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\medskip
{\bf NAME: (print!)} Dr. Z.
\quad\quad
{\bf RUID: (print!)}  None
\bigskip

{\bf SSC:} (circle) None / I / II / I and II
\bigskip

\bigskip
MATH 251 (22,23,24 ) [Fall 2020], Dr. Z. ,   Final Exam , Tue., Dec. 15, 2020
\bigskip
Email the completed test, renamed as {\tt  finalFirstLast.pdf } to DrZcalc3@gmail.com no later than
3:30pm, (or, in case of conflict, three and half hours after the start).

\bigskip

\hrule

\bigskip

WRITE YOUR FINAL ANSWERS BELOW
\medskip
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
\bigskip
1. $-18$
\bigskip
2.  $\int_0^{\frac{1}{2}} \,\int_{\frac{1}{4}}^1 \, f(x,y) \, dx \, dy \, + \,\int_{\frac{1}{2}}^{1} \,\int_{y^2}^1 \, f(x,y) \, dx \, dy \, + \,$
\bigskip
3.   $z=-\frac{1}{2} x - \frac{5}{6} y + \frac{7}{18} \pi$
\bigskip
4.   $3{\bf i} -6{\bf j} +9 {\bf k}$ OR $\langle 3, -6,9 \rangle$
\bigskip
5.   $\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}$
\bigskip
6.    $-4\, \sqrt{3}$
\bigskip
7.    $6 \cos 2$ OR $-2.496881019$
\bigskip
8.    $8\pi$ 
\bigskip
9.    $-15$
\bigskip
10.    $(\frac{3}{4}, -1)$, saddle point.
\bigskip
11.   $3 \frac{1}{3000}$ OR $\frac{9001}{3000}$ OR $3.0003333\dots$ .
\bigskip
12.   $\frac{\sqrt{2}}{6}$
\bigskip
13.    $\int_0^2 \, \int_{\frac{\pi}{2}}^{\pi} \, \int_0^{\frac{\pi}{2}} \, \rho^6 \sin^4 \phi \, \sin^2 \theta \, \cos^2 \theta \, d\rho d\theta d\phi$
\bigskip
14.    $\frac{1}{3}$
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15.    $\int_0^1 \, \int_0^u \, 2 \sqrt{u^4+4u^2v^2+v^4} \,dv \, du$
\bigskip
16.    $14$
\bigskip
17.    $0$
\hfill\eject


Sign the following declaration:
\bigskip
I  \quad \quad \quad \quad \quad \quad \quad \quad Hereby declare that all the work was done by myself. 
I was allowed to use Maple (unless  specifically told not to), calculators, the book, and all the material in the
web-page of this class but {\bf not} other resources on the internet.
\bigskip
I only spent (at most) $3$ hours on  doing the exam. The last 30 minutes were spent in checking
and double-checking the answers.
\bigskip
I also understand that I may be subject to a random short chat to verify that I actually did it all by myself.
\bigskip
Signed:

\bigskip

{\bf Possibly useful facts from school Geometry} (that you are welcome to use) : (i) The area of a circle radius $r$ is $\pi r^2$. 
(ii) The circumference of a circle radius $r$ is $2\pi r$ (iii) The parametric equation of an ellipse with
axes $a$ $b$ and parallel to the $x$ and $y$ axes respectively is $x=a \cos \theta$, $y=b\cos \theta$, $0<\theta<2\pi$.
(iv) The area of an ellipse with axes $a$ and $b$ is $\pi a b$ (v) The volume and surface area of a sphere radius $R$ are
$\frac{4}{3} \pi R^3$ and $4\pi R^2$ respectively (vi) The volume of a cone is the area of the base times the height over $3$.
(vii) The volume of a pyramid is the area of the base times the height over $3$. (viii) The area of a triangle is
base times height over $2$.


\bigskip

{\bf Formula that you may (or may not) need}
\bigskip
If the surface $S$ is given in {\bf explicit} notation $z=g(x,y)$, above the region of the $xy$-plane , $D$, then
$$
\int\int_S {\bf F} \cdot d {\bf S}=
$$
$$
\int\int_D \left (
-P {{\partial g} \over {\partial x}}
-Q {{\partial g} \over {\partial y}}
+R \right ) \, dA \quad .
$$


\vfill\eject
\medskip
{\bf 1}. (12 pts.)   {\bf Without using Maple (or any software)} Compute the {\bf vector-field line integral}
$$
\int_C \, (cos\, ( e^{\sin x})  +5y) \, dx \, + \, ( sin\, ( e^{\cos y} ) \,+ \, 11 x\,)\, dy \,  \quad,
$$
over the path consisiting of the  five line segments (in that order)
$$
(1,0) \rightarrow (-1,0) \rightarrow (-1,1) \rightarrow (0,2) \rightarrow (1,1) \rightarrow  (1,0) \quad .
$$

Explain!

