APEX Answers to Selected Exercises

The $x$ and $y$ axes are uncalibrated. There are five instances where the field lines run parallel to the $x$ axis. One of them is on the $x$ axis itself, other two pairs of such field lines are above and below the $x$ axis. In between the $x$ axis and the first horizontal field line for some positive $y$ value, the field lines are all northeast facing. Above the horizontal field line for some $y$ value until another with a higher $y$ value, the field lines in between are southeast facing.

Similarly below the $x$ axis till the first horizontal line with some negative $y$ value, the field lines in between are southeast facing. In between this horizontal line and another horizontal line with a higher negative $y$ value, the field lines are northeast facing.

8.1.4.16.

Answer.

$Graph showing slope field for the given differential equation.$

The $x$ and $y$ axes are uncalibrated. The field lines run almost parallel to the $x$ axis. Above the axis the field lines are slightly facing north east. Below the $x$ axis the lines are directed facing southeast.

8.1.4.18.

Answer.

$Graph showing slope field for the given differential equation.$

The $x$ and $y$ axes are uncalibrated, the field lines in the first quadrant are shown. The field lines very close to the $y$ axis are almost north facing for higher values of $y$ and almost east facing for lower values of $y\text{.}$ With smaller values of $x\text{,}$ the field lines, from left to right the lines first face northeast then east and southeast after for greater values of $x\text{.}$

A curve is drawn that starts at a point for some small value of $x$ and a high value of $y\text{.}$ The curve has a positive slope at first after reaching a peak it declines almost close to the $x$ axis.

8.1.4.19.

Answer.

$Graph showing slope field for the given differential equation.$

The $x$ and $y$ axes are uncalibrated, the field lines in the first quadrant are shown. Front left to right, a little away from the x axis the field lines are northeast facing that transition to north facing. Moving further right then again become northeast facing then transition to southeast facing, further right they become south facing then east facing. The pattern then repeats. Very close to the $x$ axis the field lines are almost parallel to it.

A wave is drawn that starts at some y intercept above the origin. It has a high positive slope, it reaches peak when the field lines change from northeast facing to southeast facing, then it declines until the point the field lines are parallel to the $x$ axis. The curve continues to form a second wave.

8.1.4.20.

Answer.

$Graph showing slope field for the given differential equation.$

The $x$ and $y$ axes are uncalibrated, the field lines in the first quadrant are shown. There are two instances where the field lines are parallel to the $x$ axis. From under the $x$ axis to the first such line the field lines transition from almost north facing to northeast facing. Between the horizontal field line for a small $y$ value and a greater $y$ value the field lines are facing southeast. Above the line with a higher $y$ value the field lines transition from northeast facing to north facing.

8.1.4.21.

Answer.

Answer.

$x$	$0.0$	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$
$y(x)$	0.5000	0.5412	0.6806	0.9747	1.5551	2.7183
$h = 0.2$	0.5000	0.5000	0.5816	0.7686	1.1250	1.7885
$h = 0.1$	0.5000	0.5201	0.6282	0.8622	1.3132	2.1788

Answer.

separable; $\displaystyle y = 1$

8.3.2.21.

Answer.

$Graph showing slope field for the given differential equation.$

The $x$ and $y$ axes are uncalibrated, the field lines in the first quadrant are shown. On the bottom right the field lines are facing northeast. On the top left the field lines transition from southeast facing to east facing moving downwards. A curve is shown that almost represents a straight line with a positive slope.

The solution will increase and begin to follow the line $y=x-1\text{.}$

$y = x-1 + e^{-x}$

8.3.2.22.

Answer.

$Graph showing slope field for the given differential equation.$

The $x$ and $y$ axes are uncalibrated, the field lines in the first quadrant are shown. The lines in the top are southeast facing, for lower values of $y$ from left to right the field lines are northeast facing then they transition to east facing. A downward sloping curve is shown on the field lines.

