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Magazine Chapter 16: Problem 7
Compute \(\int_{\partial D}(x / y) d x+(2+3 x) d y,\) where \(D\) is described by \(1 \leq x \leq 2,1 \leq y \leq x^{2}\).
Short Answer
Expert verified
The integral evaluates to \(\frac{11}{2} - \ln 2\).
Step by step solution
01
Understand the Problem
The given problem is to evaluate the line integral \(\int_{\partial D} \left( \frac{x}{y} \right) dx + \left( 2 + 3x \right) dy\), where \(D\) is the region bounded by \(1 \leq x \leq 2\) and \(1 \leq y \leq x^{2}\). This line integral can be converted to a double integral over the region \(D\) using Green's Theorem.
02
Apply Green's Theorem
According to Green's Theorem, the line integral \(\oint_{\partial D} P \, dx + Q \, dy = \int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy\). Here, \(P(x,y) = \frac{x}{y}\) and \(Q(x,y) = 2 + 3x\).
03
Differentiate Q with respect to x
Find \(\frac{\partial Q}{\partial x}\) for \(Q(x,y) = 2 + 3x\). The partial derivative with respect to \(x\) is \(\frac{\partial}{\partial x}(2 + 3x) = 3\).
04
Differentiate P with respect to y
Find \(\frac{\partial P}{\partial y}\) for \(P(x,y) = \frac{x}{y}\). Using the quotient rule, \(\frac{\partial}{\partial y}\left(\frac{x}{y}\right) = -\frac{x}{y^2}\).
05
Set up the Double Integral using Green's Theorem
Using Green's Theorem, set up the double integral as:\[\int_{1}^{2} \int_{1}^{x^2} \left(3 - \left(-\frac{x}{y^2}\right)\right) dy \, dx\]which simplifies to:\[\int_{1}^{2} \int_{1}^{x^2} \left(3 + \frac{x}{y^2}\right) dy \, dx\]
06
Compute the Inner Integral
Evaluate the inner integral \(\int_{1}^{x^2} \left(3 + \frac{x}{y^2}\right) dy\). The integral of \(3\) with respect to \(y\) is \(3y\), and the integral of \(\frac{x}{y^2}\) is \(-\frac{x}{y}\):\[\left[3y - \frac{x}{y}\right]_{1}^{x^2}\]This results in \(\left[3x^2 - \frac{x}{x^2}\right] - \left[3 \times 1 - \frac{x}{1}\right]\).
07
Simplify the Result of the Inner Integral
Simplifying further gives: \(3x^2 - \frac{1}{x} - (3 - x) = 3x^2 + x - 3 - \frac{1}{x}\).
08
Compute the Outer Integral
Now, evaluate the outer integral:\[\int_{1}^{2} \left(3x^2 + x - 3 - \frac{1}{x}\right) dx\]This requires evaluating each term separately.
09
Integrate Term by Term
Evaluating each term:- \(\int_{1}^{2} 3x^2 \, dx = \left[x^3\right]_{1}^{2}\) which gives \(8 - 1 = 7\).- \(\int_{1}^{2} x \, dx = \left[\frac{x^2}{2}\right]_{1}^{2}\) which gives \(2 - \frac{1}{2} = \frac{3}{2}\).- \(\int_{1}^{2} -3 \, dx = \left[-3x\right]_{1}^{2}\) which gives \(-6 + 3 = -3\).- \(\int_{1}^{2} -\frac{1}{x} \, dx = \left[-\ln |x|\right]_{1}^{2}\) which gives \(-\ln 2 + 0 = -\ln 2\).
10
Combine the Results
Add all results from the integration: \[7 + \frac{3}{2} - 3 - \ln 2\ = 4 + \frac{3}{2} - \ln 2\]
11
Provide Final Answer
The evaluated integral results in a final answer of \[\frac{11}{2} - \ln 2\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integral
A line integral is a type of integral where a function is evaluated along a curve. In this exercise, the line integral involves the expression \(\int_{\partial D} \left( \frac{x}{y} \right) dx + \left( 2 + 3x \right) dy\). Understanding line integrals is like measuring how a function accumulates values while moving along a path. This is different from regular integrals that accumulate over an interval for one variable determined by bounds on the real line.
- Line integrals often arise in physics and engineering, such as computing work done by a force field.
- They require the curve \(\partial D\), an oriented boundary of a region \(D\).
Double Integral
Double integrals extend the concept of a single integral to two dimensions. In the context of Green's Theorem, it evaluates a function over an entire area rather than just a curve.
Here, the line integral is converted into a double integral \(\int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy\) over the region \(D\), where \(D\) is bounded by \(1 \leq x \leq 2\) and \(1 \leq y \leq x^2\). This region \(D\) comprises all points \((x, y)\) satisfying these limits.
Here, the line integral is converted into a double integral \(\int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy\) over the region \(D\), where \(D\) is bounded by \(1 \leq x \leq 2\) and \(1 \leq y \leq x^2\). This region \(D\) comprises all points \((x, y)\) satisfying these limits.
- Double integrals allow you to compute the volume under a surface over a specific area.
- They can also be used to calculate mass, charges, and other properties distributed over a plane.
Partial Derivatives
Partial derivatives are used to analyze the rate of change of functions that have more than one variable. It focuses on a single variable at a time, treating other variables as constant. In this exercise, we compute \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\).
- For \(Q(x, y) = 2 + 3x\), \(\frac{\partial Q}{\partial x} = 3\), meaning the derivative of 3x with respect to \(x\), keeping \(y\) constant.
- For \(P(x, y) = \frac{x}{y}\), \(\frac{\partial P}{\partial y}\) uses the quotient rule to find the rate of change with respect to \(y\).
Quotient Rule
The quotient rule is a technique in calculus to find the derivative of a ratio of two functions. It's expressed as \(\left(\frac{u}{v}\right)' = \frac{v' u - u' v}{v^2}\), where \(u\) and \(v\) are functions of a variable.
In this exercise, we find \(\frac{\partial P}{\partial y}\), where \(P(x, y) = \frac{x}{y}\). Applying the quotient rule yields:
In this exercise, we find \(\frac{\partial P}{\partial y}\), where \(P(x, y) = \frac{x}{y}\). Applying the quotient rule yields:
- \(u = x\), so \(u' = 0\), since \(x\) is treated as constant when differentiating with respect to \(y\).
- \(v = y\), so \(v' = 1\).
- Therefore, \(\frac{\partial}{\partial y}\left(\frac{x}{y}\right) = -\frac{x}{y^2}\).
