Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C} \sqrt{y} d x+\sqrt{x} d y\), where \(C\) is the closed curve formed by \(y=0, x=2\), and \(y=x^{2} / 2\)

Step-by-Step Solution

Answer

The value of the line integral is \(\frac{8\sqrt{2}}{5}\).

Step 1: Understanding Green's Theorem

Green's Theorem relates the line integral around a simple closed curve to a double integral over the region it encloses. It can be stated as: \[\oint_C P \, dx + Q \, dy = \iint_S \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\]where \(C\) is the positively oriented boundary of the region \(S\). In this problem, we identify \(P=\sqrt{y}\) and \(Q=\sqrt{x}\).

Step 2: Sketching the Region

The region \(S\) is bounded by the curves \(y=0\), \(x=2\), and \(y=\frac{x^2}{2}\). Sketch these curves in the \(xy\)-plane to visualize the enclosed region. Here, \(y=0\) is the x-axis from \(x=0\) to \(x=2\), and the parabola \(y=\frac{x^2}{2}\) intersects \(x=2\) at the point \((2,2)\). The region is the area under the parabola from \(x=0\) to \(x=2\).

Step 3: Calculate Partial Derivatives

To apply Green's Theorem, calculate the partial derivatives: \[-\frac{\partial P}{\partial y} = -\frac{1}{2\sqrt{y}}\] \(\) \[\frac{\partial Q}{\partial x} =\frac{1}{2\sqrt{x}} \]Substituting these into the Green's Theorem formula gives: \[\iint_S \left( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \right) \, dA\]

Step 4: Set Up the Double Integral

Rewrite the double integral over region \(S\), which is bounded by \(y=0\) to \(y=\frac{x^2}{2}\) and \(x=0\) to \(x=2\):\[\int_{0}^{2} \int_{0}^{\frac{x^2}{2}} \left( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \right) \, dy \, dx\]

Step 5: Evaluate the Integral

Evaluate the inner integral with respect to \(y\):\[\int_{0}^{\frac{x^2}{2}} \frac{1}{2\sqrt{y}} \, dy = \left. \sqrt{y} \right|_0^{\frac{x^2}{2}} = \sqrt{\frac{x^2}{2}} = \frac{x}{\sqrt{2}}\]Now, the double integral becomes:\[\int_{0}^{2} \left( \frac{1}{2\sqrt{x}} \cdot \frac{x^2}{2} + \frac{x}{\sqrt{2}} \right) \, dx\]Combine the terms and integrate over \(x\) from \(0\) to \(2\).

Step 6: Final Calculation

Simplify and compute the integral:\[\int_{0}^{2} \left( \frac{x^2}{4\sqrt{x}} + \frac{x}{\sqrt{2}} \right) \, dx = \int_{0}^{2} \left( \frac{x^{3/2}}{4} + \frac{x}{\sqrt{2}} \right) \, dx\]Split the integral and calculate separately:\[\int_{0}^{2} \frac{x^{3/2}}{4} \, dx + \int_{0}^{2} \frac{x}{\sqrt{2}} \, dx\]Compute the antiderivatives and evaluate at the bounds.

Key Concepts

Line Integral

A line integral is a way to integrate a function over a curve. When we talk about line integrals in the context of Green's Theorem, we're considering the sum of a particular field along a specified curve, denoted as \( C \). In our exercise, the line integral is given by \( \oint_{C} \sqrt{y} \, dx + \sqrt{x} \, dy \). Here, the integral is along a closed curve.

Line integrals are particularly useful in physics to calculate things like work done by a force field. For Green's Theorem to apply, the curve must be continuously differentiable and encloses a region \( S \).

The line integral converts the field expressions into a path sum, transforming a continuous operation into a conventional sum, making certain calculations, such as enclosed area properties, easier to compute. It acts as the bridge between a continuous curve and its discrete properties such as the enclosed area.

Double Integral

Double integrals extend the concept of integration to functions of two variables, such as functions defined over a plane. They allow us to sum up values of a function over a two-dimensional area. In our context, Green's Theorem relates the line integral around a simple closed curve to a double integral over the region it encloses.

The double integral of interest here is: \[ \int_{0}^{2} \int_{0}^{\frac{x^2}{2}} \left( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \right) \, dy \, dx \] The outer integral is with respect to \(x\) and the inner one with respect to \(y\).

• For each fixed \(x\), you consider the vertical strip from \(y=0\) to \(y=\frac{x^2}{2}\).
• The double integral calculates the total value of the function across the entire specified area, which corresponds to our region \( S \).

This approach allows you to capture the function's behavior across the entire region enclosed by the curve.

Partial Derivatives

Partial derivatives are derivatives with respect to one variable while keeping others constant, offering insight into the function's behavior. When applying Green’s Theorem, you calculate the partial derivatives of the functions \(P\) and \(Q\) involved in the line integral.

In this exercise: \[ P = \sqrt{y} \quad \text{and} \quad Q = \sqrt{x} \] The partial derivatives needed are:

• \( \frac{\partial Q}{\partial x} = \frac{1}{2\sqrt{x}} \)
• \( \frac{\partial P}{\partial y} = \frac{1}{2\sqrt{y}} \)

The Green’s Theorem formula is: \[ \iint_S \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] Incorporating these derivatives helps in transforming from the line integral to a double integral representation over a region, facilitating the calculation of the enclosed area or other properties.

Region Sketching

Sketching the region enclosed by the curve is a crucial step when working with Green's Theorem as it defines the limits of integration for the double integral.

In our exercise, the region \( S \) is enclosed by the curves:

• \(y = 0\), which is the x-axis from \(x = 0\) to \(x = 2\)
• \(x = 2\), which is a vertical line at \(x = 2\)
• \(y = \frac{x^2}{2}\), a parabola that intersects \(x = 2\) at \((2, 2)\)

How to Sketch

• Start by drawing the x-axis and the line \(x = 2\).
• Sketch the parabola \(y = \frac{x^2}{2}\). This is a curve opening upwards from the origin.
• Observe the parabola intersects the x-axis at the origin and then touches \((2, 2)\).

The region \( S \) is thus the area below the parabola and above the x-axis, extending from \(x = 0\) to \(x = 2\). Visualizing this region helps in setting up the integral limits accurately.