### Vector Line Integrals

The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let

[vecs F(x,y,z)=P(x,y,z),hat{mathbf i}+Q(x,y,z),hat{mathbf j}+R(x,y,z),hat{mathbf k}]

be a continuous vector field in (ℝ^3) that represents a force on a particle, and let (C) be a smooth curve in (ℝ^3) contained in the domain of (vecs F). How would we compute the work done by (vecs F) in moving a particle along (C)?

To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve (C); such a specified direction is called an orientation of a curve. The specified direction is the positive direction along (C); the opposite direction is the negative direction along (C). When (C) has been given an orientation, (C) is called an oriented curve (Figure (PageIndex{5})). The work done on the particle depends on the direction along the curve in which the particle is moving.

A closed curve is one for which there exists a parameterization (vecs r(t)), (a≤t≤b), such that (vecs r(a)=vecs r(b)), and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain ((a,b)).

Let (vecs r(t)) be a parameterization of (C) for (a≤t≤b) such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along (C). Divide the parameter interval ([a,b]) into n subintervals ([t_{i−1},t_i]), (0≤i≤n), of equal width. Denote the endpoints of (r(t_0)), (r(t_1)),…, (r(t_n)) by (P_0),…,(P_n). Points (P_i) divide (C) into n pieces. Denote the length of the piece from (P_{i−1}) to (P_i) by (Delta s_i). For each (i), choose a value (t_i^*) in the subinterval ([t_{i−1},t_i]). Then, the endpoint of (vecs r(t_i^*)) is a point in the piece of (C) between (P_{i−1}) and (P_i) (Figure (PageIndex{6})). If (Delta s_i) is small, then as the particle moves from (P_{i−1}) to (P_i) along (C), it moves approximately in the direction of (vecs T(P_i)), the unit tangent vector at the endpoint of (vecs r(t_i^*)). Let (P_i^*) denote the endpoint of (vecs r(t_i^*)). Then, the work done by the force vector field in moving the particle from (P_{i−1}) to (P_i) is (vecs F(P_i^*)·(Delta s_i vecs T(P_i^*))), so the total work done along (C) is

[sum_{i=1}^n vecs F(P_i^*)·(Delta s_i vecs T(P_i^*))=sum_{i=1}^n vecs F(P_i^*)·vecs T(P_i^*),Delta s_i.]

Letting the arc length of the pieces of (C) get arbitrarily small by taking a limit as (n ightarrow infty) gives us the work done by the field in moving the particle along (C). Therefore, the work done by (vecs{F}) in moving the particle in the positive direction along (C) is defined as

[W=int_C vecs{F} cdot vecs{T},ds,]

which gives us the concept of a vector line integral.

DEFINITION: Line Integral of a vector field

The vector line integral of vector field (vecs{F}) along oriented smooth curve (C) is

[int_C vecs{F} cdot vecs{T}, ds=lim_{n oinfty} sum_{i=1}^{n} vecs{F}(P_i^*) cdot vecs{T}(P_i^*)Delta s_i]

if that limit exists.

With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once by the parameterization, the value of the line integral is unchanged. With vector line integrals, the orientation of the curve does matter. If we think of the line integral as computing work, then this makes sense: if you hike up a mountain, then the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, then Earth’s gravitational force does positive work on you. In other words, reversing the path changes the work value from negative to positive in this case. Note that if (C) is an oriented curve, then we let (−C) represent the same curve but with opposite orientation.

As with scalar line integrals, it is easier to compute a vector line integral if we express it in terms of the parameterization function (vecs{r}) and the variable (t). To translate the integral (displaystyle int_C vecs{F} cdot vecs{T}ds) in terms of (t), note that unit tangent vector (vecs{T}) along (C) is given by (vecs{T}=dfrac{vecs{r}′(t)}{‖vecs{r}′(t)‖}) (assuming (‖vecs{r}′(t)‖≠0)). Since (ds=‖vecs r′(t)‖,dt), as we saw when discussing scalar line integrals, we have

[vecs F·vecs T,ds=vecs F(vecs r(t))·dfrac{vecs r′(t)}{‖vecs r′(t)‖}‖vecs r′(t)‖dt=vecs F(vecs r(t))·vecs r′(t),dt.]

Thus, we have the following formula for computing vector line integrals:

[int_Cvecs F·vecs T,ds=int_a^b vecs F(vecs r(t))·vecs r′(t),dt.label{lineintformula}]

Because of Equation ef{lineintformula}, we often use the notation (displaystyle int_C vecs{F} cdot dvecs{r}) for the line integral (displaystyle int_C vecs F·vecs T,ds).

If (vecs r(t)=⟨x(t),y(t),z(t)⟩), then (dfrac{dvecs{r}}{dt}) denotes vector (⟨x′(t),y′(t),z′(t)⟩), and (dvecs{r} = vecs r'(t),dt).

Example (PageIndex{5}): Evaluating a Vector Line Integral

Find the value of integral (displaystyle int_C vecs{F} cdot dvecs{r}), where (C) is the semicircle parameterized by (vecs{r}(t)=⟨cos t,sin t⟩), (0≤t≤pi) and (vecs F=⟨−y,x⟩).

