Chapter 14: Problem 8
Evaluate \(\oint_{c}\left\\{x^{3} y^{2} \mathrm{~d} x+x^{2} y \mathrm{~d} y\right\\}\) where \(c\) is the boundary of the region enclosed by the curve \(y=1-x^{2}, x=0\) and \(y=0\) in the first quadrant.
Short Answer
Expert verified
The value of the line integral is \(\frac{1}{8}\).
Step by step solution
01
- Identify the Boundary Curve
The region to be evaluated is enclosed by the curves: the parabola given by the equation, \(y = 1 - x^2\), the vertical line \(x = 0\) and the horizontal line \(y = 0\), in the first quadrant.
02
- Parametrize the Curve
To use Green's Theorem, parameterize the curve \(C\). The boundary can be broken into three parts: 1. From \( (0,0) \) to \( (1,0) \) along \( y=0 \) 2. From \( (1,0) \) to \( (0,1) \) along the parabola \(y=1-x^2\) 3. From \( (0,1) \) to \( (0,0) \) along \( x=0 \).
03
- Apply Green's Theorem
Green's Theorem states that for a positively oriented, piecewise smooth simple closed curve \(C\) and the region \(D\) it encloses, \[\text{\oint_{c}}(P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]Here, \(P = x^3y^2\) and \(Q = x^2 y\). Compute the partial derivatives: \(\frac{\partial Q}{\partial x} = 2xy\) and \(\frac{\partial P}{\partial y} = 2x^3 y\).
04
- Calculate the Double Integral
Apply Green's theorem: \[\iint_D \left(2xy - 2x^3 y\right) dA \]The region \(D\) is bounded by 1. \(0 \leq x \leq 1\) and 2. \(0 \leq y \leq 1-x^2\). So, the double integral becomes: \[\iint_D \left(2xy - 2x^3 y\right) dy \, dx = \int_0^1 \left( \int_0^{1-x^2} (2xy - 2x^3 y) dy \right) dx\]
05
- Integrate with Respect to y
First integrate with respect to \(y\): \[\int_0^{1-x^2} (2xy - 2x^3 y) dy = \left[ x y^2 - x^3 y^2 \right]_0^{1-x^2} = x (1-x^2)^2 - x^3 (1-x^2)^2 = x (1 - 2x^2 + x^4) - x^3 (1 - 2x^2 + x^4) \]
06
- Simplify the Expression
Simplify the expression: \[ = x (1 - 2x^2 + x^4) - x^3 (1 - 2x^2 + x^4) = x - 2x^3 + x^5 - x^3 + 2x^5 - x^7 = x - 3x^3 + 3x^5 - x^7\]
07
- Integrate with Respect to x
Now, integrate with respect to \(x\): \[\int_0^1 (x - 3x^3 + 3x^5 - x^7) dx = \left[ \frac{x^2}{2} - \frac{3x^4}{4} + \frac{3x^6}{6} - \frac{x^8}{8} \right]_0^1 = \frac{1}{2} - \frac{3}{4} + \frac{1}{2} - \frac{1}{8}\]
08
- Final Calculation
Combine the final terms: \[\frac{1}{2} - \frac{3}{4} + \frac{1}{2} - \frac{1}{8} = \frac{1}{2} + \frac{1}{2} - \frac{3}{4} - \frac{1}{8} = 1 - \frac{3}{4} - \frac{1}{8} = 1 - \frac{6}{8} - \frac{1}{8} = 1 - \frac{7}{8} = \frac{1}{8}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Green's Theorem
Green's Theorem is a powerful tool in vector calculus that connects the line integral around a closed curve to a double integral over the region enclosed by the curve. It is especially useful when dealing with planar regions. The theorem is stated as follows:
For a positively oriented, piecewise smooth simple closed curve \(C\) and the region \(D\) it encloses:
\[ \oint_{C}(P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]
Here, \(P\) and \(Q\) are the components of a vector field, and the left-hand side of the equation is a line integral, while the right-hand side is a double integral. The partial derivatives \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) denote how \(Q\) changes with respect to \(x\) and how \(P\) changes with respect to \(y\), respectively. Essentially, Green's Theorem helps convert a line integral into a double integral, simplifying the calculation over difficult curves.
For a positively oriented, piecewise smooth simple closed curve \(C\) and the region \(D\) it encloses:
\[ \oint_{C}(P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]
Here, \(P\) and \(Q\) are the components of a vector field, and the left-hand side of the equation is a line integral, while the right-hand side is a double integral. The partial derivatives \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) denote how \(Q\) changes with respect to \(x\) and how \(P\) changes with respect to \(y\), respectively. Essentially, Green's Theorem helps convert a line integral into a double integral, simplifying the calculation over difficult curves.
Line Integral
A line integral, also known as a path or curve integral, is used to integrate functions along a curve. In simpler terms, it's how we sum up a function's value over a path. The line integral of a vector field \(\textbf{F} = P \hat{i} + Q \hat{j} \) around a curve \(C\) is:
\( \oint_C P \, dx + Q \, dy \)
This adds up the product of the function's value and a small segment of the curve, accommodating the curve’s orientation and parametrization. To compute it:
\( \oint_C P \, dx + Q \, dy \)
This adds up the product of the function's value and a small segment of the curve, accommodating the curve’s orientation and parametrization. To compute it:
- First, parameterize the curve by expressing \(dx\) and \(dy\) in terms of a parameter like \(t\).
- Second, substitute these parameterized values into the integral.
- Third, integrate with respect to the parameter.
Double Integral
A double integral allows us to integrate over a two-dimensional region. It is useful for finding areas, masses, and volumes where the density isn’t uniform. The typical notation for a double integral over region \(D\) is:
\( \iint_D f(x, y) \, dA \)
Here, \(f(x, y)\) is a function of two variables, and \(dA \) represents an infinitely small area element within \(D\). To compute a double integral:
\( \iint_D f(x, y) \, dA \)
Here, \(f(x, y)\) is a function of two variables, and \(dA \) represents an infinitely small area element within \(D\). To compute a double integral:
- Split the region \(D\) into tiny pieces.
- Calculate \(f(x, y)\) at each tiny piece, multiply by the tiny area, and sum them up.
Parametrization of Curves
Parametrization involves expressing a curve using a parameter, usually \(t\). This helps in simplifying computations involving curves. Specifically, for a curve \(C\), we represent \(x\) and \(y\) as functions of \(t\):
\(x = g(t)\) and \(y = h(t)\)
The parameter \(t\) typically varies over an interval, like \([a, b]\). Given a curve's parameterization, we can easily compute derivatives \(dx/dt\) and \(dy/dt\). When solving problems using Green's Theorem, correctly parameterizing your curves can make a big difference.
In the exercise:
\(x = g(t)\) and \(y = h(t)\)
The parameter \(t\) typically varies over an interval, like \([a, b]\). Given a curve's parameterization, we can easily compute derivatives \(dx/dt\) and \(dy/dt\). When solving problems using Green's Theorem, correctly parameterizing your curves can make a big difference.
In the exercise:
- We first parameterized each segment of our boundary region, simplifying each into more manageable forms.
- We used this parameterization to convert into double integrals.