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\hrule
\bigskip
{\bf ans.}  $-18$
\bigskip
\hrule
\bigskip
Since this is a {\bf closed} path, we should use {\bf Green's Theorem}. $Q_x-P_y=11-5=6$ and we need to integrate it
over the region inside the curve. Since the integrand is constant, i.e. $6$, the area-integral is simply $6$ times
the area. This is a "house" where the roof is the triangle with vertices $(-1,1),(1,1), (0,2)$ whose area is
$ 2 \cdot 1 /2=1$, and the main part of the "house" is the rectangle with vertices $(1,0),(-1,0),(-1,1),(1,1)$ whose
area is $2 \cdot 1=$, so the total area of the region is $3$. So Green's theorem tells you that the answer is $18$.

{\bf  But} notice the {\bf orientation} it is {\bf clockwise}, so we have to multiply by $-1$, getting that the
final answer is $-18$.

\vfill\eject
{\bf 2.} (12 points) Change the order of integration
$$
\int_{\frac{1}{4}}^{1} \int_{0}^{\sqrt{x}} \, f(x,y)\, dy \, dx \quad \quad .
$$
\bigskip

\hrule
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{\bf ans.} $\int_0^{\frac{1}{2}} \,\int_{\frac{1}{4}}^1 \, f(x,y) \, dx \, dy \, + \,\int_{\frac{1}{2}}^{1} \,\int_{y^2}^1 \, f(x,y) \, dx \, dy \, + \,$
\bigskip
\bigskip
\bigskip
\hrule
\bigskip
The region in Type I format is
$$
\{(x,y): \frac{1}{4}<x<1, 0<y<\sqrt{x} \} \quad.
$$
Plotting it, we see that this is the region between the vertical lines $x=\frac{1}{4}$ and $x=1$ and under the curve $y=\sqrt{x}$.

The projection of this region on the $y$ axis is the interval $0<y<1$ but now  there is no consistent "beginning" to a typical
horizontal cross-section. From  $y=0$ to $y=\frac{1}{2}$ it is the vertical line $x=\frac{1}{4}$, but starting
at $y=\frac{1}{2}$  it is the curve $y=\sqrt{x}$ which is the same as $x=y^2$.  The end is consistent, it is always $x=1$.
So in the Type II descripption we need to break it up into two parts.
$$
\{(x,y): 0<y<\frac{1}{2}, \frac{1}{4}<x<1 \} \, \cup \, \{(x,y): \frac{1}{2}<y<1, y^2<x<1 \} \quad.
$$

This explains the answer.
\vfill\eject
{\bf 3.} (12 points) Find the equation of the tangent plane at the point $( \frac{\pi}{6}, \frac{\pi}{6}, \frac{\pi}{6} )$ to the surface
given implicitly by
$$
2 \cos (x+y) + 4 \cos (x+z)+ 8 \cos (y+z) \, = \, 7 \quad \quad .
$$
Express you answer in {\bf explicit} form, i.e in the format $z=ax+by+c$.

\bigskip
\hrule
\hrule
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{\bf ans.} $z=-\frac{1}{2} -\frac{5}{6} y+ \frac{7}{18} \pi$  \quad .
\bigskip
\hrule
\bigskip
First we make sure that the point lies on the surface, by plugging-it-in. It does!

A  {\bf normal vector} at the point is the {\bf gradient} of $2 \cos (x+y) + 4 \cos (x+z)+ 8 \cos (y+z) $ that is
$$
\langle -2 \sin (x+y) - 4 \sin (x+z),-2 \sin (x+y) - 8 \sin (y+z),-4 \sin (x+z)- 8 \sin (y+z)\rangle \quad .
$$
Plugging in $x=\frac{\pi}{6}, y=\frac{\pi}{6}, z=\frac{\pi}{6}$, we get that a normal vector is
$$
\langle -2 \sin (\frac{\pi}{3}) - 4 \sin (\frac{\pi}{3}),-2 \sin ((\frac{\pi}{3}) - 8 \sin ((\frac{\pi}{3}),
-4 \sin ((\frac{\pi}{3})- 8 \sin ((\frac{\pi}{3})\rangle \quad .
$$
$$
\frac{\sqrt{3}}{2} \langle 6, -10, -12 \rangle \quad.
$$
Since we can divide by anything (non-zero), a more user-friendly normal vector is $\langle 3, 5, 6 \rangle$.