The solution will decrease and approach $y=0\text{.}$

$Computer-generated sketch of the parametric curve in this exercise.$

A curve that spirals around the origin, with several self-intersecting loops. This curve has 9 arcs and 9 loops. It is similar to the curve in the previous exercise, except that this time, the arcs bend inward toward the origin, rather than outward.

9.2.4.19.

9.2.4.19.a

Answer.

Traces the parabola $y=x^2\text{,}$ moves from left to right.

9.2.4.19.b

Answer.

Traces the parabola $y=x^2\text{,}$ but only from $-1\leq x\leq 1\text{;}$ traces this portion back and forth infinitely.

9.2.4.19.c

Answer.

Traces the parabola $y=x^2\text{,}$ but only for $0\lt x\text{.}$ Moves left to right.

9.2.4.19.d

Answer.

Answer 1.

$\frac{t+11}{6}$

Answer 2.

$\frac{t^{2}-97}{12}$

Answer 3.

$\left(2,-8\right)$

Answer 4.

$6x-11$

Answer 5.

$1$

9.2.4.36.

Answer 1.

$\ln\mathopen{}\left(t\right)$

Answer 2.

$t$

Answer 3.

$\left(0,1\right)$

Answer 4.

$e^{x}$

Answer 5.

$1$

9.2.4.37.

Answer 1.

$\cos^{-1}\mathopen{}\left(t\right)$

Answer 2.

$\sqrt{1-t^{2}}$

Answer 3.

$\left(0,0\right)$

Answer 4.

$\cos\mathopen{}\left(x\right)$

Answer 5.

$1$

9.2.4.39.

Answer 1.

$-1, 1$

Answer 2.

$\left(3,-2\right)$

9.2.4.44.

Answer 1.

$2$

Answer 2.

$\left(-4,-8\right)$

9.2.4.46.

Answer 1.

$0$

Answer 2.

$\left(1,0\right)$

9.2.4.50.

Answer.

$2\cos\mathopen{}\left(t\right);\,-2\sin\mathopen{}\left(t\right)$

9.2.4.51.

Answer.

$3\cos\mathopen{}\left(2\pi t\right)+1;\,3\sin\mathopen{}\left(2\pi t\right)+1$

9.2.4.52.

$-\frac{4}{\left(2t-1\right)^{3}}$

Answer 2.

$\left(-\infty ,0.5\right]$

Answer 3.

$\left[0.5,\infty \right)$

9.3.5.30.

Answer 1.

$\frac{2\mathopen{}\left(\sin\mathopen{}\left(t\right)\mathopen{}\left(-2\right)\sin\mathopen{}\left(2t\right)-\cos\mathopen{}\left(2t\right)\cos\mathopen{}\left(t\right)\right)}{\sin^{3}\mathopen{}\left(t\right)}$

Answer 2.

$\left[\frac{\pi }{2}, \pi \right], \left[\frac{3\pi }{2}, 2\pi \right]$

Answer 3.

$\left[0, \frac{\pi }{2}\right], \left[\pi , \frac{3\pi }{2}\right]$

9.3.5.33.

Answer.

$6\pi $

9.3.5.34.

Answer 1.

$\sqrt{101}\mathopen{}\left(e^{\frac{\pi }{5}}-1\right)$

Answer 2.

$\sqrt{101}\mathopen{}\left(e^{\frac{2\pi }{5}}-e^{\frac{\pi }{5}}\right)$

9.3.5.35.

Answer.

$2\sqrt{34}$

Problems

9.4.4.5.

Answer.

$The four points plotted in this exercise.$

On a polar grid, four points are plotted. The point $A$ is at the intersection of the initial ray and the circle of radius 2. Points $B$ and $D$ are both on the circle of radius 1. The point $B$ is on the same line as the initial ray, but in the opposite direction. The point $D$ lies above the initial ray, making an angle of $\pi/4\text{.}$ Finally, the point $C$ is at the bottom of the circle of radius $2\text{.}$

9.4.4.6.

Answer.

