Problem 30 - Thomas Calculus

Question

In Exercises \(27 - 30 ,\) integrate \(f\) over the given curve. $$ \begin{array} { l } { f ( x , y ) = x ^ { 2 } - y , \quad C : \quad x ^ { 2 } + y ^ { 2 } = 4 \text { in the first quadrant from } } \\ { ( 0,2 ) \text { to } ( \sqrt { 2 } , \sqrt { 2 } ) } \end{array} $$

Step-by-Step Solution

Answer

The integral over the curve is \( \pi - 4\sqrt{2} + 2 \).

Step 1: Parameterize the Curve

The curve \( C \) is part of a circle defined by \( x^2 + y^2 = 4 \). In the first quadrant, we can parameterize it using \( x = 2\cos t \) and \( y = 2\sin t \) where \( 0 \leq t \leq \frac{\pi}{4} \) corresponds from \((0,2)\) to \((\sqrt{2}, \sqrt{2})\).

Step 2: Substitute Parameters into Function

Substitute \( x = 2\cos t \) and \( y = 2\sin t \) into \( f(x, y) = x^2 - y \). This gives us \( f(2\cos t, 2\sin t) = (2\cos t)^2 - 2\sin t = 4\cos^2 t - 2\sin t \).

Step 3: Compute Derivatives

To find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), differentiate the parameterizations: \( \frac{dx}{dt} = -2\sin t \) and \( \frac{dy}{dt} = 2\cos t \).

Step 4: Calculate Arc Length Differential

The differential arc length is \( ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt = \sqrt{(-2\sin t)^2 + (2\cos t)^2} \, dt = 2 \, dt \).

Step 5: Set Up the Integral

The integral of \( f \) over curve \( C \) is given by \( \int_C f(x, y) \, ds = \int_0^{\frac{\pi}{4}} [4\cos^2 t - 2\sin t] \times 2 \, dt = \int_0^{\frac{\pi}{4}} [8\cos^2 t - 4\sin t] \, dt \).

Step 6: Evaluate the Integral

Using the trigonometric identity \( \cos^2 t = \frac{1 + \cos 2t}{2} \), the integral becomes \( \int_0^{\frac{\pi}{4}} [4(1 + \cos 2t) - 4\sin t] \, dt = \int_0^{\frac{\pi}{4}} [4 + 4\cos 2t - 4\sin t] \, dt \). Evaluate: \( 4t + 2\sin 2t + 4\cos t \Big|_0^{\frac{\pi}{4}} \), which computes to \( \pi - 4\sqrt{2} + 2 \).

Key Concepts

Understanding Curve Parameterization in Line Integrals

When working with line integrals, especially on curves defined by simple geometric shapes like circles, parameterization is a key step. Parameterization allows us to express a curve in terms of a single variable, often denoted as \( t \). This helps to simplify the integration process.
To parameterize a circle, we often use trigonometric functions. For the equation \( x^2 + y^2 = 4 \), which forms a circle with radius 2, a suitable parameterization within the first quadrant is:

• \( x = 2\cos t \)
• \( y = 2\sin t \)

This approach takes advantage of the trigonometric identity \( \cos^2 t + \sin^2 t = 1 \). By adjusting the curve to a parameter between \( 0 \leq t \leq \frac{\pi}{4} \), we accurately traverse the path from \((0, 2)\) to \((\sqrt{2}, \sqrt{2})\).
This simplification makes the integration over the path manageable by transforming it into an integral in terms of \( t \).

Applying Trigonometric Identities in Integration

Trigonometric identities are extremely valuable tools in calculus, especially when evaluating integrals. In this problem, one such identity \( \cos^2 t = \frac{1 + \cos 2t}{2} \) is used to simplify expressions involving \( \cos^2 t \).
These identities help to convert complex trigonometric expressions into simpler ones, which are often easier to integrate:

• For \( \cos^2 t \), use the identity \( \cos^2 t = \frac{1 + \cos 2t}{2} \) to simplify integration.

In line integrals, complex trigonometric functions can often appear once the parameterized equations are substituted back into \( f(x, y) \). Using identities to reduce these expressions significantly aids in solving the integral. In this context, understanding and applying these identities is crucial for computing integrals efficiently and accurately.

Calculating Arc Length for Line Integrals

The arc length is an essential component of setting up line integrals, as it often represents a physical path along which a function is integrated.
For a parameterized curve \( C \), the differential arc length \( ds \) is computed using the derivatives of the parameterized functions with respect to \( t \):

• \( \frac{dx}{dt} = -2\sin t \)
• \( \frac{dy}{dt} = 2\cos t \)

The expression for \( ds \) is derived from the Euclidean distance formula and takes the form:
\[ ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]
For the given problem, this simplifies to \( 2 \, dt \), which indicates the constant scaling factor by which the original parameter space is stretched or contracted. Understanding how to calculate \( ds \) accurately is crucial, as it represents the path length over which the function is integrated.

Integral Evaluation in the Context of Parameterized Curves

Evaluating an integral along a parameterized curve involves substituting the parameterized expressions into the function and multiplying by the differential arc length.
The integral for the problem is set up as:
\[ \int_0^{\frac{\pi}{4}} [8\cos^2 t - 4\sin t] \, dt \]
Applying trigonometric identities simplifies this to:
\[ \int_0^{\frac{\pi}{4}} [4 + 4\cos 2t - 4\sin t] \, dt \]
The integration is performed over the designated interval \( 0 \leq t \leq \frac{\pi}{4} \), with the length of the curve implicit in the expression \( 2 \, dt \).
This step involves careful calculation to find the definite integral, which evaluates to:
\[ 4t + 2\sin 2t + 4\cos t \Big|_0^{\frac{\pi}{4}} \]
Essentially, integral evaluation over parameterized curves combines parameterization, trigonometric simplifications, and arc length considerations to compute the aggregate effect described by the function \( f(x, y) \) along the curve.