**Solution**

We can use Equation ef{lineintformula} to convert the variable of integration from (s) to (t). We then have

[vecs F(vecs r(t))=⟨−sin t,cos t⟩ ; ext{and} ; vecs r′(t)=⟨−sin t,cos t⟩ . onumber]

Therefore,

[egin{align*} int_C vecs{F} cdot dvecs{r} &=int_0^{pi}⟨−sin t,cos t⟩·⟨−sin t,cos t⟩ ,dt [4pt] &=int_0^{pi} {sin}^2 t+{cos}^2 t ,dt [4pt] &=int_0^{pi}1 ,dt=pi.end{align*}]

See Figure (PageIndex{7}).

Example (PageIndex{6}): Reversing Orientation

Find the value of integral (displaystyle int_C vecs{F} cdot dvecs{r}), where (C) is the semicircle parameterized by (vecs r(t)=⟨cos (t+π),sin t⟩), (0≤t≤pi) and (vecs F=⟨−y,x⟩).

**Solution**

Notice that this is the same problem as Example (PageIndex{5}), except the orientation of the curve has been traversed. In this example, the parameterization starts at (vecs r(0)=⟨-1,0⟩) and ends at (vecs r(pi)=⟨1,0⟩). By Equation ef{lineintformula},

[egin{align*} int_C vecs{F} cdot dvecs{r} &=int_0^{pi} ⟨−sin t,cos (t+pi)⟩·⟨−sin (t+pi), cos t⟩dt[4pt] &=int_0^{pi}⟨−sin t,−cos t⟩·⟨sin t,cos t⟩dt[4pt] &=int_{0}^{π}(−{sin}^2 t−{cos}^2 t)dt [4pt] &=int_{0}^{pi}−1dt[4pt] &=−pi. end{align*}]

Notice that this is the negative of the answer in Example (PageIndex{5}). It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.

Let (C) be an oriented curve and let (-C) denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:

[int_{-C} vecs{F} cdot dvecs{r}=−int_Cvecs{F} cdot dvecs{r}.]

That is, reversing the orientation of a curve changes the sign of a line integral.

Exercise (PageIndex{6})

Let (vecs F=x,hat{mathbf i}+y ,hat{mathbf j}) be a vector field and let (C) be the curve with parameterization (⟨t,t^2⟩) for (0≤t≤2). Which is greater: (displaystyle int_Cvecs F·vecs T,ds) or (displaystyle int_{−C} vecs F·vecs T,ds)?

**Hint**Imagine moving along the path and computing the dot product (vecs F·vecs T) as you go.

**Answer**[int_C vecs F·vecs T ,ds]

Another standard notation for integral (displaystyle int_C vecs{F} cdot dvecs{r}) is (displaystyle int_C P,dx+Q,dy+R ,dz). In this notation, (P,, Q), and (R) are functions, and we think of (dvecs{r}) as vector (⟨dx,dy,dz⟩). To justify this convention, recall that (dvecs{r}=vecs T,ds=vecs r′(t) ,dt=leftlangledfrac{dx}{dt},dfrac{dy}{dt},dfrac{dz}{dt} ight angle,dt). Therefore,

[vecs{F} cdot dvecs{r}=⟨P,Q,R⟩·⟨dx,dy,dz⟩=P,dx+Q,dy+R,dz.]

If (dvecs{r}=⟨dx,dy,dz⟩), then (dfrac{dr}{dt}=leftlangledfrac{dx}{dt},dfrac{dy}{dt},dfrac{dz}{dt} ight angle), which implies that (dvecs{r}=leftlangledfrac{dx}{dt},dfrac{dy}{dt},dfrac{dz}{dt} ight angle,dt). Therefore

[egin{align} int_C vecs{F} cdot dvecs{r} &=int_C P,dx+Q,dy+R,dz [4pt] &=int_a^bleft(Pig(vecs r(t)ig)dfrac{dx}{dt}+Qig(vecs r(t)ig)dfrac{dy}{dt}+Rig(vecs r(t)ig)dfrac{dz}{dt} ight),dt. label{eq14}end{align} ]

Example (PageIndex{7}): Finding the Value of an Integral of the Form (displaystyle int_C P,dx+Q,dy+R,dz)

Find the value of integral (displaystyle int_C z,dx+x,dy+y,dz), where (C) is the curve parameterized by (vecs r(t)=⟨t^2,sqrt{t},t⟩), (1≤t≤4).

**Solution**

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of (t). In this case, Equation ef{eq14} allows us to make this change:

[egin{align*} int_C z,dx+x,dy+y,dz &=int_1^4 left(t(2t)+t^2left(frac{1}{2sqrt{t}} ight)+sqrt{t} ight),dt [4pt] &=int_1^4left(2t^2+frac{t^{3/2}}{2}+sqrt{t} ight),dt [4pt] &={left[dfrac{2t^3}{3}+dfrac{t^{5/2}}{5}+dfrac{2t^{3/2}}{3} ight]}_{t=1}^{t=4} [4pt] &=dfrac{793}{15}.end{align*}]

Exercise (PageIndex{7})

Find the value of (displaystyle int_C 4x,dx+z,dy+4y^2,dz), where (C) is the curve parameterized by (vecs r(t)=⟨4cos(2t),2sin(2t),3⟩), (0≤t≤dfrac{pi}{4}).

