Question

In Exercises \(5-10\), a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Evaluate the line integral \(\int_{C} f(s) d s\) $$ \begin{aligned} &C \text { is the circle with radius } 2 \text { centered at the point }(4,2) ; \text { the }\\\ &\text { surface is } f(x, y)=3 x-y \end{aligned} $$

Short answer

The value of the line integral is \(40\pi\).

Step-by-step solution

Parameterize the Circle

The circle centered at \((4,2)\) with a radius of 2 can be parameterized using the parametric equations: \( x(t) = 4 + 2\cos(t) \) and \( y(t) = 2 + 2\sin(t) \), where \( t \) ranges from \( 0 \) to \( 2\pi \).

Substitute Parametrization into f

Substitute the parametric equations into the function \( f(x, y) = 3x - y \):\[f(x(t), y(t)) = 3(4 + 2\cos(t)) - (2 + 2\sin(t)) = 12 + 6\cos(t) - 2 - 2\sin(t) = 10 + 6\cos(t) - 2\sin(t)\]

Find ds in Terms of dt

The differential \( ds \) for the parameterized path is given by the magnitude of the derivative of the position vector \( r(t) = (x(t), y(t)) \):\[dx = -2\sin(t)\,dt \dy = 2\cos(t)\,dt \ds = \sqrt{(-2\sin(t))^2 + (2\cos(t))^2} dt = \sqrt{4\sin^2(t) + 4\cos^2(t)} dt = 2 dt\]

Evaluate the Integral

Substitute \( f(x(t), y(t)) \) and \( ds \) into the line integral:\[\int_{0}^{2\pi} (10 + 6\cos(t) - 2\sin(t))(2) \, dt = \int_{0}^{2\pi} (20 + 12\cos(t) - 4\sin(t)) \, dt\]This can be split into separate integrals:\[\int_{0}^{2\pi} 20 \, dt + \int_{0}^{2\pi} 12\cos(t) \, dt - \int_{0}^{2\pi} 4\sin(t) \, dt\]The second and third integrals are zero because the integrals over a full period of \( \cos(t) \) and \( \sin(t) \) are zero. So it simplifies to:\[= 20(2\pi) = 40\pi\]

Key concepts

Parameterization of Curves

When dealing with line integrals over a curve, one of the first steps is parameterizing the curve. This means expressing the curve in terms of a single parameter, usually denoted as \( t \), which simplifies the integration process. Given a geometric shape, like a circle, parameterization involves finding equations that describe every point on the curve using \( t \).
For example, a circle with radius 2 centered at point \((4, 2)\) can be parameterized using the following equations:

• \( x(t) = 4 + 2\cos(t) \)
• \( y(t) = 2 + 2\sin(t) \)

Here, \( t \) ranges from \( 0 \) to \( 2\pi \), which corresponds to the angles around the circle. It is a compact and effective way of handling shapes that might otherwise require complex descriptions. Using parameterization, the curve \( C \) is fully described by altering \( t \).

Vector Calculus

Vector calculus is a powerful mathematical tool that plays a crucial role when dealing with integrals along curves and surfaces. In the context of a line integral over a curve, vector calculus helps us to integrate vector fields over paths. This can involve calculating the dot product of a vector field with a differential path vector, and then integrating along a defined path.
One key element is determining the path length differential \( ds \), which involves finding the differential of the parameterized curve's position function. For the parameterized circle:

• The derivative of the position vector \( r(t) = (x(t), y(t)) \)
• \( \frac{dx}{dt} = -2\sin(t) \), \( \frac{dy}{dt} = 2\cos(t) \)

Its magnitude, \( ds \), is found using the Pythagorean sum, such that:\[ ds = \sqrt{\left(-2\sin(t)\right)^2 + \left(2\cos(t)\right)^2} dt = 2 dt \]
This ensures that the function \( ds \) correctly accounts for any differential changes in the curve's position due to \( t \).

Integrals Involving Trigonometric Functions

When evaluating line integrals, especially in parametrized curves in vector calculus, we're often tasked to compute integrals that involve trigonometric functions. This occurs due to the parameterization of curves such as circles using trigonometric functions like sine and cosine.
For example, when evaluating the integral \( \int_{0}^{2\pi} (10 + 6\cos(t) - 2\sin(t))(2) \, dt \), we end up with separate components due to the presence of sine and cosine:

• \( \int_{0}^{2\pi} 12\cos(t) \, dt \)
• \( \int_{0}^{2\pi} 4\sin(t) \, dt \)

Fortunately, these types of integrals have properties that simplify calculations:

• \( \int_{0}^{2\pi} \cos(t) \, dt = 0 \)
• \( \int_{0}^{2\pi} \sin(t) \, dt = 0 \)

Thus, the trigonometric portions of the integral can often be negated, simplifying results significantly. For full circle integrations, the results are even characteristic as some of them involve complete trigonometric period evaluations, which lead to zero contributions to the final line integral value.