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Chapter 9: Problem 6

$$\begin{aligned} P=x+y^{2}, P_{y}=2 y, Q=2 x^{2} &-y, Q_{x}=4 x \\ \oint_{C}\left(x+y^{2}\right) d x+\left(2 x^{2}-y\right) d y &=\iint_{R}(4 x-2 y) d A=\int_{-2}^{2} \int_{x^{2}}^{4}(4 x-2 y) d y d x \\ &=\left.\int_{-2}^{2}\left(4 x y-y^{2}\right)\right|_{x^{2}} ^{4} d x=\int_{-2}^{2}\left(16 x-16-4 x^{3}+x^{4}\right) d x \\ &=\left.\left(8 x^{2}-16 x-x^{4}+\frac{1}{5} x^{5}\right)\right|_{-2} ^{2}=-\frac{96}{5} \end{aligned}$$

Short Answer

Expert verified
The value of the integral is \(-\frac{96}{5}\).

Step by step solution

01

Understanding the Integrals

The problem involves a line integral over a closed curve \( C \) given by \( \oint_{C}(x+y^2) \, dx + (2x^2-y) \, dy \). According to Green's Theorem, this can be converted into a double integral over the region \( R \), \( \iint_{R}(Q_x - P_y) \, dA \), where \( P = x+y^2 \) and \( Q = 2x^2-y \). The partial derivatives are \( P_y = 2y \) and \( Q_x = 4x \). Substitute these into the double integral formula to represent the line integral as \( \iint_{R}(4x - 2y) \, dA \).
02

Setting up the Double Integral

Identify the limits of integration for \( x \) and \( y \). The region \( R \) is bounded by the curves \( y = x^2 \) and \( y = 4 \) over \( x \)-interval \(-2 \leq x \leq 2\). The integral becomes \( \int_{-2}^{2} \int_{x^2}^{4}(4x - 2y) \, dy \, dx \).
03

Integrating with Respect to \( y \)

Perform the integration of \( 4x - 2y \) with respect to \( y \) while holding \( x \) constant. This gives \( \int (4x - 2y) \, dy = 4xy - y^2 \). Evaluate this from \( y = x^2 \) to \( y = 4 \). This results in \( \left[4xy - y^2\right]_{x^2}^{4} = 16x - 16 - 4x^3 + x^4 \).
04

Integrating with Respect to \( x \)

Now, integrate the resulting expression \( 16x - 16 - 4x^3 + x^4 \) with respect to \( x \) over the interval \(-2 \leq x \leq 2\). The integral becomes \( \int_{-2}^{2} (16x - 16 - 4x^3 + x^4) \, dx \). Integrate each term separately to get \( 8x^2 - 16x - x^4 + \frac{1}{5}x^5 \).
05

Evaluate the Final Integral

Evaluate the integral \( \left[8x^2 - 16x - x^4 + \frac{1}{5}x^5 \right]_{-2}^{2} \). Begin by substituting \( x = 2 \) and \( x = -2 \) into the expression separately, then subtract. The evaluation results in \( -\frac{96}{5} \). This is the value of the closed line integral over curve \( C \).