**Hint**Write the integral in terms of (t) using Equation ef{eq14}.

**Answer**(−26)

We have learned how to integrate smooth oriented curves. Now, suppose that (C) is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that (C) is a piecewise smooth curve. To be precise, curve (C) is piecewise smooth if (C) can be written as a union of n smooth curves (C_1), (C_2),…, (C_n) such that the endpoint of (C_i) is the starting point of (C_{i+1}) (Figure (PageIndex{8})). When curves (C_i) satisfy the condition that the endpoint of (C_i) is the starting point of (C_{i+1}), we write their union as (C_1+C_2+⋯+C_n).

The next theorem summarizes several key properties of vector line integrals.

Theorem: PROPERTIES OF VECTOR LINE INTEGRALS

Let (vecs F) and (vecs G) be continuous vector fields with domains that include the oriented smooth curve (C). Then

- (displaystyle int_C(vecs F+vecs G)·dvecs{r}=int_C vecs{F} cdot dvecs{r}+int_C vecs G·dvecs{r})
- (displaystyle int_C kvecs{F} cdot dvecs{r}=kint_C vecs{F} cdot dvecs{r}), where (k) is a constant
- (displaystyle int_C vecs{F} cdot dvecs{r}=int_{−C}vecs{F} cdot dvecs{r})
- Suppose instead that (C) is a piecewise smooth curve in the domains of (vecs F) and (vecs G), where (C=C_1+C_2+⋯+C_n) and (C_1,C_2,…,C_n) are smooth curves such that the endpoint of (C_i) is the starting point of (C_{i+1}). Then
[int_C vecs F·dvecs{r}=int_{C_1} vecs F·dvecs{r}+int_{C_2} vecs F·dvecs{r}+⋯+int_{C_n} vecs F·dvecs{r}.]

Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along (C), then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation (displaystyle int_a^b f(x),dx=−int_b^af(x),dx). Finally, if ([a_1,a_2]), ([a_2,a_3]),…, ([a_{n−1},a_n]) are intervals, then

[int_{a_1}^{a_n}f(x) ,dx=int_{a_1}^{a_2}f(x),dx+int_{a_1}^{a_3}f(x),dx+⋯+int_{a_{n−1}}^{a_n} f(x),dx,]

which is analogous to property iv.

Example (PageIndex{8}): Using Properties to Compute a Vector Line Integral

Find the value of integral (displaystyle int_C vecs F·vecs T ,ds), where (C) is the rectangle (oriented counterclockwise) in a plane with vertices ((0,0)), ((2,0)), ((2,1)), and ((0,1)), and where (vecs F=⟨x−2y,y−x⟩) (Figure (PageIndex{9})).

**Solution**

Note that curve (C) is the union of its four sides, and each side is smooth. Therefore (C) is piecewise smooth. Let (C_1) represent the side from ((0,0)) to ((2,0)), let (C_2) represent the side from ((2,0)) to ((2,1)), let (C_3) represent the side from ((2,1)) to ((0,1)), and let (C_4) represent the side from ((0,1)) to ((0,0)) (Figure (PageIndex{9})). Then,

[int_C vecs F·vecs T ,dr=int_{C_1} vecs F·vecs T ,dr+int_{C_2} vecs F·vecs T ,dr+int_{C_3} vecs F·vecs T ,dr+int_{C_4} vecs F·vecs T ,dr.]

We want to compute each of the four integrals on the right-hand side using Equation ef{eq12a}. Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse (C) counterclockwise):

[egin{align*} C_1&: ⟨t,0⟩,0≤t≤2[4pt] C_2&: ⟨2,t⟩, 0≤t≤1 [4pt] C_3&: ⟨2−t,1⟩, 0≤t≤2[4pt] C_4&: ⟨0,1−t⟩, 0≤t≤1. end{align*}]

Therefore,

[egin{align*} int_{C_1} vecs F·vecs T ,dr &=int_0^2 vecs F(vecs r(t))·vecs r′(t) ,dt [4pt] &=int_0^2 ⟨t−2(0),0−t⟩·⟨1,0⟩ ,dt=int_0^2 t ,dt [4pt] &=Big[ frac{t^2}{2}Big]_0^2=2. end{align*}]

Notice that the value of this integral is positive, which should not be surprising. As we move along curve (C_1) from left to right, our movement flows in the general direction of the vector field itself. At any point along (C_1), the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along (C_1), and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

[egin{align*} int_{C_2} vecs{F} cdot dvecs{r} &=int_{0}^{1}⟨2−2t,t−2⟩·⟨0,1⟩ ,dt [4pt] &=int_{0}^{1} (t−2) ,dt [4pt] &=Big[ frac{t^2}{2}−2tBig]_0^1=−dfrac{3}{2}, end{align*}]

