Problem 28 - Thomas Calculus

Question

In Exercises \(27-30,\) integrate \(f\) over the given curve. $$ \begin{array}{l}{f(x, y)=\left(x+y^{2}\right) / \sqrt{1+x^{2}}, \quad C : \quad y=x^{2} / 2 \text { from }(1,1 / 2) \text { to }} \\ {(0,0)}\end{array} $$

Step-by-Step Solution

Answer

The integral value is \( \frac{11}{20} \).

Step 1: Parameterize the Curve

The curve is given by the equation \( y = \frac{x^2}{2} \). Therefore, we can parameterize the curve as \( \mathbf{r}(t) = (t, \frac{t^2}{2}) \) where \( t \) ranges from 0 to 1, covering the points from \((0,0)\) to \( (1, \frac{1}{2}) \).

Step 2: Find the Differential Arc Length

The differential arc length along the curve parameterized by \( \mathbf{r}(t) = (t, \frac{t^2}{2}) \) is given by \( ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt \). Compute the derivatives: \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = t \), leading to \( ds = \sqrt{1 + t^2} \, dt \).

Step 3: Substitute the Parameterization into the Function

Substitute \( x = t \) and \( y = \frac{t^2}{2} \) into the function \( f(x, y) = \frac{x + y^2}{\sqrt{1+x^2}} \) to get \( f(t, \frac{t^2}{2}) = \frac{t + (\frac{t^2}{2})^2}{\sqrt{1+t^2}} = \frac{t + \frac{t^4}{4}}{\sqrt{1+t^2}} \).

Step 4: Set Up the Integral

The integral along the curve can be set up using the parameterization and differential arc length: \[ \int_C f(x,y) \, ds = \int_0^1 \frac{t + \frac{t^4}{4}}{\sqrt{1+t^2}} \cdot \sqrt{1 + t^2} \, dt \]Simplifying inside the integral, this becomes \[ \int_0^1 \left( t + \frac{t^4}{4} \right) \, dt \]

Step 5: Solve the Integral

Evaluate the integral from Step 4 by integrating each term separately:\[ \int_0^1 t \, dt + \int_0^1 \frac{t^4}{4} \, dt \]Calculate:\[ \int_0^1 t \, dt = \frac{1}{2} \left[t^2\right]_0^1 = \frac{1}{2} \]\[ \int_0^1 \frac{t^4}{4} \, dt = \frac{1}{4} \cdot \frac{1}{5} \left[t^5\right]_0^1 = \frac{1}{20} \]Add these results to get:\[ \frac{1}{2} + \frac{1}{20} = \frac{11}{20} \]

Step 6: Conclusion

Thus, the value of the curve integral of the function \( f \) over the given curve is \( \frac{11}{20} \).

Key Concepts

Parameterization

Parameterization is the process of expressing a curve or surface in terms of one or more parameters, making it easier to perform calculations on or analyze the curve. In our exercise, the curve is defined by the function \( y = \frac{x^2}{2} \) which forms a parabolic path from the point \((0,0)\) to \((1, \frac{1}{2})\). To parameterize this curve, we use the parameter \( t \), letting \( x = t \) and consequently, \( y = \frac{t^2}{2} \). Therefore, the parameterized curve can be written as \( \mathbf{r}(t) = (t, \frac{t^2}{2}) \). This expression allows us to evaluate the behavior of the curve over the interval \([0, 1]\).
Parameterization is particularly useful because it transforms the curve into a format that facilitates the integration process. It converts a potentially complex curve into a series of simpler equations dependent upon the parameter \( t \). As a result, using parameterization helps in simplifying the process of integrating functions over curves.

Differential Arc Length

The differential arc length, denoted as \( ds \), gives us a way to approximate the length of a small segment of a curve. When integrating over a curve, the arc length is crucial as it helps take into account the curve's path rather than just a straight line.
For a curve parameterized by \( \mathbf{r}(t) = (t, \frac{t^2}{2}) \), the differential arc length is determined using the formula:

• First, compute the derivatives: \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = t \).
• Then, calculate \( ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt = \sqrt{1 + t^2} \, dt \).

This formula determines the length of a small segment \( ds \) of the curve, taking into account both the horizontal and vertical changes as \( t \) varies. Using the expression for \( ds \) is vital when setting up integrals over curved paths, as it ensures we are accurately representing the curve’s geometry.

Definite Integration

Definite integration is the process of calculating the integral of a function over a specified interval, resulting in a number that represents the area under the curve within that interval. In the context of curve integration, definite integration lets us determine the total contribution of the function across the curve's path.
After parameterization and finding the differential arc length, we can form the definite integral for our problem:

• We substitute the parameterized values into the function \( f(x, y) = \frac{x + y^2}{\sqrt{1+x^2}} \), arriving at \( f(t, \frac{t^2}{2}) = \frac{t + \frac{t^4}{4}}{\sqrt{1+t^2}} \).
• Combining this with the differential arc length, our integral becomes: \[ \int_0^1 \left( t + \frac{t^4}{4} \right) \, dt \]
• This integral is then split into two simpler integrals: \[ \int_0^1 t \, dt + \int_0^1 \frac{t^4}{4} \, dt \]
• Each of these integrals can be evaluated separately, yielding \( \frac{1}{2} \) and \( \frac{1}{20} \), respectively.

Summing these results gives us the total value of the integral, \( \frac{11}{20} \). This final result represents the cumulative effect of the function \( f \) across the defined curve, taking into account not just the values of \( f \) itself but also the specific path \( C \) is following.