Problem 1 - The Calculus with Analytic Geometry

Question

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

Step-by-Step Solution

Answer

The value of the integral is \( \frac{16}{3} \).

Step 1: Understand the Curve and Limits

The line integral is given over the curve y=x from the origin (0,0) to the point (2,2). This means for the parameter t, we have x=t and y=t with t ranging from 0 to 2.

Step 2: Parametrize the Curve

To evaluate the integral, parametrize the curve C. Here, x = t and y = t, so dx=dt and dy=dt.

Step 3: Substitute and Setup the Integral

Substitute x=t, y=t, dx=dt, and dy=dt into the line integral expression: dx term: \[\int_{0}^{2}(t^2 + t^2)dt = \int_{0}^{2}2t^2 dt \]dy term: \[\int_{0}^{2}(t^2 - t^2)dt = \int_{0}^{2}0 dt \]So the integral narrows down to one term.

Step 4: Evaluate the Integral

Evaluate the integral \[\int_{0}^{2}2t^2 dt \]. Find the antiderivative of 2t^2, which is \[\frac{2t^3}{3} \]. Now, apply the limits of integration from 0 to 2: \[\left.\frac{2t^3}{3} \right|_{0}^{2} = \frac{2(2)^3}{3} - \frac{2(0)^3}{3} = \frac{16}{3}.\]

Key Concepts

parametrization

When dealing with line integrals, it's essential to convert the curve in the coordinate plane into a parametric form. This process is called parametrization. It allows us to describe the curve using a parameter, often denoted by 't'. For the given exercise, the curve is the line given by the equation y = x from the origin (0,0) to the point (2,2). Since both x and y values change simultaneously along this line, we can use t to represent them. Specifically, we can set x = t and y = t for the parameter t, ranging from 0 to 2. This parametrization simplifies the curve and allows us to express the line integral in terms of 't', making the calculations manageable.

curve integration

Curve integration involves evaluating a line integral along a given curve. A line integral sums up values of a function along a curve, taking into account both the function values and the geometry of the curve. In this exercise, we need to evaluate the integral over the curve C using the parameter 't'. The line integral expression can be split into two parts: terms associated with dx and dy.

For our parametrization, dx = dt and dy = dt. Substituting these, along with x = t and y = t into the original integral, the dx term is \(\int_{0}^{2} (t^2 + t^2) dt = \int_{0}^{2} 2t^2 dt\) while the dy term simplifies to zero: \(\int_{0}^{2}(t^2 - t^2) dt = \int_{0}^{2} 0 dt\). This decomposition into \(dx\) and \(dy\) terms is crucial for easier evaluation of the line integral.

antiderivative

An antiderivative is a function whose derivative is the given function. It's also known as the indefinite integral. In the context of evaluating line integrals, finding the antiderivative allows us to compute the definite integral by applying the limits of integration.

For the term \(\int_{0}^{2} 2t^2 dt \), we first find the antiderivative of the function 2t^2. The antiderivative of 2t^2 is \(\frac{2t^3}{3}\) because differentiating \(\frac{2t^3}{3}\) yields 2t^2. After finding this, we proceed to evaluate it within the specified limits.

limits of integration

The limits of integration define the interval over which we evaluate the integral. For the parametrized curve in this exercise, the parameter 't' ranges from 0 to 2. This interval comes from the endpoints of the line segment y = x, from (0,0) to (2,2). When we've found the antiderivative of \(2t^2\), we use these limits to compute the definite integral.

We apply the limits by evaluating the antiderivative at the upper and lower bounds and then subtracting the values: \[\left. \frac{2t^3}{3} \right|_{0}^{2} = \frac{2(2)^3}{3} - \frac{2(0)^3}{3} \] which simplifies to \(\frac{16}{3}\).

Understanding how to apply these limits allows for accurate and complete evaluation of the integral.