So an equation of a tangent plane at that point, in {\bf implicit format} is
$$
2 (x -\frac{\pi}{6}) + 5 (y -\frac{\pi}{6})+ 6 (z -\frac{\pi}{6}) \, = \, 0 \quad .
$$
Rearranging, and solving for $z$, we get 


\vfill\eject
{\bf 4.} (16 points) Let ${\bf a},{\bf b},{\bf c}$ be three vectors such that

$$
{\bf a} \times {\bf b}\,= {\bf i} \,+ \,{\bf j} - {\bf k} \quad , \quad
{\bf b} \times {\bf c} \,= {\bf i} \,- \,{\bf j} + {\bf k} \quad , \quad
{\bf a} \times {\bf c} \,= 2{\bf i} \,+ \,{\bf j} + 2{\bf k} \quad .
$$
What is
$$
({\bf a}+{\bf b}+{\bf c}) \times (2{\bf a}-{\bf b}+3 {\bf c})   \quad \quad ?
$$
\bigskip
\hrule
\hrule
\bigskip
{\bf ans.} $3{\bf i} - 6 {\bf j} + 9 {\bf k}$ OR $\langle 3, -6, 9 \rangle $
\bigskip
\hrule
\bigskip
Using the {\bf distributive property} of the {\bf cross-product} (a fancy name for "opening parantheses") we have
$$
2\,{\bf a} \times {\bf a}- {\bf a} \times {\bf b}+ 3\,{\bf a} \times {\bf c}
$$
$$
+2\,{\bf b} \times {\bf a}- {\bf b} \times {\bf b}+ 3\,{\bf b} \times {\bf c}
$$
$$
+2\,{\bf c} \times {\bf a}- {\bf c} \times {\bf b}+ 3\,{\bf c} \times {\bf c}
$$

But the cross-product of a vector with itself is the {\bf zero vector}, so we get rid of
${\bf a} \times {\bf a}$, ${\bf b} \times {\bf b}$, ${\bf c} \times {\bf c}$. 

Another important property is that
${\bf y} \times {\bf x} =  -{\bf x} \times {\bf y}$, 

so 
${\bf b} \times {\bf a}= -{\bf a} \times {\bf b}$,
${\bf c} \times {\bf a}= -{\bf a} \times {\bf c}$,
${\bf c} \times {\bf b}= -{\bf b} \times {\bf c}$.

So we get

$$
- {\bf a} \times {\bf b}+ 3\,{\bf a} \times {\bf c}
$$
$$
-2\,{\bf a} \times {\bf b}+3\,{\bf b} \times {\bf c}
$$
$$
-2\,{\bf a} \times {\bf c}+{\bf b} \times {\bf c}
$$
$$
=-3 \,{\bf a} \times {\bf b} +  {\bf a} \times {\bf c}+  4\,{\bf b} \times {\bf c} \quad .
$$
Finally, using the data, we get
$$
-3\, ({\bf i} + {\bf j} - {\bf k}) +   2\,{\bf i} \,+ \,{\bf j} + 2\,{\bf k}      +  4 \,({\bf i} \,- \,{\bf j} + {\bf k})
$$
$$
3\,{\bf i} \,- \,6 \,{\bf j}\, +\, 9 \,{\bf k} \quad .
$$
This is the answer. In  the usual notation it is $\langle 3, -6, 9 \rangle $.
\vfill\eject
{\bf 5.} (12 points) Find the three angles of the triangle $ABC$ where
$$
A=(0,0,0) \quad, \quad
B=(1,0,1) \quad, \quad
C=(1,1,0) \quad \quad .
$$
\hrule
\hrule
\bigskip
\bigskip
{\bf ans.}  The angle at $A$ is: $\frac{\pi}{3}$ radians  \quad;\quad
\bigskip
The angle at $B$ is: $\frac{\pi}{3}$ radians  \quad;\quad
\bigskip
The angle at $C$ is: $\frac{\pi}{3}$ radians  \quad;\quad
\bigskip
\hrule
\bigskip

