Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C}\left(x^{2}+4 x y\right) d x+\left(2 x^{2}+3 y\right) d y\), where \(C\) is the ellipse \(9 x^{2}+16 y^{2}=144\)

Step-by-Step Solution

Answer

The value of the line integral is 0.

Step 1: Understanding Green's Theorem

Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region S bounded by C. Specifically, \[ \oint_C P\,dx + Q\,dy = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dA \] where \( P = x^2 + 4xy \) and \( Q = 2x^2 + 3y \).

Step 2: Calculate Partial Derivatives

Calculate the partial derivative of \( Q \) with respect to \( x \) and \( P \) with respect to \( y \). \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y) = 4x \]\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2 + 4xy) = 4x \]

Step 3: Setup the Double Integral

Substitute the partial derivatives into the double integral formula from Green's Theorem. The integrand becomes:\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 4x - 4x = 0 \] Hence, the double integral is: \[ \iint_S 0\,dA = 0 \]

Step 4: Conclusion from the Integral Result

Since the integrand of the double integral over the region S is zero, the value of the line integral around C, according to Green's Theorem, is zero. Thus:\[ \oint_C \left(x^2 + 4xy\right)\,dx + \left(2x^2 + 3y\right)\,dy = 0 \]

Step 5: Sketch the Region S

The region S is an ellipse given by the equation \(9x^2 + 16y^2 = 144\). By dividing the entire equation by 144, we get the standard form: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] This ellipse has a semi-major axis of 4 along the x-axis and a semi-minor axis of 3 along the y-axis. Sketching this ellipse gives a visual understanding of the region S over which the integral is evaluated.

Key Concepts

Line Integral

The concept of a line integral extends the idea of an integral to functions along curves in a plane or space. In simple terms, a line integral allows us to integrate functions defined along a path, rather than just over an interval. Line integrals are used in many fields, including physics and engineering, for measuring quantities like work done by a force field.

• To calculate a line integral, you need a vector field and a curve over which to integrate.
• In this exercise, the line integral involves the ellipse described by the equation \(9x^2 + 16y^2 = 144\).
• The line integral around a closed curve like this one can be related to a double integral over the region it encloses, as stated by Green's Theorem.

The line integral is given by \(\int_C (P \, dx + Q \, dy)\), where \(P\) and \(Q\) are defined functions of \(x\) and \(y\). Green's Theorem provides an alternative method for evaluating such integrals.

Partial Derivatives

Partial derivatives are an essential tool in calculus, especially when dealing with functions of multiple variables. They represent the rate of change of a function with respect to one variable while holding the others constant.

• In context of the exercise, we calculate \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\).
• For the function \(Q = 2x^2 + 3y\), the partial derivative with respect to \(x\) is \(4x\).
• For the function \(P = x^2 + 4xy\), the partial derivative with respect to \(y\) is \(4x\).

These derivatives help form the double integral that arises when applying Green's Theorem. In this case, both partial derivatives yield the same result, leading to a cancellation in the integrand.

Double Integral

Double integrals are an extension of single-variable integrals to functions of two variables, typically over a specified region in a plane. When applying Green's Theorem, a double integral is key to transforming a problem of calculating a line integral into a more manageable form.

• In our problem, the double integral is over the region \(S\) enclosed by the ellipse.
• The integrand in Green's Theorem becomes \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\), which simplifies to zero.
• This results in the double integral \(\iint_S 0 \, dA = 0\), indicating no net accumulation over the area \(S\).

The simplicity of this solution arises because the integrand is zero, a unique situation that simplifies computational efforts considerably.

Ellipse

An ellipse is a type of conic section that looks like a squished circle. It's defined as the set of points for which the sum of distances to two fixed points (foci) is constant.

• The given ellipse here is described by \(9x^2 + 16y^2 = 144\).
• Converting this to the standard form, \(\frac{x^2}{16} + \frac{y^2}{9} = 1\), shows the ellipse's semi-major and semi-minor axes.
• Here, the semi-major axis is 4 along the x-axis, and the semi-minor axis is 3 along the y-axis.

In Green's Theorem, this elliptical region is where we perform the double integral, which elegantly simplifies our original problem of the line integral.