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MATH 251 (22,23,24 ) Brief indications and answers to  Dr. Z. Second Chance Club for Exam 2 Worksheet
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{\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.}: $0$.

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{\bf Problem 2a}:

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

$$
\int_{0}^{e} \, \int_{\ln y}^{e} \, f(x,y) dx \, dy \quad .
$$
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{\bf Problem 2b}:

Change the order of integration 
$$
\int_{0}^{\pi} \int_{0}^{\sin x} f(x,y) dy \, dx
$$
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{\bf Sol. of 2b}. The type I description is
$$
D=\{ (x,y): 0<x<\pi, 0 <y< \sin x \} \quad .
$$
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 at $x=\sin^{-1} y$ and ends at $x=\pi-\sin^{-1} y$. 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
$$
\int_{0}^{1} \, \int_{\sin^{-1} y}^{\pi-\sin^{-1} y} \, f(x,y) dx \, dy
$$


{\bf Problem 2c}:

Change the order of integration 
$$
\int_{0}^{1} \int_{e^y}^{e} f(x,y) dx \, dy
$$
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{\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 simplify
 ${\bf N}.\langle x-1, y - 2, z-4 \rangle =0$, and finally use algebra to get it in explicit form.




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{\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.

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{\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.


{\bf Sol. of 4a}: This is really easy. ${\bf F}=grad(f)$ so you don't have to find the potential function, it is given to to you.
By the  Fundamental  Theorem of Line Integrals the answer is $f(End)-f(Start)$. The starting point is $(0,0,0)$ the ending point is
$(3,9,27)$ so the answer is $\sin(3+9^2+27^3)-\sin(0)=\sin\, 19767$

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{\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.

{\bf Sketch}: Same way! Answer is $0$.
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{\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

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


{\bf Sketch of Sol. of 5a} : You translate to spherical language. The only challenging part are the limits.
Of course $\rho$ goes from $0$ to $1$.
Since $z<0$ it follows that  $\pi/2 <\phi<\pi$, since $x>0,y<0$ the projection on the equatoirial plane (i.e. the $xy$-plane)
is the {\bf fourth} quadrant where $\frac{3\pi}{2} < \theta< 2\pi$. The answer turns out to be $0$.

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{\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

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



{\bf Sketch of Sol. of 5b} : You translate to spherical language. The only challenging part are the limits.
Of course $rho$ goes from $0$ to $1$.
Since $z$  is no restricted, $\phi$ gets its full scope $0<\phi<\pi$.
Since $y<0$ we are talking about the lower-half-plane (under the $x$-axis) so the
$\pi \theta< 2\pi$. The answer turns out to be $0$.

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{\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

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


{\bf Sketch of Sol. of 5c} : You translate to spherical language. The only challenging part are the limits.
Of course $rho$ goes from $0$ to $\sqrt{8}$.
Since there are no restictions on $x,y,z$ $\phi$ and $\theta$ take their default limits of integration
$[0,\pi]$ and $[0,2\pi]$ respectively. The answer turns out to be $0$.

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{\bf Problem 6a}   Convert the integral to polar coordinates, do not evaluate.
$$
\int_{-3}^{0} \int_0^{\sqrt{9-x^2}} (x^2+y) \, dy \, dx
$$
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{\bf Sketch of Sol. of 6a}. You convert to polar langauge. The region is the left half of the disk $x^2+y^2<9$.
$r$ goes from $0$ to $3$, while $theta$ goes from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$ .

{\bf Problem 6b}   Convert the integral to polar coordinates, do not evaluate.
$$
\int_{0}^{4} \int_{-\sqrt{16-x^2}}^{0} \,\, (x^2+y) \, dy \, dx
$$
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{\bf Sketch of Sol. of 6b}. You convert to polar langauge. The region is the bottom half of the disk $x^2+y^2<9$.
$r$ goes from $0$ to $4$, while $\theta$ goes from $\pi$ to $2\pi$ .


{\bf Problem 6c}   Convert the integral to polar coordinates, do not evaluate.
$$
\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \int_{x}^{\sqrt{1-x^2}} \,\, (x^3+y^2) \, dy \, dx
$$
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{\bf Comment}: This problem was assigned by mistake. It is too hard for the Final. I meant to ask about

$$
\int_{0}^{\frac{\sqrt{2}}{2}} \int_{x}^{\sqrt{1-x^2}} \,\, (x^3+y^2) \, dy \, dx
$$

For this simplified problem (the one I intended) the region is one-eigth of the disk. namely one where $\theta$ goes from $0$ to $\pi/4$.
The answer to this version is $\frac{\pi}{24}$. The original problem is more challenging, you have to use trig, and the answer is
$\frac{7}{90}+\frac{\pi}{12}$.


{\bf Problem 7a}: Decide whether the following limit exists. If it does, find it, if not, explain!
$$
\lim_{(x,y)\rightarrow (1,3)} \,\, \frac{x-1}{y-3} \quad, \quad
$$
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{\bf Problem 7b}: Decide whether the following limit exists. If it does, find it, if not, explain!
$$
\lim_{(x,y,z)\rightarrow (0,0,0)} \,\, \frac{x+y+2z}{2x+y+z} \quad, \quad
$$
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{\bf Sketch of Sol.} Both limits do not exist. Pick two random lines that pass throgh the point in question and show that you get
different answers.



{\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 $(1,1,-1)$

{\bf Sketch of Sol. to 8a}: You do it directly. First you have to figure out the parametric representation of the line segment. It is
${\bf r}(t)= \langle t , t , -t \rangle$, $0<t<1$.

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

{\bf Sketch of Sol. to 8a}: You do it directly. First you have to figure out the parametric representation of the upper circle. It is
${\bf r}(t)= \langle \cos t, \sin t \rangle$, $0<t<\pi$.

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{\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.

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{\bf Sketch of So. to 9a}: This is similar to the exam problem, except that it is in the third-quadrant.


{\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.


{\bf Sketch of So. to 9b}: This is similar to the exam problem, except that it is easier. Now you are specifically given
the projection on the $xy$ plane that is $\{(x,y): 0<x<1, 0<y<1\}$

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

{\bf Sketch of Sol. to 10a}: This is similar to the exam problem, except that now we are asked for the {\bf value} not the point.
The answer turns out to be $\frac{32}{27}$.

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{\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)

{\bf Sketch of Sol. to 10b}:
$$
grad(f)= \langle y^2z, 2xyz, xy^2 \rangle \quad, \quad
grad(g)= \langle 2, 1, 1 \rangle \quad, \quad
$$
Spelling out $grad(f)=\lambda grad(g)$ and adding the constraint we get
$$
\{ y^2z=2\lambda , 2xyz=\lambda, xy^2=\lambda, 2x+y+z=4 \} \quad.
$$
Going into Maple, and typing

{\tt solve({y**2*z=2*L,2*x*y*z=L,x*y**2=L,2*x+y+z=4},{x,y,z,L});}

gives you (in addition to two trivial solutions)

{\tt L = 2, x = 1/2, y = 2, z = 1}

Discarding the $L$ we get that the desired point is $(\frac{1}{2},2,1)$.




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