$The four points plotted in this exercise.$

On a polar grid, four points are plotted. Points $A$ and $B$ are both on the same line as the initial ray, but to the left of the origin $O\text{,}$ with $A$ on the circle of radius 2, and $B$ on the circle of radius 1. The point $C$ is on the circle of radius 1, and makes an angle slightly greater than a right angle with the initial ray, placing it above and just to the left of the origin $O\text{.}$ The point $D$ lies almost immediately below the point $C\text{;}$ the circle of radius $1/2$ is not marked as part of the grid.

9.4.4.7.

Answer.

$A=P(2.5,\pi/4)$ and $P(-2.5,5\pi/4)\text{;}$

$B=P(-1,5\pi/6)$ and $P(1,11\pi/6)\text{;}$

$C=P(3,4\pi/3)$ and $P(-3,\pi/3)\text{;}$

$D=P(1.5,2\pi/3)$ and $P(-1.5,5\pi/3)\text{;}$

9.4.4.8.

Answer 1.

$\left(2, 0.523599\right), \left(-2, -2.61799\right)$

Answer 2.

$\left(1, -1.0472\right), \left(-1, 2.0944\right)$

Answer 3.

$\left(2, 2.35619\right), \left(-2, -0.785398\right)$

Answer 4.

$\left(2.5, 3.14159\right), \left(2.5, -3.14159\right)$

9.4.4.9.

Answer 1.

$\left(\sqrt{2},\sqrt{2}\right)$

Answer 2.

$\left(\sqrt{2},-\sqrt{2}\right)$

Answer 3.

$\left(\sqrt{5},\tan^{-1}\mathopen{}\left(\frac{-1}{2}\right)\right)$

Answer 4.

$\left(\sqrt{5},\pi +\tan^{-1}\mathopen{}\left(\frac{-1}{2}\right)\right)$

9.4.4.10.

Answer 1.

$\left(-3,0\right)$

Answer 2.

$\left(\frac{-1}{2},\frac{\sqrt{3}}{2}\right)$

Answer.

Answer.

Converges

Terms and Concepts

10.4.3.3.

Answer.

10.5.3.

Answer.

Many examples exist; one common example is $a_n = (-1)^n/n\text{.}$

Terms and Concepts

11.1.7.5.

Answer.

A hyperboloid of two sheets

11.1.7.6.

Answer.

A hyperboloid of one sheet

Problems

11.1.7.7.

Answer 1.

$\sqrt{6}$

Answer 2.

$\sqrt{17}$

Answer 3.

$\sqrt{11}$

Answer 4.

$\text{do}$

11.1.7.9.

Answer 1.

$\left(4,-1,0\right)$

Answer 2.

$3$

11.1.7.10.

Answer 1.

$\left(-2,1,2\right)$

Answer 2.

$\sqrt{5}$

11.1.7.15.

Answer.

Link to full-sized image

Viewed with the $y$ axis pointing out the page, the surface looks like the graph $z=x^3$ in the $xz$ plane. The rest of the surface is generated by sliding this curve back and forth along the $y$ axis. The result is a sheet that looks something like a lounge chair.

11.1.7.16.

Answer.

Link to full-sized image

In the $yz$ plane, the surface appears to be a cosine curve. This curve is translated back and forth along the $x$ axis, so that when viewed from other angles, it appears to be a wavy sheet.

11.1.7.17.

Answer.

Link to full-sized image

A “classic” cylinder, in the shape of a tube. Cross-sections parallel to the $xy$ plane are ellipses.

11.1.7.18.

Answer.

Answer.

A vector with magnitude 1.

11.2.4.

Answer.

Answer.

$\vec x = \la-1,0,-3\ra\text{.}$

11.2.17.

Answer 1.

$\sqrt{5}$

Answer 2.

$\sqrt{13}$

Answer 3.

$\sqrt{26}$

Answer 4.

$\sqrt{10}$

11.2.18.

Answer 1.

$\sqrt{17}$

Answer 2.

$\sqrt{3}$

Answer 3.

$\sqrt{14}$

Answer 4.

$\sqrt{26}$

11.2.19.

Answer 1.

$\sqrt{5}$

Answer 2.

