# Integrating vector-valued functions along curves

• Eclair_de_XII
In summary, the parametrizations given assume a counter-clockwise orientation for the unit square with bounds of ##0\leq t\leq 1##. The hypotenuse ##(C_1)## is parametrized as ##r(t)=(1-t,1-t)## and the function is defined as ##f(r(t))=(a(1-t)^2,b(1-t)^2)##. The bottom ##(C_2)## is parametrized as ##r(t)=(t,0)## with function ##f(r(t))=(0,bt^2)##, and the right side ##(C_3)## is parametrized as ##r(t)=(1,t
Eclair_de_XII
Homework Statement
Let ##\mathbf{f}=(ay^2,bx^2)## and let ##D## be region given by ##0\leq x\leq 1,0\leq y\leq x##. Compute ##\int_{\partial D}\mathbf{f}\cdot \mathbf{d}r## and ##\int_{\partial D}\mathbf{f}\cdot\mathbf{N}ds##.
Relevant Equations
Arc-length differential: ##ds=|r'|dt##
Unit vector normal to ##r'(t)##: ##\mathbf{N}##
The following parametrizations assume a counter-clockwise orientation for the unit square; the bounds are ##0\leq t\leq 1##.

Hypotenuse ##(C_1)##
%%%
##r(t)=(1-t,1-t)##
##dr=(-1,-1)\,dt##
##f(r(t))=f(1-t,1-t)=(a(1-t)^2,b(1-t)^2)##
##f\cdot dr=-(a+b)(1-t^2)\,dt##
\begin{align}
\int_{C_1} f\cdot dr&=&\int_0^1 -(a+b)(1-t)^2\,dt\\
&=&(a+b)\int_0^1 (1-t)^2(-dt)\\
&=&(a+b)\int_0^1(1-t)^2d(1-t)\\
&=&\frac{1}{3}(a+b)(1-t)^3|0^1\\
&=&-\frac{1}{3}(a+b)
\end{align}

Bottom ##(C_2)##
%%%
##r(t)=(t,0)##
##dr=(1,0)\,dt##
##f(r(t))=f(t,0)=(0,bt^2)##
##f\cdot dr=0##
##\int_{C_2}f\cdot dr=0##

Right ##(C_3)##
%%%
##r(t)=(1,t)##
##dr=(0,1)\,dt##
##f(r(t))=f(1,t)=(at^2,b)##
##f\cdot dr=b##
##\int_{C_3}f\cdot dr=\int_0^1b\,dt=b##

##\int_{\partial D} f\cdot dr=\sum_{i=1}^3\int_{C_i}f\cdot dr=-\frac{1}{3}(a+b)+b=\frac{2}{3}b-\frac{1}{3}a##

I don't know how to get started on the second part because I cannot figure out which unit normal vector to use for each side of the right triangle. For example, do I use ##N=(-1,0)## for ##C_3## or do I use ##N=(1,0)##? Likewise, do I use ##N=\sqrt{\frac{1}{2}}(1,-1)## or ##N=\sqrt{\frac{1}{2}}(-1,1)## in the integral? This part confuses me. Should I use the outward-normal-first vector, in other words, the normal vector that points outward from the given side of the triangle? I remember this being mentioned in my vector analysis course, but I do not particularly recall its significance.

Last edited:
Delta2
The convention is to use the outward normal.

Eclair_de_XII
Thank you. This is very useful to keep in mind.

