Problem 11 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. \(\int_{C}\left(x y+x^{2}\right) d x+x^{2} d y ; C:\) the parabola \(y=2 x^{2}\) from the origin to the point \((1,2)\).

Step-by-Step Solution

Answer

\(\frac{11}{6}\)

Step 1 - Parametrize the curve

The curve given is a parabola described by the equation \(y=2x^{2}\). Parametrize the curve by setting \(x=t\) for \(t \in [0,1]\). Thus, \(x=t\) and \(y=2t^2\).

Step 2 - Compute \(dx\) and \(dy\)

Differentiate the parametrization with respect to \(t\). Find \(dx/dt = 1\) so \(dx = dt\) and \(dy/dt = 4t\) so \(dy = 4t \, dt\).

Step 3 - Substitute into the integral

Substitute \(x\), \(y\), \(dx\), and \(dy\) into the integral: \[\int_{0}^{1} \left( t \(2t^2\) + t^{2} \right) dt + t^{2} (4t) dt = \int_{0}^{1} \left( 2t^3 + t^2 + 4t^3 \right) dt \]

Step 4 - Simplify the integrand

Simplify the integrand to get: \[\int_{0}^{1} 6t^3 + t^2 \, dt\right) \]

Step 5 - Integrate

Integrate each term separately: \[\int_{0}^{1} 6t^3 + t^2 \, dt = \left[ 6 \frac{t^4}{4} + \frac{t^3}{3} \right]_{0}^{1} = \left[ \frac{3}{2}t^4 + \frac{1}{3}t^3 \right]_{0}^{1} = \left( \frac{3}{2} + \frac{1}{3} \right) - (0) = \frac{11}{6} \]

Key Concepts

Parametrize the Curve

To evaluate a line integral, the first step is to parametrize the curve. This means describing the curve using a parameter, often denoted as t. For the given problem, the curve is a parabola, which can be expressed as the equation \(y=2x^2\). To parametrize, we set \(x=t\) where \(t\) ranges from 0 to 1. This helps us because now we can write \(x\) and \(y\) in terms of \(t\), so we have \(x=t\) and \(y=2t^2\). This transformation simplifies the process of evaluating line integrals.

Differentiation

Once the curve is parametrized, the next step is to find the differentials \(dx\) and \(dy\). Since we have \(x=t\), differentiating \(x\) with respect to \(t\) gives us \frac{dx}{dt}=1\(, so \)dx = dt\(. For \)y=2t^2\(, differentiating with respect to \)t\( gives us \frac{dy}{dt}=4t\), so \(dy=4t \, dt\). Differentiation is crucial because it allows us to substitute \(dx\) and \(dy\) in the integral, making it easier to work with the parameters rather than the original variables, especially in more complex integrals.

Integration

After substituting \(x\), \(y\), \(dx\), and \(dy\) into the integral, we proceed to the integration step. Here, we combine and simplify the terms under the integral to make the integral easier to solve. For the problem at hand, simplifying gives us \(\frac{11}{6}\) as the answer. Integrating involves finding the antiderivatives of the function with respect to the parameter \(t\) over the specified interval, followed by evaluating these antiderivatives at the boundaries of the interval. This is a fundamental step in solving line integrals as it allows us to find the total accumulated value along the curve.