{\bf First Way}: 

The length of $AB$ is $\sqrt{(1-0)^2+(0-0)^2+(1-0)^2}=\sqrt{2}$

The length of $AC$ is $\sqrt{(1-0)^2+(1-0)^2+(0-0)^2}=\sqrt{2}$

The length of $BC$ is $\sqrt{(1-1)^2+(1-0)^2+(0-1)^2}=\sqrt{2}$

Since all the sides are the same it is an {\bf equilateral triangle} so all the angles are the same and equal to $\frac{pi}{3}$.

\bigskip

{\bf First Way}: Let $\theta_A$ be the angle at $A$, then
$$
\cos \theta_A \, = \, \frac{AB.AC}{|AB| |AC|} \quad,
$$
The vector $AB$ is $\langle 1, 0,1 \rangle>$ .
The vector $AC$ is $\langle 1, 1,0 \rangle>$
So
$$
\cos \theta_A \, = \, \frac{2}{\sqrt{2}\sqrt{2}}= \frac{1}{2} \quad,
$$
and $\theta_A=\frac{\pi}{6}$.
Similarly for $\theta_B$ and $\theta_C$.

\vfill\eject
{\bf 6.} (12 points)  Find the directional derivative of
$$
f(x,y,z)=x^3+y^3+z^3+xyz \quad ,
$$
at the point $(1,1,1)$ in a direction pointing to the point $(-1,-1,-1)$ \quad .
\bigskip

\hrule
\bigskip
{\bf ans.} $-4 \sqrt{3}$
\bigskip
\hrule
\bigskip
$$
grad(f)= \langle \,f_x \, , \, f_y \, , \, f_z \,\rangle
$$
$$
\langle 3x^2+yz \,, \, 3y^2+xz \,, \, 3z^2+xy \rangle
$$

Plugging-in $x=1,y=1,z=1$, we get
$$
grad(f)(1,1,1)= \langle 4,4,4 \rangle
$$

The {\bf direction} is $\langle -1,-1,-1 \rangle - \langle 1,1,1 \rangle =\langle -2,-2,-2 \rangle$.
The {\bf unit direction} ${\bf u}$ is $\langle -1,-1,-1 \rangle /\sqrt{3}$.

Hence the {\bf directional derivative} $grad(f). {\bf u}$ is
$$
\langle 4,4,4 \rangle ,  \langle -1,-1, -1\rangle /\sqrt{3} = -3 \cdot 4/\sqrt{3}= -4\,\sqrt{3} \, \quad ,
$$
\vfill\eject
{\bf 7.} (12 points)  Using the Chain Rule (no credit for other methods), find 
$$
\frac{\partial g}{\partial u}
$$ 
at $(u,v)=(0,1)$, where 
$$
g(x,y)=3x^2-3y^2 \quad ,
$$ 
and
$$
x=e^u \cos v \quad, \quad y=e^u \sin v \quad .
$$

\bigskip

\hrule
\bigskip
{\bf ans.}  $6\, \cos \, 2$
\bigskip
\hrule
\bigskip
The chain rule says:
$$
g_u = g_x \cdot x_u \, + \, g_y \cdot y_u \quad
$$

In this problem
$$
g_x = 6x \quad, \quad g_y = -6y
$$
$$
x_u = e^u \cos v \quad, \quad y_u= e^u \sin v
$$
At $u=0$, $v=1$, $x=\cos 1$, $y=\sin 1$, so
$$
g_x = 6\cos\, 1 \quad, \quad g_y = -6\sin \, 1
$$
$$
x_u = \cos \,1 \quad, \quad y_u=  \sin\,  1
$$
So at this point
$$
g_u = 6 \cos \, 1 \cdot \cos \, 1 \, -6 \sin 1 \cdot \sin\, 1 = 6 (\cos^2 1 - \sin^2 1)= 6 \cos\, 2 \quad.
$$

(Here we used the famous trig identity $\cos^2 \theta - \sin^2 \theta=\cos \, 2\theta$.)

