

%begin macros
%\magnification=\magstephalf
\magnification=\magstep1
\def\g{\bigtriangledown}
\def\L{{\cal L}}
\baselineskip=14pt
\parskip=10pt
\def\Tilde{\char126\relax}
\font\eightrm=cmr8 \font\sixrm=cmr6 
\font\eighttt=cmtt8
\def\P{{\cal P}}
\def\Q{{\cal Q}}
\parindent=0pt
\overfullrule=0in
\def\frac#1#2{{#1 \over #2}}
%end macros


\bigskip
MATH 251 (22,23,24 ) Brief indications and answers to  Dr. Z. Second Chance Club for Exam 2 Worksheet
\bigskip
\bigskip

{\bf Problem 1a}.  Compute the line integral
$$
\int_C \, xe^{xyz} \, dx \, + \,  y e^{xyz} \, dy \, + \, ze^{xyz} \, dz \quad,
$$
over the path
$$
{\bf r}(t) = \langle t,  t^2, t^3 \rangle \quad, \quad 0 \leq t \leq 1 \quad .
$$
Explain!


{\bf Comment}: Thie problem was assigned by mistake. The point is to use the Fundamental Theorem of Line Integrals.
But taking the curl indicates that it is {\bf not} conservative. Of course one can do it directly, getting
a messy $t$ integral, that even Maple can't do exactly, but using {\tt evalf(Int(MONSTER, t=0..1);} one can get
the answer that I emailed you.

{\bf Problem 1b}.  Compute the line integral
$$
\int_C \, (4x^3y^2+1) \, dx \, + \,  (2x^4y+1) \, dy \, \quad,
$$
over the path
$$
{\bf r}(t) = \langle \sin t^2 ,  \cos t^2 \rangle \quad, \quad 0 \leq t \leq \sqrt{\pi/2} \quad .
$$
Explain!


{\bf Skecth of Sol. to 1b}: Here it is conservative, since $P_y=Q_x$, the potentail function is $f(x,y)=x^4y^2+x+y$ (you do it!).
The start is $(0,1)$ the end is $(1,0)$ and $f(1,0)-f(0,1)=0$. {\bf Ans.}: $0$.

\bigskip:


{\bf Problem 2a}:

Change the order of integration 
$$
\int_{0}^{1} \int_{0}^{e^x} f(x,y) dy \, dx
$$
\bigskip

{\bf  Sol. of 2a}: This is type I. If you sketch it (you do it), the area in question is above the line
segment $0<x<1$ on the $x$-axis and {\bf under} the curve $y=e^x$. 

$$
D=\{ (x,y): 0<x<1, 0<y<e^x \} \quad .
$$

From the type II perspective, the projection on
the $y$-axis is the line segment $0<y<e$ and a  typical horozontal cross-section is from where $y=e^x$ in other words
$x=\ln y$ all the way to the vertical line $x=1$. So the type II description is
$$
D=\{ (x,y): 0<y<e, \ln y <x<e \} \quad .
$$
So the integral is now

$$
\int_{0}^{e} \int_{\ln y}^{e} f(x,y) dx \, dy \quad .
$$
\bigskip

{\bf Problem 2b}:

Change the order of integration 
$$
\int_{0}^{\pi} \int_{0}^{\sin x} f(x,y) dy \, dx
$$
\bigskip

{\bf Sol. of 2b}. The type I description is
$$
D=\{ (x,y): 0<x<\pi, \ln y <x< \sin x \} \quad .
$$
If you plot it, this is the area above $0<x<\pi$ on the $x$-axis and under $y=\sin x$.
The projection on the $y$ axis is the line segment $0<y<1$ and a typical horizontal part
starts as $\sin^{-1} y$ and ends in $\pi-\sin^{-1} y$. So the type II description is
$$
D=\{ (x,y): 0<y<1, \sin^{-1} y <x< \pi -\sin^{-1} y \} \quad .
$$
And the integral is
$$
\int_{0}^{1} \int_{\sin^{-1} y}^{\sin^{-1} y} f(x,y) dx \, dy
$$


{\bf Problem 2c}:

Change the order of integration 
$$
\int_{0}^{1} \int_{e^y}^{e} f(x,y) dx \, dy
$$
\bigskip

{\bf  Sol. or 2c}: Now it is in Type II format

$$
D=\{ (x,y): 0<y<1, e^y<x<e \} \quad .
$$

If you plot it, the projection on the $x$-axis is the interval $0<x<e$ and any horizontal cross-section starts 
where $x=e^y$ and ends where $x=e$. But $x=e^y$ is the same as saying $y=\ln x$, so the required region is
under $y=\ln x$, and the type I description i
$$
D=\{ (x,y): 0<x<e, 0<y<\ln x \} \quad ,
$$
giving
$$
\int_{0}^{e} \int_{0}^{\ln x} f(x,y) dy \, dx
$$


{\bf Problem 3a.} Find the equation of the tangent plane at the point $(1,2,4)$ to the surface
given parametrically by
$$
x(u,v)= u^2  \quad , \quad  y(u,x)=uv \quad \, , \quad z(u,v)=v^2  \quad , \quad
-\infty < u < \infty \quad , \quad
-\infty < v < \infty \quad .
$$
Express you answer in {\bf explicit} form, i.e in the format $z=ax+by+c$.

