Integrating vector-valued functions along curves

[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.

 

[pasmith]

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...

 

[Eclair_de_XII]

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]

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 $-\frac{1}{3}(a+b)$.