\vfill\eject
{\bf 8.} (12 points)  Without using Maple (or any other software), compute the vector-field surface integral $\int_S {\bf F}.d{\bf S}$ 
if
$$
{\bf F}= \langle \, 3x +cos(y^3+yz) \, , \, -2y+ e^{x+z^2} \, , \, 5z+\sin(xy^3+e^x) \, \rangle \quad \quad ,
$$
and $S$ is the closed surface   in 3D space bounding  the region
$$
\{(x,y,z): x^2+y^2+z^2<4 \quad and \quad  x>0 \quad and \quad  y<0 \quad and \quad z>0 \} \quad .
$$

\bigskip
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{\bf ans.} $8 \pi$
\bigskip
\hrule
\bigskip
This calls for the {\bf divergence theorem}. 
$$
div({\bf F})=3-2+5=6 \quad.
$$
So the value that we need is the integral of $6$ over the region. But taking $6$ out, it is $6$ times the volume of
the region. The region is one-eighth of a sphere radius $2$, so the volume is
$$
\frac{1}{8} \frac{4}{3} \cdot 2^3 = \frac{4}{3} \pi \quad.
$$
Multipying by $6$ gives the answer, $8 \pi$.

\vfill\eject
{\bf 9.} (12 points) Compute the vector-field surface integral $\int \int_S {\bf F}.d{\bf S}$ 
if 
$$
{\bf F} \, = \, \langle \, 3z \, , \, 2x \, , \, y+z \, \rangle \quad, 
$$
and $S$ is the oriented surface
$$
z=2x+3y \quad, \quad 0<x<1, \quad 0<y<1  \quad ,
$$
with {\bf upward pointing} normal.


\bigskip
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\bigskip
{\bf ans.}  $-15$
\bigskip
\hrule
\bigskip

There are no short cuts here. We need the formula
$$
\int\int_D \left (
-P {{\partial g} \over {\partial x}}
-Q {{\partial g} \over {\partial y}}
+R \right ) \, dA \quad .
$$

Here
$$
P=3z \quad, \quad 
Q=2x \quad, \quad 
R=y+z
$$
$$
g(x,y)=2x+3y
$$

and the region is $D=\{(x,y): 0<x<1, 0<y<1\}$.

The {\bf integrand} is
$$
(-3z)(2) - (2x)(3)+y+z =-6x+y-5z \quad .
$$
Replacing $z$ by $2x+3y$ this gives
$$
-6x+y-5(2x+3y)= -6x+y-10x-15y=-16x-14y \quad .
$$

Integrating we get
$$
\int_0^1 \, \int_0^1 (-16x-14y) \, dx dy = 
-16 \int_0^1 \, \int_0^1 x \, dx dy  --14 \int_0^1 \, \int_0^1 y \, dx dy = 
-15 \quad .
$$


\vfill\eject
{\bf 10.} (12 points)  Without using Maple or software,
find the critical point(s) of 
$$
f(x,y)=4x-y^2-\ln(2x+y) \quad,
$$ 
and decide for each whether it
is a local maximum, local minimum, or saddle point. Explain.

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\hrule
\bigskip
{\bf ans.}  $(\frac{3}{4}, -1)$, saddle point.
\bigskip
\hrule
\bigskip
$$
f_x = 4 - \frac{2}{2x+y} \quad, \quad f_y =-2y- \frac{1}{2x+y} \quad.
$$
For future refenence
$$
f_{xx} = \frac{4}{(2x+y)^2} \quad, \quad f_{xy}= \frac{2}{(2x+y)^2} \quad, \quad f_{yy}=-2+ \frac{1}{(2x+y)^2} \quad .
$$

Solving
$$
 4 - \frac{2}{2x+y}=0 \quad, \quad -2y- \frac{1}{2x+y}=0 \quad.
$$
From the first equation $\frac{1}{2x+y}=2$. Putting it in the second equation we get $-2y-2=0$ so $y=-1$. Going back
to the first equation we get $x=\frac{3}{4}$.

So there is only one {\bf critical point} $(\frac{3}{4}, -1)$.

Plugging into $f_{xx}, f_{xy}, f_{yy}$ we get
$$
f_{xx}= 16 \quad, \quad
f_{xy}=8 \quad, \quad
f_{yy}=3
$$

So the {\bf discriminant} $D=f_{xx} f_{yy} - (f_{xy})^2$ at that point is $16 \cdot 2 - 8^2= -32$. Since it is
{\bf negative} the point is a {\bf saddle point}.