{\bf Sketch}: The relevant point is $(u,v)=(1,2)$, You find ${\bf r}_u$ and ${\bf r}_v$ ,plug-in $u=1$ and $v=2$.
Take the cross product ${\bf r}_u(1,2) \times{\bf r}_v(1,2)$ get the normal vector ${\bf N}$ and get
do ${\bf N}.\langle x-1, y - 2, z-4 \rangle =0$.



\bigskip

{\bf Problem 3b.} Find the equation of the tangent plane at the point $(-1,-1,2)$ to the surface
given parametrically by
$$
x(u,v)= u^3  \quad , \quad  y(u,x)=v^3 \quad \, , \quad z(u,v)=-2uv  \quad , \quad
-\infty < u < \infty \quad , \quad
-\infty < v < \infty \quad .
$$
Express you answer in {\bf explicit} form, i.e in the format $z=ax+by+c$.

{\bf Sol. to 3b}:  From the first two equations you get $u=-1$ and $v=-1$ but then $z=-2$ {\bf not} $z=2$.
So the point does {\bf not} lie on the surface, so the answer is N/A.

{\bf Comment}: This was a typo. I meant the point to be $(-1,-1,-2)$. But if it comes in an exam you would
get full credit. Don't try to "correct" the question.

\bigskip

{\bf Exam Problem 4.} Let $f(x,y,z)=e^{\cos x^2 \, + \, \sin xyz + \cos xz}$, and let

$$
{\bf F}= \langle \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z} \rangle
$$

Let $C$ be the  curve
$$
r(t)= \langle \cos t , t, \sin t \rangle \quad, \quad 0 \leq t \leq 2\pi \quad .
$$
Find the value of the line-integral
$$
\int_C {\bf F} . d{\bf r} \quad .
$$
Explain! Just giving the answer will give you no credit.

\bigskip

Here is what I did wrong (if applicable):

\bigskip


{\bf Problem 4a} Let $f(x,y,z)=\sin(x+y^2+z^3)$, and let

$$
{\bf F}= \langle \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z} \rangle
$$

Let $C$ be the  curve
$$
r(t)= \langle t , t^2, t^3 \rangle \quad, \quad 0 \leq t \leq 3 \quad .
$$
Find the value of the line-integral
$$
\int_C {\bf F} . d{\bf r} \quad .
$$
Explain! Just giving the answer will give you no credit.


\bigskip


{\bf Problem 4b} Let $f(x,y)=e^{cos x + 3 \sin y}$, and let

$$
{\bf F}= \langle \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}  \rangle
$$

Let $C$ be the  curve
$$
r(t)= \langle \sin 2t , \cos t \rangle \quad , \quad  0 \leq t \leq \pi \quad .
$$
Find the value of the line-integral
$$
\int_C {\bf F} . d{\bf r} \quad .
$$
Explain! Just giving the answer will give you no credit.

\bigskip

{\bf Exam Problem  5.} (10 points) Evaluate the triple integral
$$
\int_R \, (x^2+y^2+z^2)^3 \, dx \, dy \, dz \quad,
$$

where $R$ is the region in 3D space given by

$$
\{ (x,y,z) \, | \, x^2+y^2+z^2 \leq 1 \quad, \quad x,y,z \geq 0 \} \quad .
$$

Here is what I did wrong (if applicable):

\bigskip


{\bf Problem  5a} Evaluate the triple integral
$$
\int_R \, (x+y) (x^2+y^2+z^2)^2 \, dx \, dy \, dz \quad,
$$

where $R$ is the region in 3D space given by

$$
\{ (x,y,z) \, | \, x^2+y^2+z^2 \leq 1 \quad, \quad x \geq 0, y<0, z<0 \} \quad .
$$


\bigskip


{\bf Problem  5b}  Evaluate the triple integral
$$
\int_R \, z (x^2+y^2+z^2) \, dx \, dy \, dz \quad,
$$

where $R$ is the region in 3D space given by

$$
\{ (x,y,z) \, | \, x^2+y^2+z^2 \leq 1 \quad, \quad  y<0 \} \quad .
$$


\bigskip


{\bf Problem  5c} Evaluate the triple integral
$$
\int_R \, (z-x)  \, dx \, dy \, dz \quad,
$$

where $R$ is the region in 3D space given by

$$
\{ (x,y,z) \, | \, x^2+y^2+z^2 \leq 8 \} \quad .
$$


\bigskip


{\bf Exam Problem 6.}   Evaluate the double integral
$$
\int_{-3}^{0} \int_0^{\sqrt{9-x^2}} (x^2+y^2)^2 \, dy \, dx
$$

Here is what I did wrong (if applicable):

