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Eclair_de_XII

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- 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.

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.

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