$3\sqrt{5}$

Answer 3.

$2\sqrt{5}$

Answer 4.

$4\sqrt{5}$

11.2.20.

Answer 1.

$7$

Answer 2.

$35$

Answer 3.

$42$

Answer.

Terms and Concepts

11.5.4.1.

$x = -2, y = 5+t$

Answer.

$2$

Terms and Concepts

11.6.1.

Answer.

A point in the plane and a normal vector (i.e., a direction orthogonal to the plane).

11.6.2.

Answer.

A normal vector is orthogonal to the plane.

Problems

11.6.3.

Answer.

Answers will vary.

11.6.4.

Answer.

$\left(-2,9,0\right), \left(2,9,3\right)$

11.6.5.

Answer.

Answer.

$Graph of the vector valued function from the example.$

Graph of the function $\vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra$ on $[0,2\pi]\text{.}$ The graph of the function resembles a saddle centered at the origin whose height is defined by the $z$-axis. The two sides of the saddle that taper off fall into negative $z$ and lie in the second and third quadrants in the $xy$ plane. Ignoring the $z$ coordinate, the curve is a unit circle in the $xy$ plane. Ignoring the $x$ or $y$ coordinates individually, the curve looks like the $\infty$ symbol in the $yz$ and the $xz$ planes, respectively. We now describe the $z$ coordinate with respect to travelling along the unit circle in the $xy$ plane. Starting at $t=\text{,}$ the function begins at the point $(1,0,0)\text{.}$ As $t$ increases and we travel along the unit circle in the $x$ and $y$ coordinates, $z$ increases until we get to $t=\frac{\pi}{2}$ at which $z=1\text{.}$ Then, continuing along the unit circle, $z$ decreases until it reaches a minimum of $z=-1$ when $t=\frac{3\pi}{4}\text{.}$ Continuing along the circle, $z$ begins to increase once again, reaching one more maximum of $z=1$ when $t=\frac{5\pi}{4}\text{.}$ Finally, $z$ begins to decrease, reaching its last minimum of $z=1$ when $t=\frac{7\pi}{4}\text{,}$ after which $z$ increases, and the curve ends where it began, at the point $(1,0,0)\text{.}$

12.1.4.17.

Answer.

$\left|t\right|\sqrt{1+t^{2}}$

12.1.4.19.

Answer.

$\sqrt{4+t^{2}}$

12.1.4.21.

Answer.

Answer.

$\left<2\cos\mathopen{}\left(t\right),2\sin\mathopen{}\left(t\right),2t\right>$

12.1.4.31.

Answer.

$\left<1,0\right>$

$\left|t\right|\sqrt{9t^{2}-12t+8}$

Answer 2.

$0$

Answer 3.

$-1$

12.3.3.19.

Answer 1.

$\left|\sec\mathopen{}\left(t\right)\right|\sqrt{\tan^{2}\mathopen{}\left(t\right)+\sec^{2}\mathopen{}\left(t\right)}$

Answer 2.

$0$

Answer 3.

$\frac{\pi }{4}$

12.3.3.22.

Answer 1.

$\sqrt{8t^{2}+3}$

Answer 2.

$0$

Answer 3.

$1$

12.3.3.30.

Answer.

$\left<t^{2}-t+5,\frac{3t^{2}}{2}-t-\frac{5}{2}\right>$

12.3.3.32.

Answer.

$\left<10t,-16t^{2}+50t\right>$

12.3.3.34.

Answer 1.

$\left<-10,0\right>$

Answer 2.

$5\pi $

Answer 3.

$\left<\frac{-10}{\pi },0\right>$

Answer 4.

$5$

12.3.3.36.

Answer 1.

$\left<10,20,-10\right>$

Answer 2.

$10\sqrt{6}$

Answer 3.

$\left<1,2,-1\right>$

Answer 4.

$\sqrt{6}$

12.3.3.38.

12.3.3.38.a

Answer.

$t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4\text{,}$ where $n$ is an integer

12.3.3.38.b

Answer.

