Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{F}=2 z \mathbf{i}+2 y \mathbf{k}\), and let \(\partial S\) be the intersection of the cylinder \(x^{2}+y^{2}=a y\) with the hemisphere \(z=\sqrt{a^{2}-x^{2}-y^{2}}\), \(a>0\). Assuming distances in meters and force in newtons, find the work done by the force \(\mathbf{F}\) in moving an object around \(\partial S\) in the counterclockwise direction as viewed from above.

Step-by-Step Solution

Answer

The work done is zero because the force vector is perpendicular to the motion.

Step 1: Identify the expression for force

The force vector is given by \( \mathbf{F} = 2z \mathbf{i} + 2y \mathbf{k} \). This represents a vector field where the components are functions of the position coordinates \( x, y, z \).

Step 2: Understand the geometry of \( \partial S \)

\( \partial S \) is the curve of intersection between the cylinder \( x^2 + y^2 = ay \) and the hemisphere \( z = \sqrt{a^2 - x^2 - y^2} \). This intersection forms a closed loop which lies in the plane \( z = 0 \) (as it is a boundary of the hemisphere).

Step 3: Find parameterization of \( \partial S \)

Convert the cylinder equation \( x^2 + y^2 = ay \) to polar coordinates to simplify: set \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Thus, \( r^2 = ar \sin(\theta) \) implies \( r = a \sin(\theta) \).

Step 4: Parameterize the curve in terms of \( \theta \)

The intersection curve can thus be parameterized as \( x(\theta) = a \sin(\theta) \cos(\theta) \), \( y(\theta) = a \sin^2(\theta) \), and \( z = 0 \).

Step 5: Express \( \mathbf{F} \) along \( \partial S \)

At \( z = 0 \) (on \( \partial S \)), the force \( \mathbf{F} \) reduces to \( \mathbf{F} = 2y \mathbf{k} \). Along \( \partial S \), since \( y = a \sin^2(\theta) \), the force becomes zero because \( \mathbf{k} \) represents the vertical direction, and \( \partial S \) is in the horizontal plane.

Step 6: Calculate the work done

The work done by a force \( \mathbf{F} \) around a closed path is given by the line integral \( W = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} \). Here, since \( \mathbf{F} \) has no horizontal component (and is zero at \( \partial S \)), the line integral evaluates to zero: \( W = 0 \).

Step 7: Conclude about work

Since the line integral of the force \( \mathbf{F} \) along the path \( \partial S \) results in zero, the total work done by this force in moving an object around \( \partial S \) is zero.

Key Concepts

Line Integrals

In vector calculus, a line integral is an essential concept for calculating the work done by a force along a path. Imagine you've got a path and you're interested in finding how much a force works to move an object along this route. This is where line integrals come into play.
In general, the line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is expressed as \( \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) represents a small segment of the curve, and \( \cdot \) signifies the dot product. This definition informs you that the work is essentially the accumulation of the dot product of the vector field with each tiny segment along the path.
In practical terms, if the force has no component along the direction you're moving, then no work is done. Even if the vector might look daunting at first, understanding its direction on your path solves half the puzzle.
In our example, even though the vector field might seem complex initially, when converted into workable expressions on the path \( \partial S \), it simplified to zero. The vertical component \( \mathbf{k} \) did not contribute to the horizontal motion, leading to zero work done.

Vector Fields

Vector fields are mathematical constructs used to assign a vector to every point in space. They can represent multiple physical phenomena such as wind velocity, gravitational fields, or, as in our example, a force field.
Understanding vector fields requires visualizing how vectors, defined by direction and magnitude, interact with the space around them. In our case, the vector field \( \mathbf{F} = 2z \mathbf{i} + 2y \mathbf{k} \) means:

• There is both a \( z \) component, aligned with the \( i \) direction, and a \( y \) component, aligned with the \( k \) direction.
• The position of the object influences these component values along the given path.

In practice, vector fields are used by engineers and physicists to model and predict the behavior of fields, such as how forces change in a space. Zeroing in on their specific direction or path constraints often simplifies their contribution, like turning the vector field into zero effect on the curve \( \partial S \) in our solution.

Cylinders and Spheres

Cylinders and spheres are common geometric shapes used in vector calculus to describe physical boundaries or regions in space. Understanding their intersections, projections, and the space they enclose is essential for tasks like calculating line integrals or evaluating vector fields.
In this problem, we have a unique case of the intersection between a cylinder and a hemisphere. The cylinder is described by \( x^2 + y^2 = ay \), while the hemisphere is presented by \( z = \sqrt{a^2 - x^2 - y^2} \). These shapes form a kind of closed loop as they intersect, which is critical in defining \( \partial S \), the path for our line integral.

• A cylinder generally wraps around in a consistent manner based on its equation.
• A hemisphere includes a part of a sphere cut by a plane, with all points at or below a specific \( z \)-value.

Understanding these intersections provides valuable insight into relevant boundary paths and aids in using line integrals effectively. In our example, realizing that the curve of intersection was on the boundary of the hemisphere at \( z = 0 \) simplified finding the parameterization and evaluating the actual intersection. This simplification was a key step in solving our calculus exercise.