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Evaluate the following line integrals: a) \(\oint_{C} y^{2} d x+x y d y\), where \(C\) is the square with vertices \((1,1),(-1,1),(-1,-1)\), \((1,-1)\) b) \(\oint_{C} y d x-x d y\), where \(C\) is the circle \(x^{2}+y^{2}=1(\) cf. (5.24)); c) \(\oint_{C} x^{2} y^{2} d x-x y^{3} d y\), where \(C\) is the triangle with vertices \((0,0),(1,0),(1,1)\).

Short Answer

Expert verified
In summary, the three line integral evaluations are as follows: a) The line integral of \(y^{2} dx + xy dy\) over the square is equal to 2. b) The line integral of \(y dx - x dy\) over the circle is equal to \(-2\pi\). c) The line integral of \(x^{2}y^{2} dx - xy^{3} dy\) over the triangle is equal to \(\frac{1}{30}\).

Step by step solution

01

Parameterize the square

We will parameterize the square by dividing it into four line segments. 1st segment: \((x,y)=(1,t)\) with \(-1\leq t\leq1\). 2nd segment: \((x,y)=(-t,1)\) with \(-1\leq t\leq1\). 3rd segment: \((x,y)=(-1,-t)\) with \(-1\leq t\leq1\). 4th segment: \((x,y)=(t,-1)\) with \(-1\leq t\leq1\).
02

Compute the line integral for each segment

For each segment, we need to compute the line integral by substituting the parameterization and integrate with respect to the parameter. 1st segment: $$ \int_{-1}^{1}[(1)^2 dt + (1)(t)d(1)] = \int_{-1}^{1} dt = [t]_{-1}^{1}=1-(-1)=2. $$ 2nd segment: $$ \int_{-1}^{1}[(-t)^2 dt + (-t)(1)d(-t)] = \int_{-1}^{1}[t^2 - t^2 dt] = \int_{-1}^{1} 0 dt = 0. $$ 3rd segment: $$ \int_{-1}^{1}[(-1)^2 dt + (-1)(-t)d(-1)] = \int_{-1}^{1}[ dt - dt] = \int_{-1}^{1} 0 dt = 0. $$ 4th segment: $$ \int_{-1}^{1}[(t)^2 dt + (t)(-1)d(t)] = \int_{-1}^{1}[t^2 - t^2 dt] = \int_{-1}^{1} 0 dt = 0. $$
03

Sum the line integrals of each segment

The total line integral is the sum of the line integrals of each segment: $$ 2+0+0+0=2. $$ #b) Line integral of \(y dx - x dy\) over a circle#
04

Parameterize the circle

The circle \(x^{2}+y^{2}=1\) can be parameterized as \((x,y)=(\cos t,\sin t)\) with \(0\leq t\leq2\pi\).
05

Compute the line integral over the circle

Substitute the parameterization and integrate with respect to the parameter: $$ \oint_{C} y dx - x dy = \int_{0}^{2\pi} [\sin t d(\cos t) - \cos t d(\sin t)] = \int_{0}^{2\pi} (-\sin^2t-\cos^2t) dt = -\int_{0}^{2\pi} dt=-[t]_{0}^{2\pi}=-2\pi. $$ #c) Line integral of \(x^{2}y^{2} dx - xy^{3} dy\) over a triangle#
06

Parameterize the triangle

We will parameterize the triangle by dividing it into three line segments. 1st segment: \((x,y)=(t,0)\) with \(0\leq t\leq1\). 2nd segment: \((x,y)=(1,t)\) with \(0\leq t\leq1\). 3rd segment: \((x,y)=(1-t,t)\) with \(0\leq t\leq1\).
07

