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+\magnification=1200
+
+\noindent
+{\it George Weigt --- Advanced Calculus Homework \#10}
+
+\beginsection Page 312, problem 1a.
+
+Evaluate the line integral
+$$\int_C z\,dx+x\,dy+y\,dz$$
+where $C$ is the curve that for $0\le t\le 2\pi$ is parameterized as
+$$\eqalign{
+x&=\cos t\cr
+y&=\sin t\cr
+z&=t\cr
+}$$
+
+\bigskip
+\noindent
+Solution: We have
+$$\eqalign{
+dx&=-\sin t\,dt\cr
+dy&=\cos t\,dt\cr
+dz&=dt\cr
+}$$
+Hence
+$$\eqalign{
+\int_C z\,dx+x\,dy+y\,dz
+&=\int_0^{2\pi}\left(-t\sin t+\cos^2 t+\sin t\right)dt\cr
+&=\left(t\cos t-\sin t+{t\over2}+{\sin t\cos t\over2}-\cos t\right)\bigg|_0^{2\pi}\cr
+&=(2\pi-0+\pi+0-1)-(0-0+0+0-1)\cr
+&=3\pi\cr
+}$$
+
+\beginsection Page 312, problem 1b.
+
+Evaluate the line integral
+$$\int_C x^2\,dx-xz\,dy+y^2\,dz$$
+where $C$ is a straight line form $(1,0,1)$ to $(2,3,2)$.
+
+\bigskip
+\noindent
+Solution: Let $0\le t\le 1$.
+Then $C$ can be parameterized as follows.
+$$\eqalign{
+x&=1+t\cr
+y&=3t\cr
+z&=1+t\cr
+}$$
+We have
+$$dx=dt,\qquad dy=3\,dt,\qquad dz=dt$$
+Hence
+$$\eqalign{
+\int_C x^2\,dx-xz\,dy+y^2\,dz
+&=\int_0^1[(t+1)^2-3(t+1)^2+9t^2]\,dt\cr
+&=\int_0^1(7t^2-4t-2)\,dt\cr
+&=\left({7t^3\over3}-2t^2-2t\right)\bigg|_0^1\cr
+&=-{5\over3}\cr
+}$$
+
+\beginsection Page 312, problem 1d.
+
+Evaluate the line integral
+$$\int_C u_T\,ds$$
+where ${\bf u}=2xy^2z{\bf i}+2x^2yz{\bf j}+x^2y^2{\bf k}$,
+and $C$ is the circle $x=\cos t$, $y=\sin t$, $z=2$, directed by increasing $t$.
+
+\bigskip
+\noindent
+Solution: We have
+$$dx=-\sin t\,dt,\qquad dy=\cos t\,dt,\qquad dz=0$$
+hence
+$$\eqalign{
+I&=\int_C 2xy^2z\,dx+2x^2yz\,dy+x^2y^2\,dz\cr
+&=\int_0^{2\pi} 4(\cos t)(\sin^2 t)(-\sin t)\,dt+4(\cos^2 t)(\sin t)(\cos t)\,dt\cr
+&=4\int_0^{2\pi}(-\cos t\sin^3 t+\cos^3 t\sin t)\,dt\cr
+&=4\int_0^{2\pi}\sin t\cos t\,(-\sin^2 t+\cos^2 t)\,dt\cr
+}$$
+Now for some trigonometric gymnastics. We have
+$$\sin t\cos t={\sin2t\over 2}\qquad\cos^2t-\sin^2t=\cos 2t$$
+hence
+$$4\sin t\cos t\,(-\sin^2 t+\cos^2 t)=2\sin2t\cos2t
+=\sin4t$$
+Therefore
+$$I=\int_0^{2\pi}\sin4t\,dt=-{\cos 4t\over4}\bigg|_0^{2\pi}=0$$
+
+\beginsection Page 312, problem 1e.
+
+Evaluate the line integral
+$$\int_C u_T\,ds$$
+where ${\bf u}=\mathop{\rm curl}{\bf v}$, ${\bf v}=y^2{\bf i}+z^2{\bf j}+x^2{\bf k}$,
+and $C$ is the path $x=2t+1$, $y=t^2$, $z=1+t^3$, $0\le t\le 1$, directed by increasing $t$.
+
+\bigskip
+\noindent
+Solution: We have
+$$\mathop{\rm curl}{\bf v}
+=\left(\matrix{
+\partial x^2/\partial y-\partial z^2/\partial z\cr
+\partial y^2/\partial z-\partial x^2/\partial x\cr
+\partial z^2/\partial x-\partial y^2/\partial y\cr
+}\right)=\left(\matrix{-2z\cr-2x\cr-2y}\right)
+$$
+Therefore
+$$\int_C u_T\,ds=-2\int_C z\,dx+x\,dy+y\,dz$$
+From the given parameteric functions we have
+$$dx=2\,dt,\qquad dy=2t\,dt,\qquad dz=3t^2\,dt$$
+Hence
+$$\eqalign{
+\int_C u_T\,ds
+&=-2\int_0^1 (1+t^3)(2\,dt)+(2t+1)(2t\,dt)+3t^4\,dt\cr
+&=-2\int_0^1 (3t^4+2t^3+4t^2+2t+2)\,dt\cr
+&=-2\left({3t^5\over5}+{t^4\over2}+{4t^3\over3}+t^2+2t\right)\bigg|_0^1\cr
+&=-{163\over15}\cr
+}$$
+
+\beginsection Green's Theorem Strikes Back!
+
+We have
+$$P=2y-3x+5+e^y(y-1),\qquad Q=4x-3y-2+xye^y$$
+and
+$${\partial P\over\partial y}=ye^y+2,\qquad{\partial Q\over\partial x}=ye^y+4$$
+Let $C_{12}$ be the path $C$ followed by the path from $(-1,0)$ back to the
+starting point $(1,0)$ so that $C_{12}$ is
+a closed loop.
+By Green's Theorem we have
+$$\int_{C_{12}} P\,dx+Q\,dy=\int\!\!\!\int_R
+\left({\partial Q\over\partial x}-{\partial P\over\partial y}\right)\,dx\,dy
+=2\int\!\!\!\int_R dx\,dy=2$$
+Now we have to subtract out the line integral over the path that we added to make a closed loop.
+Let $C_2$ be the path from $(-1,0)$ to $(1,0)$. In addition, let
+$$x=t,\qquad y=0,\qquad dx=dt,\qquad dy=0,\qquad-1\le t\le 1$$
+Then
+$$\int_{C_2} P\,dx+Q\,dy=\int_{-1}^1 (-3t+4)\,dt
+=\left(-{3t^2\over2}+4t\right)\bigg|_{-1}^1=8$$
+Therefore the final answer is
+$$\int_C=\int_{C_{12}}-\int_{C_2}=2-8=-6$$
+
+
+\end