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\begin{document}


%%%(change to appropriate class and semester)
MATH 2222HJ Spring 2016

%%%(change to appropriate quiz type and date)
Quiz 6 \hspace{1.9in} SOLUTION
\vspace{2pc}

%%%(modify rules, time, points as appropriate)
Show all work clearly and in order, and circle your final answers.  

Justify your answers algebraically whenever possible. Unjustified work may not receive full credit.
\vspace{2pc}


\begin{problem}{2}
Find $\displaystyle\int_C y \sin(z) \mathrm{d}s$ where $C$ is the circular helix given by $\left\{ \begin{array}{ll}
x = \cos(t) \\
y=\sin(t) \\
z=t \\
0 \leq t \leq 2 \pi 
\end{array} \right.$.
\end{problem}

\textit{Solution:} Our curve is parametrized as $$\left\{ \begin{array}{ll}
\vec{r}(t) = <\cos(t), \sin(t), t> \\
0 \leq t \leq 2\pi.
\end{array} \right.$$
Hence $\vec{r}'(t) = < -\sin(t), \cos(t), 1>$ and
$$\lVert \vec{r}^{\prime}(t) \rVert = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{2}.$$
Therefore \href{http://www.wolframalpha.com/input/?i=sqrt%282%29*%28integral+from+t%3D0+to+t%3D2pi+of+sin^2%28t%29+dt%29}{compute}
$$\begin{array}{ll}
\displaystyle\int_C y \sin(z) \mathrm{d}s &= \sqrt{2} \displaystyle\int_0^{2\pi} \sin^2(t) \mathrm{d}t \\
&=\dfrac{\sqrt{2}}{2} \displaystyle\int_0^{2\pi} 1-\cos(2t) \mathrm{d}t \\
&= \dfrac{1}{\sqrt{2}} \left[ t -\dfrac{1}{2} \sin(t) \right]_0^{2\pi} \\
&= \sqrt{2} \pi.
\end{array}$$

\begin{problem}{3}
Evaluate $\displaystyle\int_C \vec{F}(x,y,z) d\vec{r}$ where $\vec{F}=<y,z,x>$ and $C=C_1 \cup C_2$ is the union of two curves, where $C_1$ is the line segment from $(2,0,0)$ to $(3,4,5)$ and $C_2$ is the line segment from $(3,4,5)$ to $(3,4,0)$.
\end{problem}

\textit{Solution:} Parametrize $C_1$ as 
$$\left\{ \begin{array}{ll}
\vec{r}_1(t) = t <3,4,5> + (1-t)<2,0,0>= <t+2,4t,5t> \\
0 \leq t \leq 1,
\end{array} \right.$$
and parametrize $C_2$ as 
$$\left\{ \begin{array}{ll}
\vec{r}_2(t) = t<3,4,0> + (1-t)<3,4,5> = <3,4,5-5t> \\
0 \leq t \leq 1,
\end{array} \right.$$
so we see
$$\vec{r}_1'(t) = <1,4,5>,$$
and
$$\vec{r}_2'(t) = <0,0,-5>.$$
Hence \href{http://www.wolframalpha.com/input/?i=%28integral+from+t%3D0+to+t%3D1+of+29t%2B10+dt%29+%2B+%28integral+from+t%3D0+to+t%3D1+of+%28-15%29+dt%29}{compute}
$$\begin{array}{ll}
\displaystyle\int_C <y,z,x> \mathrm{d}\vec{r} &= \displaystyle\int_{C_1} <y,z,x> \mathrm{d}\vec{r} + \displaystyle\int_{C_2} <y,z,x> \mathrm{d}\vec{r} \\
&= \displaystyle\int_0^1 <4t,5t,t+2> \cdot <1,4,5> \mathrm{d}t + \displaystyle\int_0^1 <4,5-5t,3> \cdot <0,0,-5> \mathrm{d}t \\
&= \displaystyle\int_0^1 4t+20t+5t+10 \mathrm{d}t + \displaystyle\int_0^1 -15 \mathrm{d}t \\
&= \displaystyle\int_0^1 29t+10 \mathrm{d}t -15 \displaystyle\int_0^1 1 \mathrm{d}t \\
&= (\dfrac{29}{2}+10) - 15 \\
&= \dfrac{19}{2}.
\end{array}$$
\showpoints
\end{document}