\vfill\eject
{\bf 11.} (12 points)  Without using Maple or software, using a {\bf Linearization} around the point $(1,1,2)$, approximate $f(1.001,0.999,2.001)$ if
$$
f(x,y,z)\,=\, \sqrt{2x^2+3y^2+z^2} \quad .
$$
\bigskip
\hrule
\bigskip
{\bf ans.}  $3 \, \frac{1}{3000}$ OR $\frac{9001}{3000}$ OR $3.0003333\dots$.
\bigskip
\hrule
\bigskip
$$
f= (2x^2+3y^2+z^2)^{\frac{1}{2}}
$$
$$
f_x = \frac{1}{2} (2x^2+3y^2+z^2)^{-\frac{1}{2}} (4x) \, = \,\frac{2x}{\sqrt{2x^2+3y^2+z^2}}
$$
$$
f_y = \frac{1}{2} (2x^2+3y^2+z^2)^{-\frac{1}{2}} (6y) \, = \,\frac{3y}{\sqrt{2x^2+3y^2+z^2}}
$$
$$
f_z = \frac{1}{2} (2x^2+3y^2+z^2)^{-\frac{1}{2}} (2z) \, = \,\frac{z}{\sqrt{2x^2+3y^2+z^2}}
$$
At $x=1$, $y=1$, $z=2$ we have
$$
grad(f)= \langle \frac{2}{3} , 1, \frac{2}{3} \rangle \quad .
$$

The {\bf linearization} is
$$
L(x,y,z)= f(1,1,2) + \frac{2}{3} (x-1)+ 1 \cdot (y-1) + \frac{2}{3} (z-2)=
= 3 + \frac{2}{3} (x-1)+ 1 \cdot (y-1) + \frac{2}{3} (z-2) \quad.
$$
Plugging in the acual values $x=1.001$, $y=0.999$, $z=2.001$ gives the approximation
$$
3 + \frac{2}{3} (\frac{1}{1000})+ 1 \cdot (-\frac{1}{1000}) + \frac{2}{3} (\frac{1}{1000})
\, = \,
3 + \frac{1}{3000} = \frac{9001}{3000} \quad .
$$


\vfill\eject
{\bf 12.} (12 points) Without using Maple (or any other software) and by using polar coordinates (no credit for doing it
directly) find
$$
\int_0^{\frac{\sqrt{2}}{2}} \, \int_0^x x \, dy \, dx \, + \,
\int_{\frac{\sqrt{2}}{2}}^1 \, \int_0^{\sqrt{1-x^2}} x \, dy \, dx \, \quad .
$$
Explain!
 
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\bigskip
\bigskip
{\bf ans.}  $\frac{\sqrt{2}}{6}$ 
\bigskip
\bigskip
\hrule
\bigskip
The region of integration is
$$
D= \{0<x<\frac{\sqrt{2}}{2}, 0<y<x \} \cup \{\frac{\sqrt{2}}{2}<x<1, 0<y<\sqrt{1-x^2} \}  \quad .
$$
If you draw a picture, this is exactly the one-eighth of the unit circle
$$
\{ (r, \theta): 0<r<1 \quad, \quad 0<\theta <\frac{\pi}{4} \} \quad .
$$
Converting to polar coordinates, we get
$$
 \int_0^{\frac{\pi}{4}} \, \int_0^1 \, \, (r\cos \theta) r \, dr \, dtheta
$$
$$
 \int_0^{\frac{\pi}{4}} \, \int_0^1 \, \, (r^2 \cos \theta) r\, dr \, dtheta
$$
$$
\left ( \int_0^1 \, r^2 \, dr \right )  \left ( \int_0^{\frac{\pi}{4}} \cos \theta \right ) 
\, = \,
\frac{1}{3} \cdot \sin \, \frac{\pi}{4}= \frac{\sqrt{2}}{6} \quad .
$$


\vfill\eject
{\bf 13.} (12 points)  Convert the triple iterated integral
$$
\int_{0}^2 \int_0^{\sqrt{4-z^2}} \int_{-\sqrt{4-z^2-y^2}}^{0} \,\,\, x^2\,y\,z \,dx\,dy\,dz 
$$
to {\bf spherical coordinates}. {\bf Do not evaluate.}

\bigskip
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\bigskip
\bigskip
{\bf ans.}  $\int_0^2 \, \int_{\frac{\pi}{2}}^{\pi} \, \int_0^{\frac{\pi}{2}} \, \rho^6 \sin^4 \phi \, \sin^2 \theta \, \cos^2 \theta \, d\rho d\theta d\phi$
\bigskip
\bigskip
\hrule
\bigskip

The challenging part is figuring out the region in spherical coordinates. Sincs $z$ is positive $\phi$ goes from $0$ to $\frac{\pi}{2}$
(the Northern hemisphere). Since $x$ is negatie and $y$ is positive in the integration region we are talking about the
{\bf second quadrant} so $\theta$ goes from $\frac{\pi}{2}$ to $\pi$.

