Chapter 6: Problem 152
Use Green’s theorem. Let \(C\) be the curve consisting of line segments from \((0,0)\) to \((1,1)\) to \((0,1)\) and back to \((0,0) .\) Find the value of \(\int_{C} x y d x+\sqrt{y^{2}+1 d y}.\)
Short Answer
Expert verified
The value of the integral is \(-\frac{1}{6}\).
Step by step solution
01
Understand Green's Theorem
Green's Theorem relates a line integral around a simple closed curve to a double integral over the region bounded by the curve. For a vector field \( \vec{F} = (P, Q) \), Green's Theorem states: \[ \int_{C} P\,dx + Q\,dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \].
02
Identify \(P\) and \(Q\)
From the integral \( \int_{C} xy\, dx + \sqrt{y^2+1}\, dy \), identify the components of the vector field: - \( P = xy \) - \( Q = \sqrt{y^2+1} \)
03
Compute Partial Derivatives
Compute the partial derivatives: - \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(\sqrt{y^2+1}) = 0 \), since the expression inside \( Q \) doesn't depend on \( x \). - \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x \).
04
Set Up the Double Integral
Substitute the partial derivatives into Green's Theorem formula: \[ \iint_{R} \left( 0 - x \right) dA = -\iint_{R} x \, dA \].
05
Determine the Region \( R \)
The region \( R \) is the triangle with vertices at \((0,0)\), \((1,1)\), and \((0,1)\). Its bounds are defined by the inequalities: \(0 \leq x \leq y \) and \( 0 \leq y \leq 1 \).
06
Evaluate the Double Integral
Evaluate the double integral:\[ -\int_{0}^{1} \int_{0}^{y} x \, dx \, dy \].First, integrate with respect to \( x \):\[ -\int_{0}^{1} \left[ \frac{x^2}{2} \right]_{0}^{y} \, dy = -\int_{0}^{1} \frac{y^2}{2} \, dy \].Then integrate with respect to \( y \):\[ -\left[ \frac{y^3}{6} \right]_{0}^{1} = -\frac{1}{6}. \]
07
Conclusion
The value of the line integral \( \int_{C} xy \, dx + \sqrt{y^2+1} \, dy \) is \(-\frac{1}{6}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integral
The concept of a line integral explores the idea of integrating a function along a curve. Imagine you are traveling along a path, and at each point on this path, the function gives the 'weight' or 'force' experienced. A line integral calculates the total effect along that path.
\[\int_C f(x, y) \, ds\]A line integral is valuable in physics and engineering where dimensions such as force or heat are not uniform throughout a path.
For a vector field \(\vec{F} = (P, Q)\), a line integral can be expressed as:
\[\int_C P\,dx + Q\,dy\].
In the exercise, this is expressed as \(\int_{C} xy \, dx + \sqrt{y^2+1} \, dy\). The components \(P\) and \(Q\) are functions affecting the path, calculated through Green's Theorem in the exercise, simplifying the line integral to a function over the enclosed region.
\[\int_C f(x, y) \, ds\]A line integral is valuable in physics and engineering where dimensions such as force or heat are not uniform throughout a path.
For a vector field \(\vec{F} = (P, Q)\), a line integral can be expressed as:
\[\int_C P\,dx + Q\,dy\].
In the exercise, this is expressed as \(\int_{C} xy \, dx + \sqrt{y^2+1} \, dy\). The components \(P\) and \(Q\) are functions affecting the path, calculated through Green's Theorem in the exercise, simplifying the line integral to a function over the enclosed region.
Double Integral
A double integral extends the idea of an integral into two dimensions. It calculates the accumulation of a quantity over a region, often providing a volume under a surface described by a function.
The generic expression for a double integral is:
\[\iint_{R} f(x, y) \, dA\].
In the area defined by the curve \(C\), Green’s Theorem enables converting the complex line integral into a manageable double integral over the triangular region with vertices \((0, 0), (1, 1), (0, 1)\).
For our problem, Green's Theorem yields:
\[\iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = -\iint_{R} x \, dA\].
This allows focusing on calculating a simpler double integral over \(x\), leading to a result of \(-\frac{1}{6}\). Double integrals elegantly handle the accumulation of diverse fields across bounded areas.
The generic expression for a double integral is:
\[\iint_{R} f(x, y) \, dA\].
In the area defined by the curve \(C\), Green’s Theorem enables converting the complex line integral into a manageable double integral over the triangular region with vertices \((0, 0), (1, 1), (0, 1)\).
For our problem, Green's Theorem yields:
\[\iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = -\iint_{R} x \, dA\].
This allows focusing on calculating a simpler double integral over \(x\), leading to a result of \(-\frac{1}{6}\). Double integrals elegantly handle the accumulation of diverse fields across bounded areas.
Vector Field
A vector field is a mathematical construction that assigns a vector to every point in space. Imagine standing in a wind tunnel and watching arrows indicate the direction and strength of the wind at each location - that's a basic representation of a vector field.
In mathematical terms:
\(\vec{F} = (P(x, y), Q(x, y))\), where each pair \((P, Q)\) provides a vector at every coordinates \((x, y)\).
For solving our exercise, the given line integral formula \(\int_{C} xy \, dx + \sqrt{y^2+1} \, dy\) represents a vector field where \(P = xy\) and \(Q = \sqrt{y^2+1}\).
Understanding this helps reveal how Green's Theorem transitions between the line integral around a curve and the double integral over the region, using vectors to summarize changes across space.
In mathematical terms:
\(\vec{F} = (P(x, y), Q(x, y))\), where each pair \((P, Q)\) provides a vector at every coordinates \((x, y)\).
For solving our exercise, the given line integral formula \(\int_{C} xy \, dx + \sqrt{y^2+1} \, dy\) represents a vector field where \(P = xy\) and \(Q = \sqrt{y^2+1}\).
Understanding this helps reveal how Green's Theorem transitions between the line integral around a curve and the double integral over the region, using vectors to summarize changes across space.