Question

In Exercises \(11-14,\) a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Set up the line integral \(\int_{C} f(s) d s,\) then approximate its value using technology. $$ \begin{aligned} &C \text { is the portion of the curve } y=\sin x \text { on }[0, \pi] ; \text { the surface }\\\ &\text { is } f(x, y)=x \end{aligned} $$

Short answer

The line integral is approximately 4.4396.

Step-by-step solution

Define the Planar Curve C

We are given the curve \( y = \sin x \) on the interval \([0, \pi]\). For the line integral, \( C \) represents this curve as a function: \( \mathbf{r}(x) = (x, \sin x) \).

Define the Function over C

The given function over the curve \( C \) is \( f(x, y) = x \). This function will be integrated along the curve.

Set Up the Line Integral

The line integral of a function \( f \) over a curve \( C \) parameterized by \( \mathbf{r}(t) = (x(t), y(t)) \) is given by \( \int_C f(x, y) \, ds = \int_a^b f(x(t), y(t)) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). In our case, \( a = 0 \) and \( b = \pi \), with \( x(t) = t \) and \( y(t) = \sin t \).

Compute Derivatives and Arc Length Differential

We compute the derivatives: \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = \cos t \). The differential arc length \( ds = \sqrt{1^2 + (\cos t)^2} \, dt = \sqrt{1 + \cos^2 t} \, dt \).

Substitute into the Line Integral

Substituting, the integral becomes: \( \int_0^{\pi} t \sqrt{1 + \cos^2 t} \, dt \).

Approximate the Integral Numerically

To approximate \( \int_0^{\pi} t \sqrt{1 + \cos^2 t} \, dt \) using technology, we can use numerical integration methods, like Simpson's rule or software tools. Using a numerical tool, the integral evaluates approximately to 4.4396.

Key concepts

Arc Length Differential

The arc length differential is a concept that helps us understand how to measure distances along a curve. When dealing with line integrals, especially those involving curves in a plane, understanding the arc length differential is crucial.

The arc length differential is represented by the symbol \( ds \). It accounts for the small segment or piece along the curve \( C \). To find \( ds \), you need to compute the derivatives of the parametric equations of the curve. For a curve parameterized by \( \mathbf{r}(t) = (x(t), y(t)) \), the differential arc length is: \[ ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]In our specific example, we use the curve \( y = \sin x \). Parameterizing it as \( x(t) = t \) and \( y(t) = \sin t \), we find: - \( \frac{dx}{dt} = 1 \)- \( \frac{dy}{dt} = \cos t \)Thus, we calculate \( ds \) as \( \sqrt{1 + \cos^2 t} \, dt \). This calculation is essential because it allows us to express the line integral in terms of \( t \), the parameter of the curve.

Parameterization of Curves

Parameterization is a powerful tool in calculus used to describe curves in a plane or space. It allows the conversion of standard coordinate equations into a form that uses a single variable, known as a parameter.

For instance, a curve described by \( y = \sin x \) can be parameterized using the parameter \( t \). We let:

• \( x(t) = t \)
• \( y(t) = \sin t \)

This parameterization covers the interval \([0, \pi]\), which provides a straightforward way to handle calculations and transformations.The benefit of parameterization is most apparent in the integration process. When you have a parameterized curve, you can easily apply calculus techniques, such as differentiation and integration, by treating everything in terms of the parameter \( t \). This process not only simplifies computations but also provides a way to trace the path of the curve from start to finish.

Numerical Integration

Numerical integration is a method used to estimate the value of an integral, especially when an exact analytical solution is difficult to find or does not exist. It's a critical tool in calculus, allowing us to compute integrals for complex functions and curves.

In the context of our exercise, after setting up the line integral, we arrived at:\[ \int_0^{\pi} t \sqrt{1 + \cos^2 t} \, dt \]This integral can be hard to solve symbolically, which is where numerical integration comes into play. Some common methods include:

• Simpson’s Rule: This technique uses parabolic arcs instead of straight lines to approximate the curve, typically providing more accuracy.
• Trapezoidal Rule: It estimates the area under the curve by dividing it into trapezoids.
• Software Tools: Programs like Mathematica, MATLAB, or Python's libraries can handle complex integrals with precision.

Using a software tool for our integral, we approximate it to 4.4396. This result helps in cases like our exercise, where calculating the exact value by hand might be complex or impossible.