\bigskip


{\bf Problem 6a}   Convert the integral to polar coordinates, do not evaluate.
$$
\int_{-3}^{0} \int_0^{\sqrt{9-x^2}} (x^2+y) \, dy \, dx
$$
\bigskip



{\bf Problem 6b}   Convert the integral to polar coordinates, do not evaluate.
$$
\int_{0}^{4} \int_{-\sqrt{16-x^2}}^{0} \,\, (x^2+y) \, dy \, dx
$$
\bigskip



{\bf Problem 6c}   Convert the integral to polar coordinates, do not evaluate.
$$
\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \int_{x}^{\sqrt{1-x^2}} \,\, (x^3+y^2) \, dy \, dx
$$
\bigskip





{\bf Exam Problem 7.} (10 points altogether) Decide whether the following limits exist. If it does find them.
If it does not {\bf Explain} why not?
$$
(a) \,\, (2 \,\, points) \, \lim_{(x,y)\rightarrow (\pi/2, \pi/2)} \,\, \frac{\cos x + \sin x}{x+y} \quad, \quad
(b)\,\, (2 \,\, points) \, \lim_{(x,y)\rightarrow (0,0)} \,\, \frac{x^2-y^2}{x-y} \quad, \quad
$$
$$
(c) \,\, (2 \,\, points) \, \lim_{(x,y)\rightarrow (0,0)} \,\, \frac{x-y}{x^2-y^2} \quad, \quad
(d)\,\, (4 \,\, points) \,\lim_{(x,y)\rightarrow (1,1)} \,\, \frac{x+y-2}{2x+y-3} \quad, \quad
$$
\bigskip

Here is what I did wrong (if applicable):
\bigskip

{\bf Problem 7a}: Decide whether the following limit exists. If it does, find it, if not, explain!
$$
\lim_{(x,y)\rightarrow (1,3)} \,\, \frac{x-1}{y-3} \quad, \quad
$$
\bigskip


{\bf Problem 7b}: Decide whether the following limit exists. If it does, find it, if not, explain!
$$
\lim_{(x,y,z)\rightarrow (0,0,0)} \,\, \frac{x+y+2z}{2x+y+z} \quad, \quad
$$
\bigskip


{\bf Exam Problem 8.}   Compute the line integral $\int_C f \, ds$ where
$$
f(x,y,z)=xyz \quad
$$
and $C$ is the line segment starting at  $(0,0,0)$ and ending at $(1,2,-3)$

Here is what I did wrong (if applicable):


{\bf Problem 8a}   Compute the line integral $\int_C f \, ds$ where
$$
f(x,y,z)=xy^2+yz^2+z \quad
$$
and $C$ is the line segment starting at  $(0,0,0)$ and ending at $(1,1,-1)$

\bigskip

{\bf Problem 8b}   Compute the line integral $\int_C f \, ds$ where
$$
f(x,y)=x+y \quad
$$
and $C$ is the upper circle $\{(x,y): x^2+y^2=1, y>0\}$.

\bigskip

{\bf Exam Problem 9.}  Compute the vector-field surface integral $\int \int_S {\bf F}.d{\bf S}$ 
if ${\bf F}$ is 
$$
{\bf F} \, = \, \langle z,z,x \rangle \quad, 
$$
and $S$ is the oriented surface
$$
z=9-x^2-y^2 \quad, x \geq 0, y \geq 0 , z \geq 0
$$
with {\bf downward pointing} normal.

\bigskip

Here is what I did wrong (if applicable):

\bigskip

{\bf  Problem 9a}  Compute the vector-field surface integral $\int \int_S {\bf F}.d{\bf S}$ 
if ${\bf F}$ is 
$$
{\bf F} \, = \, \langle x+z,y+z,-x \rangle \quad, 
$$
and $S$ is the oriented surface
$$
z=9-x^2-y^2 \quad, x < 0, y < 0 , z \geq 0
$$
with {\bf upward pointing} normal.

\bigskip


{\bf  Problem 9b}  Compute the vector-field surface integral $\int \int_S {\bf F}.d{\bf S}$ 
if ${\bf F}$ is 
$$
{\bf F} \, = \, \langle x+z,y+z,-x \rangle \quad, 
$$
and $S$ is the oriented surface
$$
z=9-x^2-y^2 \quad, 0<x<1 , 0<y < 1 , z \geq 0
$$
with {\bf downward pointing} normal.

\bigskip



{\bf Exam Problem 10.}  Find the {\bf point } on the plane $x+2y+3z=18$ where the function $f(x,y,z)=xyz$ is 
{\bf as large as possible}.

Here is what I did wrong (if applicable):

\bigskip
{\bf Problem 10a}  Find the {\bf maximum value } of the function $f(x,y,z)=xyz$ 
on the plane $2x+y+z=4$

\bigskip
{\bf Problem 10b}  Find the point on the  plane $2x+y+z=4$ where $f(x,y,z)=xy^2z$ is as large as possible.
(You can use Maple)




\end