[egin{align*} int_{C_3} vecs F·vecs T ,ds &=int_0^2⟨(2−t)−2,1−(2−t)⟩·⟨−1,0⟩ ,dt [4pt] &=int_0^2t ,dt=2, end{align*}]

and

[egin{align*} int_{C_4} vecs{F} cdot dvecs{r} &=int_0^1⟨−2(1−t),1−t⟩·⟨0,−1⟩ ,dt [4pt] &=int_0^1(t−1) ,dt [4pt] &=Big[ frac{t^2}{2}−tBig]_0^1=−dfrac{1}{2}. end{align*}]

Thus, we have (displaystyle int_C vecs{F} cdot dvecs{r}=2).

Exercise (PageIndex{8})

Calculate line integral (displaystyle int_C vecs{F} cdot dvecs{r}), where (vecs F) is vector field (⟨y^2,2xy+1⟩) and (C) is a triangle with vertices ((0,0)), ((4,0)), and ((0,5)), oriented counterclockwise.

**Hint**Write the triangle as a union of its three sides, then calculate three separate line integrals.

**Answer**0

### Applications of Line Integrals

Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.

Suppose that a piece of wire is modeled by curve *C* in space. The mass per unit length (the linear density) of the wire is a continuous function (
ho(x,y,z)). We can calculate the total mass of the wire using the scalar line integral (displaystyle int_C
ho(x,y,z) ,ds). The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by (
ho(x^*,y^*,z^*) ,Delta s) for some point ((x^*,y^*,z^*)) in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral (displaystyle int_C
ho(x,y,z) ,ds).

Example (PageIndex{9}): Calculating the Mass of a Wire

Calculate the mass of a spring in the shape of a curve parameterized by (⟨t,2cos t,2sin t⟩), (0≤t≤dfrac{pi}{2}), with a density function given by ( ho(x,y,z)=e^x+yz) kg/m (Figure (PageIndex{10})).

**Solution**

To calculate the mass of the spring, we must find the value of the scalar line integral (displaystyle int_C(e^x+yz),ds), where (C) is the given helix. To calculate this integral, we write it in terms of (t) using Equation ef{eq12a}:

[egin{align*} int_C left(e^x+yz ight) ,ds &=int_0^{ frac{pi}{2}} left((e^t+4cos tsin t)sqrt{1+(−2cos t)^2+(2sin t)^2} ight),dt[4pt] &=int_0^{ frac{pi}{2}}left((e^t+4cos tsin t)sqrt{5} ight) ,dt [4pt] &=sqrt{5}Big[e^t+2sin^2 tBig]_{t=0}^{t=pi/2}[4pt] &=sqrt{5}(e^{pi/2}+1). end{align*}]

Therefore, the mass is (sqrt{5}(e^{pi/2}+1)) kg.

Exercise (PageIndex{9})

Calculate the mass of a spring in the shape of a helix parameterized by (vecs r(t)=⟨cos t,sin t,t⟩), (0≤t≤6pi), with a density function given by ( ho (x,y,z)=x+y+z) kg/m.

**Hint**Calculate the line integral of ( ho) over the curve with parameterization (vecs r).

**Answer**(18sqrt{2}{pi}^2) kg

When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the **work done by a vector field** representing a force is a standard use of vector line integrals. Recall that if an object moves along curve (C) in force field (vecs F), then the work required to move the object is given by (displaystyle int_C vecs{F} cdot dvecs{r}).

Example (PageIndex{10}): Calculating Work

How much work is required to move an object in vector force field (vecs F=⟨yz,xy,xz⟩) along path (vecs r(t)=⟨t^2,t,t^4⟩,, 0≤t≤1?) See Figure (PageIndex{11}).

**Solution**

Let (C) denote the given path. We need to find the value of (displaystyle int_C vecs{F} cdot dvecs{r}). To do this, we use Equation ef{lineintformula} :

[egin{align*}int_C vecs{F} cdot dvecs{r} &=int_0^1(⟨t^5,t^3,t^6⟩·⟨2t,1,4t^3⟩) ,dt [4pt] &=int_0^1(2t^6+t^3+4t^9) ,dt [4pt] &={Big[dfrac{2t^7}{7}+dfrac{t^4}{4}+dfrac{2t^{10}}{5}Big]}_{t=0}^{t=1}=dfrac{131}{140}; ext{units of work}. end{align*}]

### Key Concepts

- Line integrals generalize the notion of a single-variable integral to higher dimensions. The domain of integration in a single-variable integral is a line segment along the (x)-axis, but the domain of integration in a line integral is a curve in a plane or in space.
- If (C) is a curve, then the length of (C) is (displaystyle int_C ,ds).
- There are two kinds of line integral: scalar line integrals and vector line integrals. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field.
- Scalar line integrals can be calculated using Equation ef{eq12a}; vector line integrals can be calculated using Equation ef{lineintformula}.
- Two key concepts expressed in terms of line integrals are flux and circulation. Flux measures the rate that a field crosses a given line; circulation measures the tendency of a field to move in the same direction as a given closed curve.

