Problem 23 - Thomas Calculus

Question

Evaluate \(\int_{C} x y d x+(x+y) d y\) along the curve \(y=x^{2}\) from \((-1,1)\) to \((2,4)\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{193}{12}\).

Step 1: Understand the Problem

We are asked to evaluate the line integral \(\int_{C} x y d x+(x+y) d y\) along the curve given by \(y = x^2\) from the point \((-1, 1)\) to the point \((2, 4)\). This involves finding a parameterization of the curve and substituting into the integral.

Step 2: Parameterize the Curve

Since the curve is given by \(y = x^2\), we can use \(x\) as the parameter directly. Thus, the parameterization is \(\mathbf{r}(t) = (t, t^2)\), where \(t\) ranges from \(-1\) to \(2\).

Step 3: Compute Derivatives

Find the derivative \(\frac{d\mathbf{r}}{dt} = \left(1, 2t\right)\). This represents \(dx\) and \(dy\) as \(dx = dt\) and \(dy = 2t dt\).

Step 4: Substitute and Simplify the Integral

Substitute \(x = t\), \(y = t^2\), \(dx = dt\), and \(dy = 2t dt\) into the integral. The integrand becomes \[\int_C (t(t^2) \cdot dt) + (t + t^2)(2t dt) = \int_{-1}^{2} (t^3 + (t + t^2)2t) dt.\] Simplifying, this becomes \[\int_{-1}^{2} (t^3 + 2t^2 + 2t^3) dt = \int_{-1}^{2} (3t^3 + 2t^2) dt.\]

Step 5: Evaluate the Integral

Integrate \(\int_{-1}^{2} (3t^3 + 2t^2) dt\). The integral of \(3t^3\) is \(\frac{3}{4}t^4\) and the integral of \(2t^2\) is \(\frac{2}{3}t^3\). Therefore, evaluate \[\left[\frac{3}{4}t^4 + \frac{2}{3}t^3\right]_{-1}^{2}.\]

Step 6: Calculate the Definite Integral

Substituting the limits into the evaluated integral: - At \(t = 2\), the value is \(\frac{3}{4}(16) + \frac{2}{3}(8) = 12 + \frac{16}{3}\).- At \(t = -1\), the value is \(\frac{3}{4}(-1)^4 + \frac{2}{3}(-1)^3 = \frac{3}{4} - \frac{2}{3}\).Subtract the two results:\( (12 + \frac{16}{3}) - (\frac{3}{4} - \frac{2}{3})\). Simplify this to calculate the final integral value.

Step 7: Simplify the Result

Combine terms:\(12 + \frac{16}{3} - \frac{3}{4} + \frac{2}{3} = \frac{36}{3} + \frac{16}{3} + \frac{8}{12} = \frac{36+16}{3} + \frac{2}{3} \cdot 4 = \frac{52}{3} - \frac{1}{12}\). Convert to a common denominator and find the difference to finally compute the value of the integral.

Key Concepts

Parameterization of Curves

The parameterization of curves is a crucial concept in evaluating line integrals. It involves representing a curve using a continuous mapping from a parameter, typically denoted as \(t\), into the coordinate space. In this particular problem, we deal with the curve described by the equation \(y = x^2\), which means each point on the curve is determined by its \(x\)-coordinate.

To parameterize this curve, we choose \(x\) as the parameter itself, denoted by \(t\). This choice is straightforward because for each \(x\) value, there's a corresponding \(y\) value, specifically \(t^2\). Therefore, the parametric equations are:

• \(x = t\)
• \(y = t^2\)

Where \(t\) ranges from -1 to 2 to trace the curve from the starting point \((-1, 1)\) to the endpoint \((2,4)\). This parameterization simplifies the integration process by allowing easy substitution into the integral expression.

Calculus

Calculus plays a central role in understanding and solving problems involving line integrals. Specifically, calculus allows us to compute the integral of expressions over a path, which is useful for determining physical properties like work done by a force field along a path.

Once the curve is parameterized, we differentiate the parametric equations to find \(dx\) and \(dy\). These are crucial for performing the integration. For our parameterization \(\mathbf{r}(t) = (t, t^2)\), we find:

• \(\frac{dx}{dt} = 1\) implies \(dx = dt\)
• \(\frac{dy}{dt} = 2t\) implies \(dy = 2t dt\)

With these derivatives, we can substitute into the line integral \(\int_C x y dx + (x + y) dy\) and express it entirely in terms of the parameter \(t\). Calculus provides the tools for differentiating, substituting, and ultimately simplifying and solving these expressions through integration.

Definite Integrals

Definite integrals are used to calculate the area under a curve or the accumulation of quantity across a certain interval. They are crucial when evaluating line integrals, as in this exercise.

The parameterized and simplified line integral expression becomes \(\int_{-1}^{2} (3t^3 + 2t^2) \, dt\). To solve this, we apply the fundamental theorem of calculus, which involves finding the antiderivative (also known as the indefinite integral) and then evaluating it at the specified bounds.

Here, the antiderivatives are:

• \( \int 3t^3 \, dt = \frac{3}{4}t^4\)
• \( \int 2t^2 \, dt = \frac{2}{3}t^3\)

By computing these antiderivatives and evaluating them from \(t = -1\) to \(t = 2\), we arrive at the numerical value for the original line integral. Definite integrals, therefore, provide the final solution, converting our parametric and differentiated expressions into a tangible numerical result.