$\norm{\vrp(t)} = 24\pi\approx 51.4$ ft/s

12.3.3.38.c

Answer.

$0.27$ radians, or $15.69^\circ$

12.3.3.39.

12.3.3.39.a

Answer.

$0.013\ {\rm radians}$

Answer 1.

$\left<\frac{1}{\sqrt{1+\sin^{2}\mathopen{}\left(t\right)}},\frac{-\sin\mathopen{}\left(t\right)}{\sqrt{1+\sin^{2}\mathopen{}\left(t\right)}}\right>$

Answer 2.

$\left<\sqrt{\frac{2}{3}},\frac{-1}{\sqrt{3}}\right>$

12.4.4.8.

Answer 1.

$\left<-\sin\mathopen{}\left(t\right),\cos\mathopen{}\left(t\right)\right>$

Answer 1.

$\frac{\frac{-2}{t^{5}}}{\sqrt{1+\frac{1}{t^{4}}}}$

Answer 2.

$\frac{\frac{2}{t^{3}}}{\sqrt{1+\frac{1}{t^{4}}}}$

Answer 3.

$-\sqrt{2}$

Answer 4.

$\sqrt{2}$

Answer 5.

$\frac{-1}{4\sqrt{17}}$

Answer 6.

$\frac{1}{\sqrt{17}}$

12.4.4.28.

Answer 1.

$2$

Answer 2.

$4t^{2}$

Answer 3.

$2$

Answer 4.

$2\pi $

Answer 5.

$2$

Answer 6.

$4\pi $

12.4.4.30.

Answer 1.

$0$

Answer 2.

$5$

Answer 3.

$0$

Answer 4.

$5$

Answer 5.

$0$

Answer 6.

$5$

Terms and Concepts

12.5.4.3.

Answer.

Answers may include lines, circles, helixes

12.5.4.4.

Answer.

Answer should mention the circle is tangent to the curve and has the same curvature as the curve at that point.

12.5.4.5.

Answer.

$a_\text{T}$ is not affected by curvature; the greater the curvature, the larger $a_\text{N}$ becomes.

Problems

12.5.4.7.

Answer 1.

Answer.

Answers will vary. Interior point: $(1,0)$ (any point with $y\neq x^2$ will do). Boundary point: $(1,1)$ (any point with $y=x^2$ will do).
$S$ is an open set.
$S$ is unbounded.

13.2.5.9.

Answer.

Answers will vary. interior point: none boundary point: $(0,-1)$
$S$ is a closed set, consisting only of boundary points
$S$ is bounded

13.2.5.11.

Answer.

$D = \left\{(x,y)\, |\, 9-x^2-y^2\geq 0\right\}\text{.}$
$D$ is a closed set.
$D$ is bounded.

13.2.5.12.

Answer.

$D = \left\{(x,y)\, |\, y\geq x^2\right\}\text{.}$
$D$ is a closed set.
$D$ is unbounded.

13.2.5.13.

Answer.

$D = \left\{(x,y)\, |\, y \gt x^2\right\}\text{.}$
$D$ is an open set.
$D$ is unbounded.

13.2.5.14.

Answer.

$D = \left\{(x,y)\, |\, (x,y)\neq (0,0) \right\}\text{.}$
$D$ is an open set.
$D$ is unbounded.

Problems

13.3.7.6.

$5e^{x}\cos\mathopen{}\left(y\right)$

Answer 3.

$5e^{x}\sin\mathopen{}\left(y\right)$

Answer 4.

$5e^{x}\cos\mathopen{}\left(y\right)$

Answer 5.

$5e^{x}\cos\mathopen{}\left(y\right)$

Answer 6.

$-5e^{x}\sin\mathopen{}\left(y\right)$

13.3.7.28.

Answer.

$\frac{1}{2}x^{2}+xy+\frac{1}{2}y^{2}$

13.3.7.30.

Answer.

$\ln\mathopen{}\left(x^{2}+y^{2}\right)$

13.3.7.32.

Answer 1.