### Key Equations

**Calculating a scalar line integral**

(displaystyle int_C f(x,y,z) ,ds=int_a^bf(vecs r(t))sqrt{{(x′(t))}^2+{(y′(t))}^2+{(z′(t))}^2} ,dt)**Calculating a vector line integral**

(displaystyle int_C vecs F·dvecs{r}=int_C vecs F·vecs T ,ds=int_a^bvecs F(vecs r(t))·vecs r′(t),dt)

or

(displaystyle int_C P,dx+Q,dy+R,dz=int_a^b left(Pig(vecs r(t)ig)dfrac{dx}{dt}+Qig(vecs r(t)ig)dfrac{dy}{dt}+Rig(vecs r(t)ig)dfrac{dz}{dt} ight) ,dt)**Calculating flux**

(displaystyle int_C vecs F·dfrac{vecs n(t)}{‖vecs n(t)‖},ds=int_a^b vecs F(vecs r(t))·vecs n(t) ,dt)

## Glossary

**circulation**- the tendency of a fluid to move in the direction of curve (C). If (C) is a closed curve, then the circulation of (vecs F) along (C) is line integral (∫_C vecs F·vecs T ,ds), which we also denote (∮_Cvecs F·vecs T ,ds).

**closed curve**- a curve for which there exists a parameterization (vecs r(t), a≤t≤b), such that (vecs r(a)=vecs r(b)), and the curve is traversed exactly once

**flux**- the rate of a fluid flowing across a curve in a vector field; the flux of vector field (vecs F) across plane curve (C) is line integral (∫_C vecs F·frac{vecs n(t)}{‖vecs n(t)‖} ,ds)

**line integral**- the integral of a function along a curve in a plane or in space

**orientation of a curve**- the orientation of a curve (C) is a specified direction of (C)

**piecewise smooth curve**- an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves

**scalar line integral**- the scalar line integral of a function (f) along a curve (C) with respect to arc length is the integral (displaystyle int_C f,ds), it is the integral of a scalar function (f) along a curve in a plane or in space; such an integral is defined in terms of a Riemann sum, as is a single-variable integral

**vector line integral**- the vector line integral of vector field (vecs F) along curve (C) is the integral of the dot product of (vecs F) with unit tangent vector (vecs T) of (C) with respect to arc length, (∫_C vecs F·vecs T, ds); such an integral is defined in terms of a Riemann sum, similar to a single-variable integral

## 15.2: Line Integrals - Part 2 - Mathematics

8. Evaluate ( displaystyle intlimits_

- (C) is the line segment from (left( <3,6> ight)) to (left( <0,0> ight)) followed by the line segment from (left( <0,0> ight)) to (left( <3, - 6> ight)).
- (C) is the line segment from (left( <3,6> ight)) to (left( <3, - 6> ight)).

a (C) is the line segment from (left( <3,6>
ight)) to (left( <0,0>
ight)) followed by the line segment from (left( <0,0>
ight)) to (left( <3, - 6>
ight)). Show All Steps Hide All Steps

Start Solution

Let’s start off with a quick sketch of the curve for this part of the problem.

So, this curve comes in two pieces and we’ll need to parameterize each of them so let’s take care of that next.

(

(

For each of these curves we just used the vector form of the line segment between two points. In the first case we needed to because the direction was in the decreasing (y) direction and recall that the integral limits need to be from smaller to larger value. In the second case the equation would have been just as easy to use but we just decided to use the line segment form for the slightly nicer limits.

Okay, we now need to compute the line integral along each of these curves. Unlike the first few problems in this section where we found the magnitude and the integrand prior to the integration step we’re just going to just straight into the integral and do all the work there.

Here is the integral along each of the curves.

Okay to finish this problem out all we need to do is add up the line integrals over these curves to get the full line integral.

Note that we put parenthesis around the result of each individual line integral simply to illustrate where it came from and they aren’t needed in general of course.

b (C) is the line segment from (left( <3,6>
ight)) to (left( <3, - 6>
ight)). Show All Steps Hide All Steps

Start Solution

Let’s start off with a quick sketch of the curve for this part of the problem.

So, what we have in this part is a different curve that goes from (left( <3,6> ight)) to (left( <3, - 6> ight)). Despite the fact that this curve has the same starting and ending point as the curve in the first part there is no reason to expect the line integral to have the same value. Therefore we’ll need to go through the work and see what we get from the line integral.

We’ll need to parameterize the curve so let’s take care of that.

(C:,,vec rleft( t ight) = left( <1 - t> ight)leftlangle <3,6> ight angle + tleftlangle <3, - 6> ight angle = leftlangle <3,6 - 12t> ight angle hspace<0.25in>,,,,,,,0 le t le 1)

Because this curve is in the direction of decreasing (y) and the integral needs its limits to go from smaller to larger values we had to use the vector form of the line segment between two points.

Now all we need to do is compute the line integral.

