Problem 76 - Calculus Volume 3

Question

Find \(\int_{C} y^{2} d x+\left(x y-x^{2}\right) d y\) along \(C: y=3 x\) from (0,0) to (1,3).

Step-by-Step Solution

Answer

The line integral evaluates to 5.

Step 1: Parametrize the Curve

The curve is given by the equation \(y = 3x\). Parametrize this curve using \(x\) directly because the curve is expressed in terms of \(x\). As \(x o y = 3x\), we can write: \(r(t) = (t, 3t)\), where \(t\) ranges from 0 to 1.

Step 2: Differentiate the Parametrization

Differentiate the parametric equations to find \(dx\) and \(dy\). Since \(x = t\), we have \(dx = dt\). Since \(y = 3t\), differentiate with respect to \(t\) to get \(dy = 3dt\).

Step 3: Substitute in the Line Integral

Substitute \(x = t\), \(y = 3t\), \(dx = dt\), and \(dy = 3dt\) into the line integral: \[\int_C y^2 \, dx + (xy-x^2) \, dy = \int_0^1 (3t)^2 \, dt + (t \cdot 3t - t^2) \, (3dt) \]

Step 4: Simplify the Integrand

Calculate each part of the integral separately.For the \(y^2 dx\) term:\[\int_0^1 9t^2 \, dt\]Calculate:\[\int_0^1 9t^2 \, dt = 9 \int_0^1 t^2 \, dt = 9 \left[ \frac{t^3}{3} \right]_0^1 = 3\]For the \((xy-x^2) dy\) term:\[\int_0^1 (3t^2 - t^2) \, (3dt) = \int_0^1 6t^2 \, dt\]Calculate:\[\int_0^1 6t^2 \, dt = 6 \left[ \frac{t^3}{3} \right]_0^1 = 2\]

Step 5: Combine the Results

Sum the results from the components of the integral. \(\int_C y^2 \, dx + (xy - x^2) \, dy = 3 + 2 = 5\).

Key Concepts

Parametrization of Curves

When working with line integrals, it's often necessary to express a curve analytically before you can integrate across it. In our problem, the curve is described by the equation \(y = 3x\). A useful way to handle such a curve is to parametrize it. Parametrization involves expressing the variables \(x\) and \(y\) as functions of a new variable, often called \(t\), which makes the integration process more manageable.

In this case, since \(y\) is already expressed in terms of \(x\), we can let \(x = t\), which naturally gives \(y = 3t\). This parametrization makes sense because the points on the curve will trace from the origin (0,0) to (1,3) as \(t\) varies from 0 to 1.

The parametric representation \(r(t) = (t, 3t)\) simplifies integration. The integral could then be performed along this path by considering \(t\) in the specified range.

Differential Calculus

Differential calculus comes into play when dealing with parametric equations in line integrals. Once the curve has been parametrized as \(r(t) = (t, 3t)\), we need to determine \(dx\) and \(dy\) for integration purposes. Here, differentiation of the parametric equations with respect to \(t\) is key.

Since \(x(t) = t\), differentiating gives us \(dx = dt\). Similarly, from \(y(t) = 3t\), we differentiate to find \(dy = 3dt\). These differential elements, \(dx\) and \(dy\), are crucial because they tell us how changes in \(t\) affect the coordinates and thus how they contribute to the overall integral.

• You differentiate \(x(t)\) to get \(dx = dt\).
• You differentiate \(y(t)\) to get \(dy = 3dt\).

These expressions make substituting into the integral straightforward, paving the way for evaluating the integral over the given path.

Integrating with Respect to a Parameter

After parametrization and differentiation, we translate our original line integral into a parameterized line integral. This process transforms the original Cartesian integral into an integral with respect to the parameter \(t\), making it easier to evaluate.

In our scenario, we are dealing with the integral \[\int_C y^2 \, dx + (xy-x^2) \, dy\]Using the derived parametrization \(x = t\), \(y = 3t\), and their differentials \((dx = dt, dy = 3dt)\), we substitute into the original equation and simplify:

• For \(y^2 dx\), we substitute and integrate \((3t)^2 dt\).
• For \((xy-x^2) dy\), we have \((t \cdot 3t - t^2) \, (3dt)\) as the expression to integrate.

Now, integrate these simplified expressions over the range of \(t\) from 0 to 1:
- Evaluate \( \int_0^1 9t^2 \, dt \) yielding 3.
- Evaluate \( \int_0^1 6t^2 \, dt \) resulting in 2.
Finally, summing these results gives the value of the line integral as 5. This method is efficient and allows for systematic integration along complex paths.