Compute the line integral for each segment

For each segment, we need to compute the line integral by substituting the parameterization and integrating with respect to the parameter. 1st segment: $$ \int_{0}^{1}[t^2(0)^2 dt - (t)(0)^3 d(t)] = \int_{0}^{1} 0 dt = 0. $$ 2nd segment: $$ \int_{0}^{1}[(1)^2(t)^2 dt - (1)(t)^3 d(1)] = \int_{0}^{1} t^2 dt = \frac{1}{3}t^3 \big|_{0}^{1} = \frac{1}{3}. $$ 3rd segment: $$ \int_{0}^{1}[(1-t)^2(t)^2 d(1-t) - (1-t)(t)^3 d(t)] = -\int_{0}^{1}[2t^3 - 3t^4 + t^5 dt] = -[t^4/2 - 3t^5/5 + t^6/6] \big|_{0}^{1} = -\frac{11}{30}. $$
08

Sum the line integrals of each segment

The total line integral is the sum of the line integrals of each segment: $$ 0+\frac{1}{3}-\frac{11}{30}=\frac{1}{30}. $$

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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 result in vector calculus that relates a line integral around a simple closed curve to a double integral over the plane region it encloses. Imagine you have a curve bounding a nice, flat surface. Instead of calculating the line integral directly, Green's Theorem allows you to see the same result by integrating over the area inside the curve.

Here's how it works: for a vector field \(\mathbf{F} = P(x, y)\mathbf{i} + Q(x, y)\mathbf{j}\), Green's Theorem states that:

$$ \oint_{C} (P\,dx + Q\,dy) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA $$

\(C\) is the positively oriented, simple closed curve, and \(R\) is the region it encloses. This concept simplifies many line integrals, especially in exercises where directly integrating over segments seems cumbersome.
Parameterization
Parameterization is a method to represent a curve using a single variable, typically \(t\), instead of \(x\) and \(y\) directly. It makes calculating integrals easier since all locations on the curve can be described with just one parameter.

When dealing with specific shapes like squares, circles, or triangles, you can segment the shape into simpler, linear sections. For example, a square can be broken into four line segments like this:
  • The first segment: \(x = 1,\ y = t,\ -1 \leq t \leq 1\)
  • The second segment: \(x = -t,\ y = 1\)
This way, each segment is a linear path from one point to another. Once in this form, you can easily substitute into the integral and evaluate with respect to \(t\). This is especially helpful for closed shapes.
Path Integrals
Path integrals, or line integrals, extend regular integrals to include paths in two or three dimensions. They measure things like work done by a force along a path, or the flow of a fluid along a pipe.

To compute a path integral for a vector field \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j}\) along a curve \(C\), we use:

$$ \oint_{C} P dx + Q dy $$

The idea is to integrate the vector field along the path, using parameterization to handle the curve. In exercises involving squares, circles, or triangles, you'll parameterize the path and evaluate the integral in sections, summing the results. Green's Theorem can sometimes offer a shortcut, connecting the path integral to an area integral.
Calculus
Calculus provides the tools for understanding change and accumulation, central to evaluating integrals. It involves differentiating and integrating functions to analyze real-world problems.

In terms of line integrals, calculus helps by:
  • Providing definitions of derivatives and integrals
  • Helping us understand curves via parameterization methods
  • Allowing us to apply the fundamental theorems, like the Fundamental Theorem of Calculus and Green's Theorem
Each of these involves manipulating and interpreting mathematical expressions to solve problems involving changes along paths or over areas. In exercises like these, calculus is the backbone that supports every step, from breaking down curves into simpler parts to applying theorems for simpler computation.

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Most popular questions from this chapter

Find the Newtonian potential \(U\) of the solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\) with density \(\mu=k \sqrt{x^{2}+y^{2}+z^{2}}\). Verify that \(\nabla^{2} U=-4 \pi \mu\) inside the sphere.