So, as noted at the start of this part the value of the line integral was not the same as the value of the line integral in the first part despite the same starting and ending points for the curve. Note that it is possible for two line integrals with the same starting and ending points to have the same value but we can’t expect that to happen and so need to go through and do the work.

## Parametrized Lines and Intersections

Download the video from iTunes U or the Internet Archive.

DAVID JORDAN: Hello, and welcome back to recitation. The problem I'd like to work with you today is the intersection of two parametrized lines.

So we have two lines here. L_1, given with the parametrization in terms of the variable t. And L_2, also given with the parametrization in terms of the variable t. So the first question that I want us to answer is do these lines intersect? And if so, then we want to find out where do they intersect.

So why don't you pause the video and work on this. And we can check back in a moment and we'll see how I solved it.

OK, welcome back. Let's get started. So we have these two lines in space. Before we start doing any computations, I find it useful to draw a picture. So let's see what's going on.

OK. So we have these two lines. We can just find some common points on the lines.

So, well, if we put in t equals 0 here, then it looks like we get the point 2 comma 1. OK. And now if we plug in, let's say, t is minus 1 here, then we get-- if we plug in t is minus 1 here, we get x is 3 and y is 0. So there's our line L_1.

And now let's see, L_2. If we plug in t equals 0, we get 2-- 1, 2, 3-- 4. And if we plug in, let's say, t equals minus 1 again, then we get 1 and 2. OK? So there is L_2.

And indeed, it does look like they intersect. We could have probably guessed that they intersect by looking back over here at the formulas for L_1 and L_2, because we see that the sort of direction that this is moving in, we can take derivatives in t. And we see that L_2 is, this line is moving in the direction 1 comma 2. And L_1 is moving in the direction minus 1 comma 1. And so those directions are not parallel.

And so we know that the only way the two lines could fail to intersect is if they're parallel. So actually, even without drawing this, we could have guessed that these lines do intersect. So now we know that these lines intersect.

And in fact, it even looks, you know, it kind of looks like they intersect-- from our sketch-- at the point-- it looks like-- 1 comma 2. It seems to be the point of intersection. But, you know, we got a little bit lucky with our sketch here. So let's see if that's actually true. Let's see if we can verify this in the general way that we discussed in lecture.

So, now there is one place where we have to be careful. We have two lines here, and we parametrized both of these with the variable t. But we need to keep in mind that t is what's called a dummy variable. It doesn't have any geometric meaning to the problem.

And in particular, what I want to caution you about is if we just start solving these two equations algebraically as they're given to us with the variable t, the problem that we could run into is that, you know, we're sort of moving-- as we vary t-- we're moving along this line, and as we vary t again, we're moving along this line. And you see we're moving at the same time. And so that's really solving a different problem. That's not asking about when did these lines intersect, but that would be asking about when do two particles on these lines collide, which is a harder problem.

So instead, what we need to do is we need to give a change of variables for the line L_2. So what I want to do is I'm going to write L_2-- I'm just going to write the same equations, but I'm going to introduce a new variable u.

So x is 2 plus u, and y is 4 plus 2u. OK. So now once we've done that, to find the point of intersection, well, the point of intersection is going to precisely be a point on L_2 where the x-coordinate and the y-coordinate agree with another point on L_1 with the same x- and y-coordinate. So that is, we have the-- what we want to do is we want to set the x-coordinate for L_2.

We want to set this equal to the x-coordinate for L_1, which is 2 minus t. So this was for L_1. And similarly here, we want to set the y-coordinate for L_2 equal to the y-coordinate for L_1.

So now if you think about it, if we can solve this system of equations, then what we've done is we've simultaneously found a point which is on L_1 and on L_2. And that's our goal. So that will be a point of intersection. OK, so now we just have this system of two linear equations and two variables, and we just need to solve it.

Now, we could do, you know, the-- in general, with an equation like this, we might try to add or subtract the equations. But this one is so simple, that I see that the top equation is just the same thing as t equaling to minus u. That's what the top equation says if we cancel the 2's.

And so if we plug that into the next equation, then we get 4 plus 2u equals 1 minus u. And so then we can solve this, and we get, so it looks like 3 equals minus 3u, which tells us that u equals minus 1, and then that tells us that t equals plus 1. OK?

So we found our parameters t and u. And we're not quite done yet. What we need to do is we need to go back to our parametrization. So let me go back over to our original parametrization here, and we have L_1 was 2 minus t and 1 plus t. And over here, we found that t equals 1 was the value that we're after.

So that tells us that x is 1-- 2 minus 1-- and y is 2. Excuse me. I wrote that in a funny way. x is 1 and y is 2.

Now, just as a reality check. We also found that if we solved for L_2, we wanted the variable u to be equal to minus 1. So we had 2 plus u, and 4 plus 2u.

And so let's see what happens when we plug in u equals minus 1 here. We again get x equals 2 plus minus 1 is 1. And y equals 4 plus minus 2 is 2. So we just double check that this is a point of intersection of both lines. And I'll leave it at that.