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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 fascinating part of calculus that involve integrating a function along a curve.
They look at the sum of values taken by a function across a path.
This type of integral is not limited to straight paths; it can follow any curved or winding path you like!Here's a breakdown of line integrals:
  • They are usually written as \( \oint_{C} P\, dx + Q\, dy \), where \( C \) is the curve.
  • The integral involves two components, \( P \) and \( Q \), which are functions of coordinates along the curve.
  • Line integrals are essential for fields such as physics and engineering, particularly in calculating work done by a force over a path.
  • Green's Theorem simplifies a complex line integral into a more manageable double integral, under specific conditions.
As you study Green's Theorem, pay careful attention to how a line integral is transformed into a double integral. This transformation makes certain calculations simpler and reveals deeper insights into vector fields and area integrations.
Double Integrals
Double integrals allow us to calculate the volume under a surface. They work over a region instead of just a line.Key points about double integrals include:
  • They are typically expressed as \( \iint_R f(x, y)\, dA \), where \( R \) is the region of integration.
  • Double integrals can be used to find the area of a region, calculate the volume under a surface, or even to solve complex equations like those in Green's Theorem.
  • The integral is performed in two steps: first integrating with respect to one variable while treating others as constants, then integrating the result with respect to the remaining variables.
  • When using double integrals with Green's Theorem, you're calculating over a planar region defined by the boundaries specified within the problem.
In the context of the exercise you explored, setting up the limits correctly based on the region's boundaries, such as \( y = x^2 \) and \( y = 4 \), is crucial to solving the integral accurately.
Partial Derivatives
Partial derivatives are a fundamental tool in calculus for functions of multiple variables. They represent the change of a function concerning one variable while holding others constant.Important aspects of partial derivatives include:
  • They are noted as \( \frac{\partial f}{\partial x} \) or \( \frac{\partial f}{\partial y} \), where \( f \) is a multivariable function.
  • In Green鈥檚 Theorem, partial derivatives help evaluate the curl of a vector field, giving clues about the behavior within the region \( R \).
  • Specifically, terms \( P_{y} \) and \( Q_{x} \) involve taking partial derivatives of functions \( P \) and \( Q \) with respect to \( y \) and \( x \), respectively.
  • Finding these derivatives correctly is crucial as they directly influence the outcome of converting a line integral to a double integral.
Understanding partial derivatives provides insight into how variables independently affect multi-dimensional systems, crucial for solving higher-dimensional calculus problems.
Calculus
Calculus is the mathematical study of continuous change and is split into differential and integral calculus. It is where concepts like derivatives and integrals are foundational. Here's what you should know about calculus in this context:
  • Differential calculus focuses on the rate of change and slopes of curves via derivatives.
  • Integral calculus, contrastingly, concerns itself with accumulation of quantities and the areas under/potential over curves, using integrals.
  • Green鈥檚 Theorem is an application of calculus combining these two ideas to convert a line integral to a double integral.
  • Grasping calculus well can demystify and streamline solutions to complex real-world problems, such as those involving fluid flow or electric fields.
By seeing how calculus principles underpin the problem, like the one reviewed, you gain a deeper comprehension of its capabilities in mathematics and its applications widely found within scientific disciplines.

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Most popular questions from this chapter

We use polar coordinates in the \(y z\) -plane. The density is \(\rho(x, y, z)=k z .\) By symmetry, \(\bar{y}=\bar{z}=0\). $$\begin{aligned} &\begin{aligned} m &=\int_{0}^{2 \pi} \int_{0}^{4} \int_{0}^{5} k x r d x d r d \theta=\left.k \int_{0}^{2 \pi} \int_{0}^{4} \frac{1}{2} r z^{2}\right|_{0} ^{3} d r d \theta=\frac{k}{2} \int_{0}^{2 \pi} \int_{0}^{4} 25 r d r d \theta \\ &=\left.\frac{25 k}{2} \int_{0}^{2 \pi} \frac{1}{2} r^{2}\right|_{0} ^{4} d \theta=\frac{25 k}{2} \int_{0}^{2 \pi} 8 d \theta=200 k \pi \\ M_{y z} &=\int_{0}^{2 \pi} \int_{0}^{4} \int_{0}^{5} k x^{2} r d x d r d \theta=\left.k \int_{0}^{2 \pi} \int_{0}^{4} \frac{1}{3} r x^{3}\right|_{0} ^{5} d r d \theta=\frac{1}{3} k \int_{0}^{2 \pi} \int_{0}^{4} 125 r d r d \theta \\ &=\left.\frac{1}{3} k \int_{0}^{2 \pi} \frac{125}{2} r^{2}\right|_{0} ^{4} d \theta=\frac{1}{3} k \int_{0}^{2 \pi} 1000 d \theta=\frac{2000}{3} k \pi \end{aligned}\\\ &\bar{x}=M_{y z} / m=\frac{2000 k \pi / 3}{200 k \pi}=10 / 3 . \text { The center of mass of the given solid is }(10 / 3,0,0). \end{aligned}$$

