91Ó°ÊÓ

A vector field is a function that assigns a vector to each point in a space. In the context of two-dimensional space, a vector field can be thought of as a set of arrows where each arrow represents both a direction and a magnitude at different points in the plane. In this exercise, the vector field is given by \( \mathbf{F}(x, y) = (x^{3/2} - 3y) \mathbf{i} + (6x + 5\sqrt{y}) \mathbf{j} \).
To better understand, imagine a wind map showing wind direction and speed across a region. The vector field is similar: at each point \( (x, y) \), the vector field specifies a "wind" vector with components \( P = x^{3/2} - 3y \) and \( Q = 6x + 5\sqrt{y} \).
This concept of vector fields is essential while applying Green's Theorem, which relates the line integral around a closed path to a double integral over the region enclosed by the path.

In physics, the work done by a force field on a moving particle along a path is the integral of the force over the path. In vector calculus, this involves integrating the vector field along a chosen path.
Green's Theorem simplifies this task by converting the line integral into a double integral over the region enclosed by the path. For this exercise, the path \( C \) is the boundary of a triangle. Applying Green's Theorem, the work done by the force \( \mathbf{F} \) is expressed as:

• The line integral \( \oint_C \mathbf{F} \cdot \mathbf{dr} \)
• Transformed into the double integral of \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \) over the region \( R \)

This conversion allows us to break down the problem into simpler operations involving partial derivatives, making calculations more straightforward.

Partial derivatives express how a function changes as only one of the input variables changes, keeping the others constant. In the context of Green's Theorem, partial derivatives help determine the rate of change of vector field components.
For this problem, understanding and calculating the partial derivatives of the function components \( P \) and \( Q \) is vital:

• The partial derivative of \( Q \) with respect to \( x \) is computed as \( \frac{\partial Q}{\partial x} = 6 \).
• The partial derivative of \( P \) with respect to \( y \) is \( \frac{\partial P}{\partial y} = -3 \).

By using these derivatives, Green's Theorem forms the relation:

• \( \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \), which simplifies to \( \iint_R 9 \, dA \).

This shows how partial derivatives help shift the focus from a line integral around a path to a simpler area integral, leveraging the properties of the vector field.

The area of the region over which the integral is evaluated plays a crucial role in Green's Theorem. In this exercise, the triangle with vertices \( (0,0), (5,0), \) and \( (0,5) \) is the region of concern.
This triangle is a right triangle, with base and height aligning with axes, making the area calculation straightforward:

• The base length is 5 units, and the height is 5 units.

The formula for the area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \), which for this triangle gives an area of \( 12.5 \) square units.
This area is then used to evaluate the double integral \( \iint_R 9 \, dA \), further calculating the work done by using \( 9 \times 12.5 = 112.5 \). Hence, understanding the geometric properties of the region simplifies evaluating the work done over the path.