## Comments

**Kausin Kausin wrote 7 years ago.**

Very helpful video

**mohammad sami wrote 9 years ago.**

i need this lectures as i am a lecturer in the technical

institute

**shaney wrote 9 years ago.**

waooo osum

**john renenl wrote 9 years ago.**

nice.

**carlos wrote 9 years ago.**

hi

**prakash wrote 9 years ago.**

have you gonna do the maths

**arsalan wrote 9 years ago.**

Basic Integration Formulas

**mukesh wrote 9 years ago.**

none

**pratik wrote 10 years ago.**

thanks!

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## 15.2: Line Integrals - Part 2 - Mathematics

MA 26100 Summer 2020

Video Lectures and Lecture Notes

Monday, 6/15--Course Introduction: Video, Slides

Monday, 6/15--Lesson 1 (13.1-13.4: Review of Vectors): Video, Notes

Tuesday, 6/16--Lesson 2 (13.5, 13.6 part 1: Lines, Planes, and Quadric Surfaces): Video, Notes

Thursday, 6/18--Lesson 3 (13.6,part 2, 14.1 Quadric Surfaces, Vector-Valued Functions): Video, Notes

Friday, 6/19--Lesson 4 (14.2, 14.3 part 1: Calculus of Vector-Valued Functions, Motion in Space): Video, Notes

Monday, 6/22--Lesson 5 (14.3 part 2, 14.4, 14.5: Motion in Space, Length of Curves, Curvature): Video, Notes

Tuesday, 6/23--Lesson 6 (15.1, 15.2: Functions of Several Variables, Limits and Continuity): Video, Notes

Thursday, 6/25--Lesson 7 (15.3, 15.4: Partial Derivatives, Chain Rule): Video, Notes

Friday, 6/26--Lesson 8 (15.5, 15.6: Directional Derivative and the Gradient, Tangent Plane and Linear Approximation): Video, Notes

Monday, 6/29--Lesson 9 (15.7 both parts: Maximum and Minimum Problems): Video, Notes

Monday, 6/29--Exam 1 Review: Video, Notes

Thursday, 7/2--Lesson 10 (15.8: Lagrange Multipliers): Video, Notes

Monday, 7/6--Lesson 11 (16.1: Double Integrals over Rectangular Regions): Video, Notes

Tuesday, 7/7--Lesson 12 (16.2: Double Integrals over General Regions): Video, Notes

Thursday, 7/9--Lesson 13 (16.3: Double Integrals in Polar Coordinates): Video, Notes

Friday, 7/10--Lesson 14 (16.4: Triple Integrals): Video, Notes

Monday, 7/13--Lesson 15 (16.5 both parts: Triple Integrals in Cylindrical and Spherical Coordinates): Video, Notes

Tuesday, 7/14--Lesson 16 (16.6: Integrals in Mass Calculations, 17.1: Vector Fields): Video, Notes

Thursday, 7/16--Lesson 17 (17.2 both parts: Line Integrals of Functions and Vector Fields): Video, Notes

Friday, 7/17--Lesson 18 (17.3: Conservative Vector Fields and the Fundamental Theorem of Line Integrals): Video, Notes

Friday, 7/17--Exam 2 Review: Video, Notes

Tuesday, 7/21--Lesson 19 (17.4: Green's Theorem, 17.5: Curl and Divergence): Video, Notes

Thursday, 7/23--Lesson 20 (17.6 part 1: Surface Integrals): Video, Notes

Friday, 7/24--Lesson 21 (17.6 part 2: Surface Integrals): Video, Notes

Monday, 7/27--Lesson 22 (17.6 part 3: Surface Integrals): Video, Notes

Tuesday, 7/28--Lesson 23 (17.7 part 1: Stokes' Theorem): Video, Notes

Thursday, 7/30--Lesson 24 (17.7 part 2: Stokes' Theorem): Video, Notes

Friday, 7/31--Lesson 25 (17.8 part 1: The Divergence Theorem): Video, Notes

Monday, 8/3--Lesson 26 (17.8 part 2: The Divergence Theorem): Video, Notes

## Analytic Geometry and Calculus III, MATH 222, Fall 2018

Course information <------ Click here. __ Free Textbooks.__ (Unless you donate)

**1. Marsden and Weinstein**

**2. Openstax**

I will refer to the first one as "Mars" and the second one as "Open" in the table below.

Mars has the advantage of being much shorter than Open.

Open has the advantage of being much longer than Mars.

If you can do the HW and the practice tests then you are good to go.

I don't expect you to read more than you feel that you need to solving problems is

**much**more important.

MATLAB Projects.

0. Start here: A gentle introduction

This part is never due, but I strongly encourage you to at least skim it as it may save you a lot of time.

1. First project, Due TBA (Not before the first test): Graphing

2. Second project, Due TBA: Root Finding and Optimization

3. Third project, Due TBA: Integration

Final Exam Locations.

