Problem 28 - Thomas Calculus

Question

Apply Green's Theorem to evaluate the integrals. \(\oint_{C}(3 y d x+2 x d y)\) \(C\): The boundary of \(0 \leq x \leq \pi, 0 \leq y \leq \sin x\)

Step-by-Step Solution

Answer

The integral evaluates to -2.

Step 1: Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the region \(R\) it encloses. It states \(\oint_{C} (P\, dx + Q\, dy) = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dA\). Here, \(P(x, y) = 3y\) and \(Q(x, y) = 2x\).

Step 2: Compute Partial Derivatives

First, compute the partial derivatives needed for Green's Theorem. \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2x) = 2\) and \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3y) = 3\).

Step 3: Substitute into Green's Theorem

Substitute the partial derivatives into the expression \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\) to find \(2 - 3 = -1\). This expression will be integrated over the region \(R\).

Step 4: Set Up Double Integral Over Region R

The region \(R\) is defined as the area under the curve \(y = \sin x\) from \(x=0\) to \(x=\pi\). The double integral is set up as \(\iint_{R} -1 \, dA = \int_{0}^{\pi}\int_{0}^{\sin x} -1 \, dy \, dx\).

Step 5: Evaluate the Inner Integral

Integrate with respect to \(y\). \(\int_{0}^{\sin x} -1 \, dy = -y\) evaluated from \(y=0\) to \(y=\sin x\) gives \(-\sin x\).

Step 6: Evaluate the Outer Integral

Integrate with respect to \(x\) to find \(\int_{0}^{\pi} -\sin x \, dx\). The antiderivative is \(\cos x\), so the result is \([\cos x]_{0}^{\pi} = [-1 - 1] = -2\).

Step 7: Conclusion

The line integral of \(3y\, dx + 2x\, dy\) around \(C\) is \(-2\).

Key Concepts

Line Integral

In mathematics, a line integral is used to find the integral of a function along a curve. Imagine a line integral as adding up values of a function along a path. You take into account both the magnitude, or value, of the function, and the direction of the path. Think about it like walking along a windy road and measuring the incline at each small step.

The exercise above asks you to compute a particular type of line integral:

• We use differential forms like \(P\, dx + Q\, dy\) to represent the function along the curve \(C\).
• Here, \(P(x, y) = 3y\) and \(Q(x, y) = 2x\).

When we apply a line integral, we're focusing on how these changes relate to the curve itself, capturing changes in the function as defined by \(dx\) and \(dy\). For Green’s Theorem, line integrals help describe the physical properties of the region \(R\) enclosed by the curve \(C\).

Double Integral

A double integral extends the concept of a single integral to functions of two variables over a region. In our context, the double integral helps calculate areas or accumulate quantities over a 2D region. For example, finding the area under a curve can be performed using a double integral.

In relation to the exercise:

• The double integral involves integrating over a region \(R\) described under the curve \(y = \sin x\) from \(x = 0\) to \(x = \pi\).
• Green's theorem links the line integral around this region to a double integral over the region.

The double integral \(\iint_{R} -1 \, dA\) is set up to evaluate this relationship. It reflects calculating the total accumulation (area, in this case) that follows from the integrals of each small element of area in \(dA\), across the entire region \(R\).

Partial Derivatives

Partial derivatives measure how a multivariable function changes as its variables are varied. They capture the rate of change of the function with respect to one variable while keeping the others constant. Think of each variable having its own mini-"slope" reflecting its influence.

For Green's theorem, the partial derivatives are crucial:

• \(\frac{\partial Q}{\partial x} = 2\) considers the change of \(Q\) with respect to \(x\).
• \(\frac{\partial P}{\partial y} = 3\) looks at the change of \(P\) with respect to \(y\).
• These partial derivatives help express the circulation or curve behavior in the region \(R\).

By substituting these derivatives into the theorem's framework, we connect the local behavior of \(P\) and \(Q\) to the global properties of the region they define.

Simple Closed Curve

A simple closed curve is a looped path that does not cross itself. You can imagine it as a fence enclosing an area, where the start and end points meet seamlessly, and no part of the fence crosses over itself. In two dimensions, it's like drawing a perfect loop on paper.

In the exercise, the boundary \(C\) forms a simple closed curve:

• It is described using the boundary conditions \(0 \leq x \leq \pi\) and \(0 \leq y \leq \sin x\).
• This curve encloses the region over which we apply both line and double integrals.

Understanding this curve is important because it defines the limits for applying Green's Theorem. It's the closed path around which we compute the line integral and the area that is filled within defines the region for the double integral.