Hypotenuse ##(C_1)##
%%%
##N=\sqrt{\frac{1}{2}}(-1,1)##
##f=(1-t)^2(a,b)##
##r'=-(1,1)##
##ds=|r'|\,dt=\sqrt{2}\,dt##

##f\cdot N=\sqrt{\frac{1}{2}}(1-t)^2(-a+b)\,dt##
\begin{align}
\int_{C_1} f\cdot N&=&\int_0^1 \sqrt{\frac{1}{2}}(b-a)(1-t)^2\sqrt{2}\,dt\\
&=&(a-b)\int_0^1(1-t)^2(-1\cdot dt)\\
&=&(a-b)\int_0^1(1-t)^2\,d(1-t)\\
&=&\frac{1}{3}(a-b)(1-t)^3|_0^1\\
&=&-\frac{1}{3}(a-b)
\end{align}

Bottom ##(C_2)##
%%%
##N=(0,-1)##
##ds=dt##
##f=(0,bt^2)##
##f\cdot N=-bt^2##
##\int_{C_2} f\cdot N\,ds=\int_0^1 -bt^2\,dt=-\frac{1}{3}bt^3|_0^1=-\frac{1}{3}b##

Right ##(C_3)##
%%%
##N=(1,0)##
##ds=dt##
##f=(at^2,b)##
##f\cdot N=at^2##
##\int_{C_3} f\cdot N\,ds=\int_0^1at^2=\frac{1}{3}a##

\begin{align}
\int_{\partial D} f\cdot N&=&\sum_{i=1}^3\int_{C_i} f\cdot N\,ds\\
&=&\left(-\frac{1}{3}a+\frac{1}{3}b\right)+\left(-\frac{1}{3}b\right)+\left(\frac{1}{3}a\right)\\
&=&0
\end{align}

Delta2
Why the hypotenuse ##(C_1)## has been parameterized as ##r(t)=(1-t,1-t)## and not as ##r(t)=(-t,-t)##?
EDIT: NVM I found out myself, the slope would have been the same and direction counterclockwise but ##f(r(t))## would be wrong... ##r(t)=(t,t)## won't do it either cause it is clock wise oriented...

Last edited:
I think it is possible to use the parametrization ##r(t)=(-t,-t)##, but you'd have to set the bounds to ##-1\leq t\leq 0##. But with the bounds, ##0\leq t\leq1##, the function parametrizes a line from the origin to the point ##(-1,-1)##.

Delta2
Eclair_de_XII said:
But with the bounds, 0≤t≤1, the function parametrizes a line from the origin to the point (−1,−1).
Yes but now that I think of it it is ##f(t,t)=f(-t,-t)## because ##f(x,y)=(ay^2,bx^2)## so I think we can use ##r(t)=(-t,-t)## without changing the bounds.

EDIT: I did it with ##r(t)=(-t,-t)## and the result remains the same that is ##-\frac{1}{3}(a+b)##.

## 1. What is the purpose of integrating vector-valued functions along curves?

Integrating vector-valued functions along curves allows us to calculate the total change of a vector quantity as it moves along a given path. This is useful in many applications, such as calculating work done by a force or finding the displacement of an object.

## 2. How is the integration of vector-valued functions along curves different from regular integration?

Regular integration deals with scalar functions, while integrating vector-valued functions along curves involves dealing with vector quantities. This means that instead of finding the area under a curve, we are finding the change in a vector quantity over a given path.

## 3. Can vector-valued functions be integrated along any type of curve?

Yes, vector-valued functions can be integrated along any type of curve, as long as the curve is smooth and well-defined. This means that the curve cannot have any sharp corners or discontinuities.

## 4. What are some real-life applications of integrating vector-valued functions along curves?

Integrating vector-valued functions along curves is used in physics and engineering to calculate work done by forces, displacement of objects, and fluid flow. It is also used in computer graphics to create smooth and realistic animations.

## 5. How is the integration of vector-valued functions along curves related to line integrals?

The integration of vector-valued functions along curves is a type of line integral, specifically a path integral. It is a generalization of the line integral of a scalar function, where the function is a vector instead of a scalar. Both types of integrals involve integrating over a path or curve, but with different types of functions.

Replies
12
Views
1K
Replies
2
Views
772
Replies
2
Views
784
Replies
1
Views
756
Replies
8
Views
1K
Replies
9
Views
464
Replies
10
Views
966
Replies
3
Views
1K
Replies
9
Views
1K
Replies
2
Views
1K