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 \(C\). $$ \begin{aligned} &\mathbf{F}(x, y)=\left(x^{3 / 2}-3 y\right) \mathbf{i}+(6 x+5 \sqrt{y}) \mathbf{j}\\\ &C: \text { boundary of the triangle with vertices }(0,0),(5,0), \text { and }\\\ &(0,5) \end{aligned} $$

Short answer

First, calculate the curl of \( \mathbf{F} \) to get \(6 - \frac{3}{2}\sqrt{x}\). Then, use Green's theorem to compute the double integral of this curl over the triangle provided. Finally, evaluate the integral to find the numerical value.

Step-by-step solution

Recognize the type of problem

The exercise is about calculating the work done by a force \(F\) on a particle moving along a path \(C\). This is fundamentally an application of Green's Theorem in physics.

Setup

According to Green’s Theorem, work done by the force F along the closed path C is given by the double integral of the curl of \(\mathbf{F}\) over the region inside of \(C\), where curl is given as: \( curl~ \mathbf{F} = \frac{\partial}{\partial x}(6x + 5\sqrt{y}) - \frac{\partial}{\partial y}(x^{3/2} - 3y)\). Calculate this curl.

Calculate the Curl

Using the given force vector \( \mathbf{F}(x, y)=\left(x^{3 / 2}-3 y\right) \mathbf{i}+(6 x+5 \sqrt{y}) \mathbf{j} \), we find: \( curl~\mathbf{F} = \frac{\partial}{\partial x}(6x + 5\sqrt{y}) - \frac{\partial}{\partial y}(x^{3/2} - 3y) = (6 - 0) - (3/2 \sqrt{x} - 3) = 6 - \frac{3}{2}\sqrt{x}\).

Use Green’s Theorem to perform double integration

Now the work done by the force is given by: \( W = \int_{triangle} curl~\mathbf{F} ~dA \). Where the limits of integration over the triangle from (0,0) to (5,0) to (0,5) gives: \( W = \int_0^5 \int_0^{5-x} (6 - \frac{3}{2}\sqrt{x}) ~dy~dx \).

Evaluate the integral

Evaluate the double integral to find the work. The result is a numerical value that represents the work done by the force on the particle.

Key concepts

Curl of a Vector Field

Understanding the curl of a vector field is crucial when dealing with fluid flow, electromagnetism, and the work done by force fields. It's a vector operation that measures the rotation at a point in the field. Imagine setting a tiny paddlewheel at any point in a flowing fluid; if the paddlewheel begins to rotate, there's a curl at that point.

The formal definition uses partial derivatives, as seen in the exercise. To compute the curl of the given vector field \( \mathbf{F} \), you subtract the rate of change across one component from the rate of change across another. In our case, curl is expressed as \( curl~ \mathbf{F} = \frac{\partial}{\partial x}(6x + 5\sqrt{y}) - \frac{\partial}{\partial y}(x^{3/2} - 3y) \). When visualizing this, think of it as measuring how much the vector field is trying to rotate around a particular point.

Line Integrals

Next, let's explore line integrals, which come into play when we want to calculate things like the work done by a force field along a path. A line integral adds up a series of infinitesimally small quantities along a curve. It's as if you were walking along a path, collecting or spending energy based on the force you encounter at every tiny step.

In our exercise, the path is the boundary of a triangle, and the force field affects the particle traveling along this path. By integrating the force field along the curve, we calculate the total work done. Green's Theorem wonderfully simplifies the process by allowing us to evaluate the line integral as a double integral over the region enclosed by the curve.

Work Done by a Force

The work done by a force in physics is calculated by taking the dot product of the force vector with the displacement vector and integrating it along the path of motion. Work can also be computed as an integral of a force field along a path taken by an object, and it represents the energy transferred to or from the object.

In the context of Green's Theorem, we're looking at a closed path; hence the work done by the force vector field \( \mathbf{F} \) is equivalent to the double integral of the curl of \( \mathbf{F} \) over the area enclosed by the path. It translates the intricate path integral into an area integral, often simplifying the calculation, as seen in our solution steps.

Double Integrals

Finally, the concept of double integrals extends the notion of integration to two dimensions. When we perform a double integral, we're essentially adding up values over a two-dimensional area. Picture sweeping out an entire surface and accumulating a value, like volume or work, across this expanse.

The exercise's final steps involve setting up and evaluating a double integral to find the work done by the force. We integrate the curl function over the triangular region, which, thanks to Green's Theorem, corresponds to the work that would otherwise be calculated with a more complex line integral. This is where understanding the bounds of integration is key - we're summing values within the triangular path from \( (0,0) \) to \( (5,0) \) to \( (0,5) \) to get the final answer for the work done.