a) Let a surface \(S: z=f(x, y)\) be defined by an implicit equation \(F(x, y, z)=0\). Show \(\mathrm{T}\) that the surface integral \(\iint H d \sigma\) over \(S\) becomes $$ \iint_{R_{c y}} \sqrt{\left(\frac{\partial F}{\partial x}\right)^{2}+\left(\frac{\partial F}{\partial y}\right)^{2}+\left(\frac{\partial F}{\partial z}\right)^{2}} \frac{H}{\left|\frac{\partial F}{\partial z}\right|} d x d y $$ provided that \(\frac{\partial F}{\partial z} \neq 0\). i b) Prove that for the surface of part (a) with \(\mathbf{n}=\nabla F /|\nabla F|\). $$ \iint_{S}(\mathbf{v} \cdot \mathbf{n}) d \sigma=\iint_{R_{s t}}(\mathbf{v} \cdot \boldsymbol{\nabla} F) \frac{1}{\left|\frac{\partial F}{\partial z}\right|} d x d y . $$ c) Prove (5.81). d) Prove that (5.81) reduces to (5.80) when \(x=u, y=v, z=f(u, v)\).

(The solid angle) Let \(S\) be a plane surface, oriented in accordance with a unit normal \(\mathbf{n}\). The solid angle \(\Omega\) of \(S\) with respect to a point \(O\) not in \(S\) is defined as $$ \Omega(O, S)=\pm \text { area of projection of } S \text { on } S_{1}, $$ where \(S_{1}\) is the sphere of radius 1 about \(O\) and the \(+\) or - sign is chosen according to whether \(\mathbf{n}\) points away from or toward the side of \(S\) on which \(O\) lies. This is suggested in Fig. 5.37. a) Show that if \(O\) lies in the plane of \(S\) but not in \(S\), then \(\Omega(O, S)=0\). it b) Show that if \(S\) is a complete (that is, infinite) plane, then \(\Omega(O, S)=\pm 2 \pi\). c) For a general oriented surface \(S\) the surface can be thought of as made up of small elements, each of which is approximately planar and has a normal \(\mathbf{n}\). Justify the following definition of element of solid angle for such a surface element: $$ d \Omega=\frac{\mathbf{r} \cdot \mathbf{n}}{r^{3}} d \sigma $$ where \(\mathbf{r}\) is the vector from \(O\) to the element. d) On the basis of the formula of (c), one obtains as solid angle for a general oriented surface \(S\) the integral $$ \Omega(O, S)=\iint_{S} \frac{\mathbf{r} \cdot \mathbf{n}}{r^{3}} d \sigma . $$ Show that for surfaces in parametric form, if \(O\) is the origin, $$ \Omega(O, S)=\iint_{R_{x r}}\left|\begin{array}{ccc} x & y & z \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{array}\right| \frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}} d u d v . $$ This formula permits one to define a solid angle for complicated surfaces that interac themselves. e) Show that if the normal of \(S_{1}\) is the outer one, then \(\Omega(O, S)=4 \pi\). f) Show that if \(S\) forms the boundary of a bounded, closed, simply connected region \(R\). then \(\Omega(O, S), \pm 4 \pi\), when \(O\) is inside \(S\) and \(\Omega(O, S)=0\) when \(O\) is outside \(S\). g) If \(S\) is a fixed circular disk and \(O\) is variable, show that \(-2 \pi \leq \Omega(O, S) \leq 2 \pi\) and that \(\Omega(O, S)\) jumps by \(4 \pi\) as \(O\) crosses \(S\).

Find the temperature distribution in a solid whose boundaries are two parallel planes, \(d\) units apart, kept at temperatures \(T_{1}, T_{2}\), respectively. (Hint: Take the boundaries to be the planes \(x=0, x=d\) and note that, by symmetry, \(T\) must be independent of \(y\) and \(z\), .)

If \(\mathbf{v}=\left(x^{2}+y^{2}\right) \mathbf{i}+(2 x y) \mathbf{j}\), evaluate \(\int_{C} v_{T} d s\) for the following paths: a) from \((0,0)\) to \((1,1)\) on the line \(y=x\), b) from \((0,0)\) to \((1,1)\) on the line \(y=x^{2}\); c) from \((0,0)\) to \((1,1)\) on the broken line with corner at \((1,0)\).

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