Problem 23 - Calculus Early Transcendentals

Question

Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C}\left(x^{2}+y^{2}\right) d x-x d y} \\ {C: x^{2}+y^{2}=1, \text { counterclockwise from }(1,0) \text { to }(0,1)}\end{array} $$

Step-by-Step Solution

Answer

The line integral evaluates to \( 1 - \frac{\pi}{4} \).

Step 1: Parametrize the Curve C

Given the curve is the unit circle \( x^2 + y^2 = 1 \) with the starting point \((1,0)\) and going counterclockwise to \((0,1)\). The parametrization of the unit circle can be \( x = \cos(t) \) and \( y = \sin(t) \), where \( t \) ranges from \( 0 \) to \( \frac{\pi}{2} \).

Step 2: Express dx and dy in Terms of dt

Differentiate the parametric equations to get \( dx = -\sin(t) \, dt \) and \( dy = \cos(t) \, dt \). Substitute these into the integral.

Step 3: Substitute into the Integral

Substitute \( x = \cos(t) \), \( y = \sin(t) \), \( dx = -\sin(t) \, dt \), and \( dy = \cos(t) \, dt \) into the integral:\[\int_{0}^{\frac{\pi}{2}} \left(\cos^2(t) + \sin^2(t)\right)(-\sin(t))\,dt - \cos(t) \cos(t)\,dt\]

Step 4: Simplify the Integral

Since \(\cos^2(t) + \sin^2(t) = 1\), the integral becomes:\[\int_{0}^{\frac{\pi}{2}} (-\sin(t) - \cos^2(t))\, dt\] which simplifies to \[-\int_{0}^{\frac{\pi}{2}} (\sin(t) + \cos^2(t))\, dt\]

Step 5: Solve the Integral

Evaluate the integral term by term:\[-\int_{0}^{\frac{\pi}{2}} \sin(t) \, dt - \int_{0}^{\frac{\pi}{2}} \cos^2(t) \, dt\]\[= -[-\cos(t)]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \frac{1+\cos(2t)}{2} \, dt\]\[= -[0 - (-1)] - \left(\frac{1}{2} \left[ t \right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2t) \, dt \right)\]

Step 6: Evaluate Remaining Integrals

Evaluate \( \int_{0}^{\frac{\pi}{2}} \cos(2t) \, dt \):\[\frac{1}{2}\sin(2t)\Big|_{0}^{\frac{\pi}{2}} = \frac{1}{2}(0 - 0) = 0\]Thus, the total value:\[= 1 - \left(\frac{\pi}{4} + 0 \right) = 1 - \frac{\pi}{4}\]

Step 7: Final Result

After evaluating, the line integral is calculated to be \( 1 - \frac{\pi}{4} \).

Key Concepts

Parametrization for Line Integrals

Parametrization is a powerful technique when working with line integrals, especially on curves like circles. To parametrize a curve means to express the coordinates of the curve in terms of a single parameter, often denoted as \( t \). This helps in converting the problem into a manageable form.
For example, if you consider our exercise, the curve is a unit circle \( x^2 + y^2 = 1 \). We start at the point \((1,0)\) and move counterclockwise to \((0,1)\).
In such cases, a common parametrization uses trigonometric functions due to their periodic nature, which fits well with circular paths.

• Assign \( x = \cos(t) \).
• Assign \( y = \sin(t) \).

As \( t \) moves from \(0\) to \( \frac{\pi}{2} \), the path follows the desired quarter circle. Realize that this approach effectively captures the essence of the circular path, making the integration process much simpler.

Understanding the Unit Circle

The unit circle is not just a concept in geometry, but a fundamental building block in calculus, especially with trigonometric identities.
In a unit circle, the radius equals 1, which simplifies many calculations. Every point on the unit circle represents coordinates \( (x,y) \) that satisfy \( x^2 + y^2 = 1 \).
This equation becomes even more relevant in integral calculus due to its inherent use of trigonometric identities.

• It illustrates how sine and cosine functions relate through \( \sin^2(t) + \cos^2(t) = 1 \).
• This identity simplifies expressions within the integral, making it more manageable.

By working within the framework of the unit circle, you not only resolve trigonometric expressions easily but also develop an intuitive understanding of motion along circular paths.

Integral Calculus in Evaluating Line Integrals

Integral calculus, particularly when dealing with line integrals, requires careful substitution and evaluation. In our context, after parametrization, we express \( dx \) and \( dy \) using derivatives of the parameterized functions.
This is crucial because the line integral often includes terms like \( dx \) and \( dy \). By differentiating:

• \( dx = -\sin(t) \, dt \)
• \( dy = \cos(t) \, dt \)

Now, substitute these into the integral to transform it into a function of \( t \). After simplification, recognizing identities, such as \( \cos^2(t) + \sin^2(t) = 1 \), can further simplify the integral.
Once simplified, the integral becomes a series of solvable terms:

• Evaluate each term independently: \( \int \sin(t) \, dt \) and \( \int \cos^2(t) \, dt \).
• Solve these integrals within proper limits for \( t \).

The result, in this exercise, was achieved by careful application of integral calculus principles, yielding \( 1 - \frac{\pi}{4} \). This methodical approach shows how integral calculus solves real-world problems involving paths and orientations.