Chapter 6: Problem 146
Evaluate the line integrals by applying Green’s theorem. \(\int_{C} 2 x y d x+(x+y) d y,\) where \(C\) is the path from \((0,\) 0) to \((1,1)\) along the graph of \(y=x^{3}\) and from \((1,1)\) to \((0,\) 0) along the graph of \(y=x\) oriented in the counterclockwise direction.
Short Answer
Expert verified
-\frac{1}{60}
Step by step solution
01
Define the Curve C
The curve \(C\) consists of two parts: \(C_1\) from \((0,0)\) to \((1,1)\) along \(y = x^3\) and \(C_2\) back from \((1,1)\) to \((0,0)\) along \(y = x\). This forms a closed curve that can be evaluated using Green's theorem.
02
Identify the Functions M and N
For the line integral \(\int_{C} 2xy \,dx + (x+y) \,dy\), assign \(M(x,y) = 2xy\) and \(N(x,y) = x+y\). Green's theorem relates these functions over a region \(D\) to a double integral.
03
Apply Green's Theorem Formula
Green's theorem states that \(\int_{C} M\,dx + N\,dy = \int\int_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \,dx \,dy\). Compute the partial derivatives: \(\frac{\partial N}{\partial x} = 1\) and \(\frac{\partial M}{\partial y} = 2x\).
04
Set Up the Double Integral
Since \(\int\int_{D} \left(1-2x\right)\, dx \, dy\) is needed, identify that \(D\) is bounded by \(y=x^3\) and \(y=x\) for \(x\) in \([0,1]\). Thus, the double integral becomes \(\int_{0}^{1} \int_{x^3}^{x} (1 - 2x) \, dy \, dx\).
05
Evaluate the Inner Integral
Integrate \(1 - 2x\) with respect to \(y\): \(\int_{x^3}^{x} (1-2x) \, dy = (1-2x)(y) \Big|_{x^3}^{x} = (1-2x)(x - x^3)\).
06
Simplify the Expression
Simplify the expression: \((1-2x)(x - x^3) = x - x^3 - 2x^2 + 2x^4\).
07
Evaluate the Outer Integral
Integrate with respect to \(x\): \(\int_{0}^{1} (x - x^3 - 2x^2 + 2x^4) \, dx\). This results in: \[\frac{x^2}{2} - \frac{x^4}{4} - \frac{2x^3}{3} + \frac{2x^5}{5} \Big|_{0}^{1}\].
08
Simplify the Result
Evaluate the expression at the upper and lower bounds: \( \frac{1}{2} - \frac{1}{4} - \frac{2}{3} + \frac{2}{5} = -\frac{1}{60} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integrals
Line integrals are a way to integrate functions along a curve, rather than over an area or a volume. Imagine having a path in a field, where the field's intensity can vary at each point. A line integral helps you find the total effect (like work done by a force field) along that path. In general, a line integral can be written as \[ \int_{C} P(x, y) \, dx + Q(x, y) \, dy \]where the path \(C\) is described in terms of \(x\) and \(y\). This type of integral is particularly useful in physics to compute things like the circulation of a field along a closed loop.
- Calculation involves breaking the curve into small segments.
- The integral sums up contributions from each segment.
- It is independent of the path direction, if the field is conservative.
Partial Derivatives
Partial derivatives give us the rate of change of a multivariable function regarding one of its variables, while keeping the other variables constant. In calculus, they serve as a foundation for multivariable optimization and are key in many fields of science and engineering.Let's say you have a function \(N(x, y)\). The partial derivative of \(N\) with respect to \(x\), denoted \(\frac{\partial N}{\partial x}\), tells us how \(N\) changes as \(x\) changes, keeping \(y\) fixed. Similarly, \(\frac{\partial M}{\partial y}\) tells how \(M(x, y)\) changes as \(y\) is varied.In Green's Theorem, partial derivatives help relate the curl of a vector field to a line integral around a closed curve by measuring local rotational tendency of the field:\[ \int_{C} M \, dx + N \, dy = \int\int_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dx \, dy \]It's essential for understanding the transformations involved in switching between line integrals and double integrals.
Double Integral
Double integrals allow you to calculate the volume under a surface defined by a function \(f(x, y)\) over a region \(D\). This extends the concept of an integral by summing all the infinitesimal values over a two-dimensional area.The double integral is represented generally as:\[ \int\int_{D} f(x, y) \, dx \, dy \]Here is how it works:
- The region \(D\) is split into infinite small patches.
- For each patch, calculate the small volume \(f(x, y) \, \Delta x \, \Delta y\).
- Sum up all these small volumes to find the total volume under the surface.
Closed Curve
A closed curve is a path that returns to its starting point without crossing itself. Think of a circle or an ellipse—paths that loop without end. In mathematics, closed curves are significant because they define regions that can enclose areas, offering a domain for integrations like the one involved in Green’s Theorem.Green’s Theorem specifically applies to closed curves because it relates a line integral around the curve to a double integral over the region it encloses. Importantly, the curve must be oriented correctly. Typically, for Green's Theorem, you'll orient your curve in the counterclockwise direction, which follows a standard mathematical convention:
- Ensures the region \(D\) is to the left of the path.
- Helps with establishing positive orientation.
- Distinguishes between regions that are enclosed versus exterior to the path.