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\centerline
{
\bf Dr. Z's Math251 Handout \#16.2 [Line Integrals]
}

By Doron Zeilberger

{\bf Problem Type 16.2a}: 
Evaluate the line integral,
$$
\int_{C} f(x,y) \, ds \quad,
$$
where $C$ is some curve that the problem 
gives you in parametric form, or you have to
represent yourself (typically circles, line-segments, semicircles etc.).

{\bf Example Problem 16.2a}: Evaluate the line integral,
$$
\int_{C} x^2 y \, ds \quad,
$$
where $C$ is top half of the circle $x^2+y^2=9$.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the parametric equation of the curve
$(x(t),y(t))$, $a \leq t \leq b$,
unless it is given by the problem.

&
{\bf 1.} The parametric equation of a circle of the form
$x^2+y^2=r^2$ is 
$$
x=r\cos t \, , \, y=r \sin t \, \quad .
$$
So in our case we have $r=3$ and 
$$
x=3\cos t \, , \, y=3 \sin t \, \quad .
$$
Since it is the {\it top} half,
$t$ goes from $0$ to $\pi$, so $0 \leq t \leq \pi$.

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{\bf 2.} Compute
$$
\sqrt{x'(t)^2+y'(t)^2} \quad .
$$

&
{\bf 2.} $x'(t)=-3\sin t$, $y'(t)=3\cos t$,
so
$$
\sqrt{x'(t)^2+y'(t)^2}=
\sqrt{(-3\sin t)^2+(3 \cos t)^2}
$$
$$
=\sqrt{9\sin^2 t+9 \cos ^2 t}=
\sqrt{9(\sin^2 t+ \cos ^2 t)}=
\sqrt{9}=3 \quad.
$$
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{\bf 3.} The line integral is
$$
\int_{a}^{b} f(x(t), y(t)) \sqrt{x'(t)^2+y'(t)^2} \, dt \quad.
$$
Convert everything to the $t$-language and evaluate
the $t$-integral from $t=a$ to $t=b$.
&
{\bf 3.} 
$$
\int_{C} x^2 y \, ds=
$$
$$
\int_{0}^{\pi} (3 \cos t)^2 \, (3\sin t) \cdot 3 \, dt 
$$
$$
81 \int_{0}^{\pi} \cos^2 t \, \sin t \, dt =
81 \left ( {{-\cos^3 t} \over {3}} \Bigl \vert_{0}^{\pi} \right )
$$
$$
=(-27) (\cos^3 \pi - \cos^3 0)=54 \quad .
$$
{\bf Ans.}: $54$.

\cr}

{\bf Problem Type 16.2b}: 
Evaluate the line integral
$$
\int_{C} P(x,y,z)\,dx+ Q(x,y,z) dy +R(x,y,z)  \,dz \quad ,
$$
where $C: x=x(t) \, , \,  y=y(t) \, ,\, z=z(t) \, ,\,  a \leq t \leq b$.


{\bf Example Problem 16.2b}: Evaluate the line integral
$$
\int_{C} y\,dx+ x \, dy + x^2y\sqrt{z} \,dz \quad ,
$$
where $C: x=t^3 \, , \,  y=t \, , \, z=t^2 \, , \, 0 \leq t \leq 1$.

%%Specific Problem


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Get a (single variable) definite
integral, in $t$, from $t=a$ to $t=b$, by
changing $x,y,z$ to their expressions in terms of $t$
and $dx $, $dy$, $dz$ to $x'(t)dt$, $y'(t)dt$, $z'(t)dt$,
respectively,
$$
\int_{C} P(x,y,z)\,dx+ Q(x,y,z) dy +R(x,y,z) \,dz \quad .
$$
$$
=\int_{a}^{b} 
[P(x(t),y(t),z(t))x'(t) \, + \,
$$
$$
Q(x(t),y(t),z(t))y'(t) \, + \,
$$
$$
R(x(t),y(t),z(t))z'(t)] \, dt \quad .
$$

&
{\bf 1.} 
$$
\int_{C} y\,dx+ x \, dy + x^2y\sqrt{z} \,dz \quad
$$
$$
=\int_{0}^{1} t (3t^2) dt + t^3 dt + (t^3)^2 t \sqrt{t^2} (2t) dt \quad
$$
$$
=\int_{0}^{1} [4t^3  + 2t^9] dt \quad .
$$
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{\bf 2.} Evaluate the $t$-integration.
&
{\bf 2.} 
$$
=t^4 + 
{{t^{10}} \over {5}} \Bigl \vert_{0}^{1}=
$$
$$
=1+{{1} \over {5}}-0=
{{6} \over {5}} \quad.
$$
{\bf Ans.}: ${{6} \over {5}}$.

\cr}

{\bf Problem Type 16.2c}: 
Evaluate the line integral
$$
\int_C {\bf F} \cdot d{\bf r} \quad,
$$
where $C$ is given by the vector function ${\bf r}(t)$.
$$
{\bf F} (x,y,z)= P(x,y,z) {\bf i} + Q(x,y,z) {\bf j} + R(x,y,z) {\bf k} \quad,
$$
$$
{\bf r} (t)= x(t) {\bf i} +y(t) {\bf j} +z(t) {\bf k} \quad,
\quad a \leq t \leq b \quad .
$$


{\bf Example Problem 16.2c}:  Evaluate the line integral
$$
\int_C {\bf F} \cdot d{\bf r} \quad,
$$
where $C$ is given by the vector function ${\bf r}(t)$.
$$
{\bf F} (x,y,z)= yz {\bf i} + xz {\bf j} +xy {\bf k} \quad,
$$
$$
{\bf r} (t)= t {\bf i} + t^2 {\bf j} +t^3 {\bf k} \quad,
\quad 0 \leq t \leq 2 \quad .
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} The desired line-integral equals
$$
\int_{C} P \, dx + Q \, dy + R \, dz \quad .
$$
Set-it up.
&
{\bf 1.} Our integral is
$$
\int_{C} yz \, dx + xz \, dy + xy \, dz \quad ,
$$
where $x=t$, $y=t^2$, $z=t^3$, $0 \leq t \leq 2$.

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{\bf 2.} Evaluate this line integral like we did above (16.2b).

&
{\bf 2.} 
$$
=\int_{0}^{2} (t^2)(t^3) \, dt + (t)(t^3)(2t) \, dt + 
(t)(t^2)(3t^2) \, dt \quad ,
$$
$$
=\int_{0}^{2} [t^5 +2t^5+3t^5] \, dt 
$$
$$
=\int_{0}^{2} 6t^5 \, dt =t^6 \Bigl \vert_{0}^{2}=2^6-0^6=64 \quad.
$$
{\bf Ans.}: $64$.
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\end

