Problem 24 - Calculus: Early Transcendental Functions

Question

Evaluate the line integral. \(\int_{C} z d s,\) where \(C\) is the intersection of \(x^{2}+y^{2}=4\) and \(z=0\) (oriented clockwise as viewed from above)

Step-by-Step Solution

Answer

The value of the line integral \(\int_{C} z ds\) is 0.

Step 1: Identify the Curve

We're looking for the intersection of the 2 surfaces: \(x^{2}+y^{2}=4\) and \(z=0\). This is a circle in the xy-plane with a radius of 2.

Step 2: Parametrize the Curve

Given this is a circle oriented clockwise when viewed from above, a suitable parametrization could be \(x=2 \cos(t)\), \(y=-2 \sin(t)\), and \(z=0\) for \(t\) in the interval \([0, 2 \pi]\).

Step 3: Compute the Differential ds

The magnitude of vector's \(C\) derivative is \(ds\). For a parameter \(t\), the differential \(ds = ||\textbf{r}'(t) ||dt = |-2\sin(t)|dt + |2\cos(t)|dt = 2 dt\).

Step 4: Compute the Line Integral

Now, substitute the function and differential ds into the line integral \(\int_{C} z ds\). Because \(z = 0\), our integral becomes \(\int_{C} 0 \cdot ds = 0\).

Key Concepts

Parametrization

When dealing with line integrals, the concept of parametrization is essential. Parametrization allows us to describe curves and paths in a floorless manner using parameters rather than coordinates directly. In the given problem, the curve of interest is a circle in the xy-plane, defined by the intersection of two surfaces: a cylinder, represented by the equation \(x^2 + y^2 = 4\), and a plane, \(z = 0\).

For our circle, we choose a parameter \(t\) that varies over an interval, mapping out each point on the circle. To parametrize the circle given the equation \(x^2 + y^2 = 4\):

• \(x\) is defined as \(2 \cos(t)\), turning the parameter \(t\) into the angular position around the circle's center (0, 0) with a radius of 2.
• \(y\) is set as \(-2 \sin(t)\), aligning with the negative sine function because the circle is traversed clockwise when viewed from above.
• \(z\) remains 0, consistent with the circle lying entirely in the xy-plane.

This parametrization forms the basis for computing line integrals over complex paths by translating them into integrals over an interval of the parameter, \(t\), which runs from 0 to \(2\pi\).

Integration on Curves

Integration on curves, or line integrals, involve integrating a function along a specified path or curve. The integral evaluates how the function accumulates over the curve, taking into consideration factors like direction and segment length.

In our exercise, we evaluate the line integral \(\int_C z \ ds\) along the circle \(C\) that is defined by \(x^2 + y^2 = 4\) and \(z=0\). The differential \(ds\) represents the arc length segment of the curve, accounting for the path's geometry. Here's how we computed it:

• Calculate the velocity vector: Derive from the parametrization \(\vec{r}(t) = (2 \cos(t), -2 \sin(t), 0)\) to get \(\vec{r}'(t) = (-2 \sin(t), -2 \cos(t), 0)\).
• Find the magnitude of the velocity vector: Use \( ||\vec{r}'(t)|| = \sqrt{(-2 \sin(t))^2 + (-2 \cos(t))^2} = 2 \).
• Express \(ds\) in terms of \(t\): Thus, \(ds = 2 \, dt\).

Since the \(z\) component of the function is zero, the line integral boils down to zero, because multiplying any value of the path by zero results in zero contribution.

Circle in the xy-plane

A circle in the xy-plane is a basic geometric entity, frequently encountered in both pure geometry and in multivariable calculus problems. In our exercise, the circle arises from the intersection of the cylinder \(x^2 + y^2 = 4\) (a vertical right circular cylinder with radius 2) and the plane \(z = 0\).

This intersection yields a circle located entirely within the xy-plane, centered at the origin (0,0) with a radius of 2:

• The equation \(x^2 + y^2 = 4\) assures every point on the circle is equidistant (2 units) from the origin.
• Since \(z = 0\), the circle lies flat in the xy-plane, contributing no height (\(z\) value) to the integral.

Understandably, this makes circle parametrizations simple, typically leveraging trigonometric functions like sine and cosine to encode rotational symmetry. This facilitates computations across various applications, including calculating line integrals, where the circular path engagement is direct and often simplified due to symmetrical properties.