Regarding the integrand, you use the `dictionary'
$$
x=\rho \sin \phi \cos \theta \quad , \quad
y=\rho \sin \phi \sin \theta \quad , \quad
z=\rho \cos \phi  \quad , \quad
dx \, dy \, dz \, = \, \rho^2 \sin \phi , d \rho \, d\phi \, d\theta \quad.
$$


\vfill\eject
{\bf 14.} (12 points)  Find the curvature of the curve
$$
{\bf r}(t)= \langle 5, 3 \sin t, 3 \cos t \rangle 
$$
at the point where $t=\frac{\pi}{3}$.
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{\bf ans.}  $\frac{1}{3}$ 
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{\bf Shortcut way}: This is a circle of radious $3$ so the curvature is $\frac{1}{3}$ (everywhere, not just at $t=\pi/3$).


{\bf Usual way}: Find ${\bf r}'(t)$, ${\bf r}''(t)$. Plug-in $t=\pi/3$, and use the formula for the curvature

$$
\frac{|{\bf r}'(t) \times {\bf r}''(t)| }{ |{\bf r}'(t)|^3 } \quad .
$$




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{\bf 15.} (12 points)   Set-up an iterated double integral, in type I format, but do not compute, for the surface area of the
surface given parameterically by
$$
{\bf r}(u,v) \,= \, \langle u^2 \, ,\,  uv \, , \, v^2 \rangle \quad, \quad  0<u<v<1 \quad .
$$
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{\bf ans.}  $\int_0^1 \, \int_0^u \, 2 \sqrt{u^4+4u^2v^2+v^4} \,dv \, du$
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$$
{\bf r}_u \,= \, \langle 2u \, ,\,  v \, , \, 0 \rangle \quad, \quad  0<u<v<1 \quad .
$$
$$
{\bf r}_v \,= \, \langle 0 \, ,\,  u \, , \, 2v \rangle \quad, \quad  0<u<v<1 \quad .
$$

$$
{\bf r}_u \times {\bf r}_v  \, = \,
$$
$$
\langle 2v^2, -4uv, 2u^2 \rangle \quad.
$$
So
$$
|{\bf r}_u \times {\bf r}_v  | \, = \, \sqrt{4u^4+16u^2v^2+4u^4}= 2 \sqrt{2u^4+4u^2v^2+u^4}= 
$$

Now we integrate over the region $\{ (u,v): 0<u<1, 0<v<u \}$.

\vfill\eject
{\bf 16.} (12 points)  Let 
$$
f(x,y,z)\,=\, xy^2z^3  \quad ,
$$
and let 
$$
g(x,y,z)\,=\, x+y^2+z^3 \quad.
$$

compute   the dot-product
$$
grad(f)\,.\,grad(g)    \quad \quad \quad .
$$ 

at the point $(1,1,1)$.

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{\bf ans.}  $14$
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$$
grad(f)= \langle y^2z^2 , 2xyz^3, 3xy^2z^2 \rangle \quad
$$
$$
grad(g)= \langle 1 , 2y, 3z^2 \rangle \quad
$$
$$
grad(f)(1,1,1)= \langle 1 , 2, 3 \rangle \quad
$$
$$
grad(g)(1,1,1)= \langle 1 , 2, 3 \rangle \quad
$$
So the dot-product is $1^2+2^2+3^2=14$.

\vfill\eject



{\bf 17.} (8 points) Decide whether the following limit exists. If it does ,find it. If it does not,  explain why it does
not exist.
$$
\lim_{(x,y,z,w) \rightarrow (0,0,0,0) } \, \frac{(x+y)^2-(z+w)^2}{x+y-z-w} \quad .
$$  
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{\bf ans.}  $0$
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When you plug-it-in you get $0/0$, but after you simplify you get, using $(a^2-b^2)/(a-b)=a+b$, that the function
equals $x+y+z+w$. Now you plug-it-in and get $0$.

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