Problem 10 - Calculus Early Transcendentals

Question

Use Green’s Theorem to evaluate the integral. In each exercise, assume that the curve C is oriented counterclockwise. \(\oint_{C} x^{2} y d x-y^{2} x d y,\) where \(C\) is the boundary of the region in the first quadrant, enclosed between the coordinate axes and the circle \(x^{2}+y^{2}=16\)

Step-by-Step Solution

Answer

The integral evaluates to \(-32\pi\).

Step 1: Identify the Vector Field Components

The line integral given is \( \oint_{C} x^{2} y \, dx - y^{2} x \, dy \). We can recognize this as an integral of the form \( \oint_{C} P \, dx + Q \, dy \) with \( P = x^2 y \) and \( Q = -y^2 x \).

Step 2: Set Up 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. The theorem is given by: \[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]

Step 3: Compute Partial Derivatives

Calculate \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \):\[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(-y^2 x) = -y^2 \]\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2 y) = x^2 \]

Step 4: Set Up the Double Integral

Insert the partial derivatives into the Green's Theorem formula:\[ \iint_{R} \left( -y^2 - x^2 \right) \, dA \]The region \( R \) is the first quadrant sector of the circle \( x^2 + y^2 = 16 \).

Step 5: Convert to Polar Coordinates

To evaluate the double integral, convert to polar coordinates where \( x = r\cos\theta \) and \( y = r\sin\theta \). The limits for \( r \) are 0 to 4 (since the circle has radius 4), and for \( \theta \), the limits are from 0 to \( \frac{\pi}{2} \). The Jacobian for polar coordinates is \( r \), hence the integral becomes:\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} \left( -r^2 \right) r \, dr \, d\theta \]

Step 6: Evaluate the Integral

Evaluate the integral:\[ \int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{4} -r^3 \, dr \]Integrate with respect to \( r \):\[ \int_{0}^{4} -r^3 \, dr = \left[-\frac{r^4}{4}\right]_0^4 = -\frac{256}{4} = -64 \]Now integrate with respect to \( \theta \):\[ \int_{0}^{\frac{\pi}{2}} -64 \, d\theta = -64[\theta]_0^{\frac{\pi}{2}} = -64 \times \frac{\pi}{2} = -32\pi \]

Step 7: Final Calculation

The value of the integral using Green's Theorem over the given region is \(-32\pi\).

Key Concepts

Vector Field

A vector field is a mathematical concept where each point in space is associated with a vector. Imagine arrows pointing in various directions on a map, where each arrow has a direction and a length. These arrows represent a vector field. The arrows illustrate how values change across space.

In the exercise, we were provided with a line integral \( \oint_{C} x^{2} y \, dx - y^{2} x \, dy \). From this, we identify the components of the vector field: \( P = x^2 y \) and \( Q = -y^2 x \). These expressions determine the direction and magnitude of the vector at each point \((x, y)\) in the vector field.

Understanding vector fields is crucial in physics and engineering, where they help describe things like magnetic fields, gravitational fields, and fluid flow. They essentially let us see how something like a force varies over an area.

Double Integral

A double integral allows us to calculate the volume under a surface over a region in two-dimensional space. It is like adding up little pieces over a defined region, often referred to as \( R \).

Green’s Theorem, which connects a line integral to a double integral, was used in the solution. In the theorem, we need to calculate \( \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \), representing a double integral over region \( R \).

• The double integral answers the question: What is the total quantity of interest (e.g., area) over this two-dimensional region?
• In our problem, it tells us the evaluated integral across the sector bounded by the circle \( x^2 + y^2 = 16 \) in the first quadrant.

By evaluating the double integral, you aggregate the quantities of interest over the specified area.

Polar Coordinates

Polar coordinates provide a way to express points in space using radius and angle \((r, \theta)\), unlike Cartesian coordinates \((x, y)\) that use horizontal and vertical distances.

When converting to polar coordinates, each point is represented as:\[ x = r\cos\theta \] \[ y = r\sin\theta \]

In this exercise, converting to polar coordinates made sense as the region is a sector of a circle. The transformation simplifies the calculation of the double integral. The circle has a radius of 4 and is bounded by angles from 0 to \( \frac{\pi}{2} \).

• Using polar coordinates can simplify calculations, especially when regions have circular symmetry.
• It transforms complicated region boundaries into more straightforward limits for integration.

Partial Derivatives

Partial derivatives help us understand how functions change when one variable changes, while the other variables are held constant.

In math and physics, partial derivatives show how a multi-variable function changes when only one of its input variables changes. In our exercise, finding \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) was a crucial step:

• \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(-y^2 x) = -y^2 \)
• \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2 y) = x^2 \)

By plugging these derivatives into Green's Theorem, we translated the line integral into a more manageable double integral. Partial derivatives are the building blocks for understanding changes in vector fields and are essential in calculus, especially when dealing with multi-variable functions.