 + y\) and \(Q = 4xy^2\). These are the components of the force.

Apply Green's Theorem

According to Green's Theorem, the work done by the force on a particle moving counterclockwise around the closed path \(C\) is given by \(\int\int_D (Q_x - P_y)dA\), where \(D\) is the region enclosed by \(C\), \(P_y\) is the partial derivative of \(P\) with respect to \(y\), and \(Q_x\) is the partial derivative of \(Q\) with respect to \(x\). Calculate the partial derivatives \(P_y\) and \(Q_x\). We get \(P_y = 1\) and \(Q_x = 4y^2\).

Evaluate the Double Integral

Substitute these results into the equation from Green's theorem and then calculate the double integral \(\int\int_D (Q_x - P_y)dA\). Define the limits of the double integral based on the closed path \(C\), which is defined as the boundary between the graphs of \(y=\sqrt{x}, y=0,\) and \(x=9\). Therefore, the limits for \(x\) are from \(0\) to \(9\) and for \(y\) are from \(0\) to \(\sqrt{x}\). The double integral therefore becomes \(\int_0^9 \int_0^\sqrt{x} (4y^2 - 1) dy dx\). Evaluating this integral yields a numerical result for the work done by the force.

Key concepts

Work Done by Force

Imagine pushing a shopping cart along a specific path in a supermarket. The effort you exert in moving the cart against certain resistances, like friction, is analogous to the work done by a force in the field of physics. In a more technical sense, when a force is applied to an object, causing the object to move, work is done on the object.

In the context of vector calculus, this work can be calculated using a tool called a line integral. The line integral along a curve adds up the work done by a force field when an object travels along a certain path. For the given exercise, the force acting on a particle is moving it around a closed path defined by the graphs of certain functions. Here, Green's Theorem is used as a clever means to turn this typically complex line integral problem into an easier double integral over the region enclosed by the path.

The work done by a force in this case is not only conceptual; it can carry tangible results in fields such as engineering and physics, answering questions like how much energy a field imparts on a moving charge or the amount of work required to move an object in a gravitational field.

Line Integrals

The concept of line integrals is crucial for students entering the realm of multivariable calculus. Just as we sum up values along an interval on a number line in a regular integral, a line integral extends this concept to adding up values along a curve in the plane or space.

A key application of line integrals is calculating work, as seen in our exercise. The line integral of a vector field along a curve gives us the total work done by the force represented by the vector field as the object moves along the curve. The computation can be a bit more involved since we are dealing with curves in multiple dimensions, which is why the introduction of Green's Theorem simplifies the process by converting a line integral around a closed curve into a double integral over the area it encloses.

Vector Fields

A vector field is visualized as a grid where at each point in space a vector is assigned. Imagine standing on a weather map, where each point has an arrow showing the wind direction and strength; this is the essence of a vector field in meteorology.

For our textbook exercise, we're dealing with a vector field created by a force, where each vector represents the force acting on a particle at that point in space. The vectors in our specific example took the form \(\mathbf{F}(x, y)=(3x^2+y)\mathbf{i}+4xy^2\mathbf{j}\). Understanding vector fields is key to many areas of science and engineering, including fluid dynamics and electrodynamics, as they allow us to model physical phenomena that vary across space.

Double Integrals

If we want to add up values over a two-dimensional region, we turn to double integrals. They are the two-dimensional counterpart to single integrals and a foundational brick in the building of multivariable calculus. Imagine painting a floor; the amount of paint needed would be found by multiplying the length and width -- this simple multiplication is like a double integral over a rectangle.

In the exercise, Green's Theorem allows the conversion of a line integral into a double integral, enabling us to cumulatively add up values over the region enclosed by the curve. Executing a double integral involves calculating the integral over one variable and then integrating that result over the second variable, in this case over the domain defined by \(y=\sqrt{x}\), \(y=0\), and \(x=9\). This provides a powerful way to solve complex problems involving areas, volumes, and, as with our exercise, work done by forces.