Question

The area of a region \(R\) in the plane, whose boundary is the closed curve \(C,\) may be computed using line integrals with the formula $$\text { area of } R=\int_{C} x d y=-\int_{C} y d x$$ These ideas reappear later in the chapter. Let \(R=\\{(r, \theta): 0 \leq r \leq a, 0 \leq \theta \leq 2 \pi\\}\) be the disk of radius \(a\) centered at the origin and let \(C\) be the boundary of \(R\) oriented counterclockwise. Use the formula \(A=-\int_{C} y d x\) to verify that the area of the disk is \(\pi r^{2}.\)

Short answer

Question: Using Green's theorem, find the area of a disk with radius \(a\). Answer: The area of the disk is \(\pi a^2\).

Step-by-step solution

Parametrize the boundary curve C

First, we need to parametrize the boundary curve \(C\) of the region \(R\) using polar coordinates \((r, \theta)\). Since \(C\) is a circle with radius \(a\), we can parametrize it as follows: $$x = a \cos(\theta)$$ $$y = a \sin(\theta)$$ where \(0 \leq \theta \leq 2\pi\).

Compute dx and dy

Now we compute \(dx\) and \(dy\) with respect to \(\theta\). Taking the derivatives of \(x\) and \(y\) with respect to \(\theta\), we get: $$\frac{dx}{d\theta} = -a \sin(\theta)$$ $$\frac{dy}{d\theta} = a \cos(\theta)$$

Compute the line integral

Now we compute the line integral of \(-y\) along \(C\) using the formula \(A = -\int_C y dx\). This can also be written as the integral of \(-yt\) with respect to \(\theta\): $$A = - \int_0^{2\pi} (a \sin(\theta))(-a \sin(\theta) d\theta)$$ Simplify the expression and compute the integral: $$A = \int_0^{2\pi} a^2 \sin^2(\theta) d\theta$$

Evaluate the integral

To evaluate the integral, use the trigonometric identity \(\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}\): $$A = a^2 \int_0^{2 \pi} \frac{1 - \cos(2\theta)}{2} d\theta $$ Now, we split the integral into two: $$A = a^2 \left(\frac{1}{2} \int_0^{2 \pi} 1 d\theta - \frac{1}{2} \int_0^{2 \pi} \cos(2\theta) d\theta \right)$$ Integrating each term, we get: $$A = a^2 \left(\frac{1}{2} \left[\theta \right]_0^{2 \pi} - \frac{1}{2} \left[\frac{\sin(2\theta)}{4} \right]_0^{2 \pi} \right)$$ Evaluate these terms: $$A = a^2 \left(\frac{1}{2} (2\pi - 0) - \frac{1}{2} \left(\frac{\sin(4\pi)}{4} - \frac{\sin(0)}{4}\right) \right)$$ Since \(\sin(4\pi) = 0\) and \(\sin(0) = 0\), the expression simplifies to: $$A = a^2 \left(\frac{1}{2} (2\pi)\right) = \pi a^2$$ Hence, the area of the disk is \(\pi a^2\), which is the expected result.

Key concepts

Area Calculation Using Line Integrals

The technique of using line integrals to calculate the area of a region is a powerful tool in multivariable calculus. For regions bounded by a closed curve, the calculation can be streamlined by integrating along the curve.

When a curve C encloses a region R, the area of R can be obtained using the line integral formula: \[\text{area of } R = \int_C x \, dy = -\int_C y \, dx\]. In this scenario, the use of \( -y \, dx \) integral leads to the area of a disk with radius \( a \) being \( \pi a^2 \), which closely aligns with the familiar formula for the area of a circle. This approach elegantly demonstrates the convergence of geometric intuition with the power of calculus.

Interestingly, the negative sign in the formula is a consequence of the orientation of the curve with respect to the chosen axes. If we were to flip the curve's orientation, the same approach would yield an equivalent positive integral of \( x \, dy \) instead.

Parametrization of Curves

Parametrization is the process of expressing the coordinates of points on a curve using a single parameter. In our disk example, the boundary curve C is a circle which is best described using polar coordinates and the parameter \( \theta \).

For the circle of radius \( a \), the parametric equations \( x = a \cos(\theta) \) and \( y = a \sin(\theta) \) offer a description of every point along the boundary as \( \theta \) varies from \( 0 \) to \( 2\pi \). This illustrates the essence of parametrization: to simplify the curve's representation into a form that's convenient for further calculations, such as differentiation and integration.

Why Parametrize?

Parametrization plays a key role in calculus because it provides a means to convert complex shapes and curves into manageable equations that can be integrated or differentiated with respect to a single variable, greatly facilitating the calculation of line integrals and other analysis.

Trigonometric Integrals

Dealing with trigonometric integrals, such as those involving \( \sin^2(\theta) \) or other trigonometric functions squared, often requires using trigonometric identities to simplify the expression before integration.

In our disk area problem, the integral \( \int \sin^2(\theta) \, d\theta \) is simplified by applying the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \). This step is crucial; it transforms a challenging integral into simpler terms that are easier to integrate. The resulting split into two separate integrals — one for the constant term and another for the \( \cos(2\theta) \) term — typifies strategies employed to handle trigonometric integrals.

Power of Identities

These identities are not mere algebraic curiosities; they are essential tools in calculus that turn complex, intractable expressions into forms that comply with standard integration techniques.

Polar Coordinates

Polar coordinates offer a more natural framework for describing points in a plane when dealing with circular or spiral shapes. A point's location is determined by its distance from the origin, denoted as \( r \), and its angle from the positive x-axis, denoted as \( \theta \).

In the context of our disk area problem, the entire disk can be expressed as \( R = \{ (r, \theta) : 0 \leq r \leq a, 0 \leq \theta \leq 2\pi \} \), which simplifies the conceptualization of the circle and its enclosed area. Polar coordinates not only visualize shapes traditionally challenging in Cartesian coordinates but also make certain integrals, particularly those around circular paths, more straightforward to compute.

Integration in Polar Coordinates

When integrating in polar coordinates, we often find that radial symmetry and the natural alignment of the parameter \( \theta \) with the path of integration simplify the calculations — a perfect example of choosing the right tool for the task at hand in mathematical problem-solving.