The area of the hemisphere is \(A(s)=2 \pi a^{2} .\) By symmetry, \(\bar{x}=\bar{y}=0\) \(z_{x}=-\frac{x}{\sqrt{a^{2}-x^{2}-y^{2}}}, z_{y}=-\frac{y}{\sqrt{a^{2}-x^{2}-y^{2}}}\) \(d S=\sqrt{1+\frac{x^{2}}{a^{2}-x^{2}-y^{2}}+\frac{y^{2}}{a^{2}-x^{2}-y^{2}}} d A=\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}} d A\) Using polar coordinates. $$\begin{aligned} z &=\iint_{S} \frac{z d S}{2 \pi a^{2}}=\frac{1}{2 \pi a^{2}} \iint_{R} \sqrt{a^{2}-x^{2}-y^{2}} \frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}} d A=\frac{1}{2 \pi a} \int_{0}^{2 \pi} \int_{0}^{a} r d r d \theta \\ &=\left.\frac{1}{2 \pi a} \int_{0}^{2 \pi} \frac{1}{2} r^{2}\right|_{0} ^{a} d \theta=\frac{1}{2 \pi a} \int_{0}^{2 \pi} \frac{1}{2} s^{2} d \theta=\frac{a}{2} \end{aligned}$$The centroid is \((0,0, a / 2)\)

$$4 z^{2}=3 x^{2}+3 y^{2}+3 z^{2} ; 4 \rho^{2} \cos ^{2} \phi=3 \rho^{2} ; \cos \phi=\pm \sqrt{3} / 2 ; \phi=\pi / 6 \text { or equivalently, } \phi=5 \pi / 6$$

\(R 1: y=x^{2} \Longrightarrow v+u=v-u \Rightarrow u=0\) \(\begin{aligned} R 2: y=4-x^{2} & \Longrightarrow v+u=4-(v-u) \\ & \Longrightarrow v+u=4-v+u \Longrightarrow v=2 \end{aligned}\) \(R 3: x=1 \Longrightarrow v-u=1 \Rightarrow v=1+u\) \(\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{cc}-\frac{1}{2 \sqrt{v-u}} & \frac{1}{2 \sqrt{v-u}} \\ 1 & 1\end{array}\right|=-\frac{1}{\sqrt{v-u}}\) $$\begin{aligned} \iint_{R} \frac{x}{y+x^{2}} d A &=\iint_{S} \frac{\sqrt{v-u}}{2 v}\left|-\frac{1}{\sqrt{v-u}}\right| d A^{\prime}=\frac{1}{2} \int_{0}^{1} \int_{1+u}^{2} \frac{1}{v} d v d u=\frac{1}{2} \int_{0}^{1}[\ln 2-\ln (1+u)] d u \\ &=\frac{1}{2} \ln 2-\left.\frac{1}{2}[(1+u) \ln (1+u)-(1+u)]\right|_{0} ^{1}=\frac{1}{2} \ln 2-\frac{1}{2}[2 \ln 2-2-(0-1)]=\frac{1}{2}-\frac{1}{2} \ln 2 \end{aligned}$$

We are given \(\rho(x, y, z)=k x\). $$\begin{aligned} I_{y} &=\int_{0}^{1} \int_{0}^{2} \int_{z}^{4-z} k x\left(x^{2}+z^{2}\right) d y d x d z=\left.k \int_{0}^{1} \int_{0}^{2}\left(x^{3}+x z^{2}\right) y\right|_{z} ^{4-z} d x d z \\ &=k \int_{0}^{1} \int_{0}^{2}\left(x^{3}+x z^{2}\right)(4-2 z) d x d z=\left.k \int_{0}^{1}\left(\frac{1}{4} x^{4}+\frac{1}{2} x^{2} z^{2}\right)(4-2 z)\right|_{0} ^{2} d z \end{aligned}$$ $$=k \int_{0}^{1}\left(4+2 z^{2}\right)(4-2 z) d z=4 k \int_{0}^{1}\left(4-2 z+2 z^{2}-z^{3}\right) d z=\left.4 k\left(4 z-z^{2}+\frac{2}{3} z^{3}-\frac{1}{4} z^{4}\right)\right|_{0} ^{1}=\frac{41}{3} k$$
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