Instructor | Reference Number | Room: |

Arash Goligerdian | 12053 | T 101 - Thompson 101 |

Jared Hoppis | 12055 or 12051 | AK 120 - Ackert 120 |

Luke Langston | 12047 or 12056 | W 114 - Willard 114 |

John Pratt | 12054 | KG 004 - King 004 |

Jon Rehmert | 15960 or 16270 | TH 1018 - Throckmorton 1018 |

Alex Thomson | 15565 or 16695 | BB 1088 - Business Building 1088 |

Ilia Zharkov | 12050 | BB 1070 - Business Building 1070 |

#### Week of:

#### Homework:

#### Due 7:15 PM on Wednesday of the week in question (unless otherwise indicated)

#### Final Exam: Wednesday December 12th, 6:20 PM - 8:10

Comments:

1. "40-2" means "40, 41, and 42." "30-38 even" means "30, 32, 34, 36, and 38."

2. The first assignment is due September 2nd.

Other Important Dates:

September 10th: Last day for a full refund for a drop.

September 14th: Last day to change grading option to A/Pass/F.

September 17th: Last day for a 50% refund for a drop.

September 24th: Last day to drop without a "W."

October 26th: Last day to drop. (No refund at this date.)

## Chapter 16 -- Vector Calculus

*These lesson videos are my in-class lectures for Chapter 16 during the Fall 2013 semester of Calculus III. These were prepared as a resource to be used during my maternity leave in Spring 2014. Please note that these are unedited videos filmed by my student T.A. For some days we used different recording resolutions, so the videos may not be consistent in quality. Each lesson is a playlist of the lecture recorded in shorter segments.*

- Lesson Videos: 16.1 & 16.5 Vector Fields, Curl, and Divergence (playlist)
- Mathematica Activity: Mathematica "Vector Fields and Mathematica" (PDF)
- HW: Do 16.1 & 16.5 on Assignment Sheet
- Print: Scalar/Vector Flash Cards to have ready for class (same as before)
- Optional Practice: 16.1 Vector Fields Worksheet (solutions)
- Preview of Chapter: Integration Summary for Vector Calculus (last page of notes)
- HW Video: 16.1 & 16.5 HW Video

- Lesson Videos: 16.3 The Fundamental Theorem of Line Integrals
- HW: Do 16.3 on Assignment Sheet
- Extra HW Help: 16.3 & 16.4 HW Hints
- HW Video: 16.3 HW Video

- Lesson Videos: 16.4 Green's Theorem
- HW: Do 16.4 on Assignment Sheet
- Extra HW Help: 16.3 & 16.4 HW Hints
- HW Video: 16.4 HW Video

- Lesson Videos: 16.6 Parametric Surfaces
- Mathematica Activity: Parametric Surfaces (PDF)
- HW: do 16.6 on Assignment Sheet
- HW Video: 16.6 HW Video

- Lesson Videos: 16.7 Surface Integrals

Note: Surface Integrals Flow Chart added to notes in 2015 - HW: Do 16.7 on Assignment Sheet
- HW Video: 16.7 HW Video
- Extra (keeping link here for my own reference): Another Summary of Integrals (in 16.7 notes too)

- Integration Summary for Vector Calculus (last page of notes)
- Chapter 16 Summary

- Activity. Will be posted via Google Classroom

- Course Outline Calculus D

- Calc D Final Study Questions and Test Info

Note: Test structure still TBD.

The above course notes, assignments, and quizzes are for the Calculus D (SDSU Math 252) *classes I teach at Torrey Pines High School. I wrote and* *modified these notes over several semesters. The* *explanations are my own however, I borrowed several* *examples and diagrams from the textbooks* my classes used* *while I taught the course. Over time, I have changed some* *examples and have forgotten which ones came from which* *sources. Also, I have chosen to keep the notes in my own* *handwriting rather than type to maintain their informality* *and to avoid the tedious task of typing so many formulas,* *equations, and diagrams. These notes are free for use by my* *current and former students. If other calculus students and* *teachers find these notes useful, I would be happy to know* *that my work was helpful. - Abby Brown*

* *

*** Calculus: Early Transcendentals, 6th & 4th editions, James Stewart, ©2007 & 1999 Brooks/Cole Publishing Company, ISBN 0-495-01166-5 & 0-534-36298-2. (Chapter, section, page, and formula numbers refer to this text.)*

**i>Calculus*, 5th edition, Roland E. Larson, Robert P. Hostetler, & Bruce H. Edwards, ©1994 D. C. Heath and Company, ISBN 0-669-35335-3.

## Syllabus

**This is a distance class, all the students enrolled in this class should be highly responsible in managing their schedule. This course moves very fast. If you fall behind, even by one section, you may not be able to catch up, since each section generally depends very heavily on the ones before. A student enrolled in this class has to be capable to read and understand the textbook. If in the past you struggled in self-lecturing mathematics, then this is not the class for you and it is highly recommended you switch to a face-to-face class. The instructor expects for the student to read each section of the textbook, watch the videos and read the class-notes before attempting to solve the homework problems. When asking for help you need to show all your work, by typing it on the email (better) or by attaching a scanned copy of your work. When asking for help for a WebWork problem it is recommended you use the button email to the instructor at the bottom of the screen, otherwise you may not get any answer.**

## Syllabus