$3x^{2}y^{2}+3x^{2}z$

Answer 2.

$2x^{3}y+2yz$

Answer 3.

$x^{3}+y^{2}$

Answer 4.

$2y$

Answer 5.

$2y$

13.3.7.34.

Answer 1.

$\frac{1}{x}$

Answer 2.

$\frac{1}{y}$

Answer 3.

$\frac{1}{z}$

Answer 4.

$0$

Answer 5.

$0$

Problems

14.2.3.6.

Answer.

$\frac{dz}{dt} = 3(2t)+4(2) = 6t+8\text{.}$
At $t=1\text{,}$ $\frac{dz}{dt} = 14\text{.}$

14.2.3.7.

Answer 1.

$2x-4yt$

Answer 2.

$2$

14.2.3.8.

Answer.

$\displaystyle \frac{dz}{dt} = 5(-2\sin(t) )+2(\cos(t) ) = -10\sin(t) +2\cos(t)$
At $t=\pi/4\text{,}$ $\frac{dz}{dt} = -4\sqrt{2}\text{.}$

14.2.3.9.

Answer 1.

$\frac{-\sin\mathopen{}\left(t\right)}{1+y^{2}}-\frac{2xy\cos\mathopen{}\left(t\right)}{\left(y^{2}+1\right)^{2}}$

Answer 2.

$-{\frac{1}{2}}$

14.2.3.10.

Answer.

$\ds\frac{dz}{dt} = 2x(\cos(t) ) + 4y(3\cos(t) )\text{.}$
At $t=\pi/4\text{,}$ $x=\sqrt{2}/2\text{,}$ $y=3\sqrt{2}/2\text{,}$ and $\frac{dz}{dt} = 19\text{.}$

14.2.3.11.

Answer 1.

$-\sin\mathopen{}\left(x\right)\sin\mathopen{}\left(y\right)\pi +\cos\mathopen{}\left(x\right)\cos\mathopen{}\left(y\right)\cdot 2\pi $

Answer 2.

$0$

14.2.3.13.

Answer 1.

$2x\cos\mathopen{}\left(t\right)+2y\sin\mathopen{}\left(t\right)$

Answer 2.

$-2xs\sin\mathopen{}\left(t\right)+2ys\cos\mathopen{}\left(t\right)$

Answer 3.

$4$

Answer 4.

$0$

14.2.3.21.

Answer 1.

$-2yt^{2}e^{-\left(x^{2}+y^{2}\right)}$

Answer 2.

$-2xe^{-\left(x^{2}+y^{2}\right)}-4stye^{-\left(x^{2}+y^{2}\right)}$

Answer 3.

$\frac{-2}{e^{2}}$

Answer 4.

$\frac{-6}{e^{2}}$

14.2.3.23.

$0$

Problems

14.4.5.6.

Answer 1.

$x = \frac{\pi }{3}+t, y = \frac{\pi }{6}, z = \frac{3}{4}-\frac{3\sqrt{3}}{4}t$

Answer 2.

$x = \frac{\pi }{3}, y = \frac{\pi }{6}+t, z = \frac{3}{4}+\frac{3\sqrt{3}}{4}t$

Answer 3.

$x = \frac{\pi }{3}+\frac{t}{\sqrt{5}}, y = \frac{\pi }{6}+\frac{2t}{\sqrt{5}}, z = \frac{3}{4}+\frac{3\sqrt{\frac{3}{5}}}{4}t$

14.4.5.8.

Answer 1.

$x = 1+t, y = 2, z = 3$

Answer 2.

$x = 1, y = 2+t, z = 3$

$0.666667y-2x+4.47214z = 0$

14.4.5.24.

$\left(\frac{5}{14},\frac{25}{196}\right)$

Answer 3.

$-12$

Answer 4.

Answer.

$16\pi $

15.3.8.

Answer.

$\frac{\pi }{2}$

15.3.12.

$\ds SA = \int_{-5}^{5}\int_{0}^{1} \sqrt{1+ \frac{4x^2e^{2x^2}}{\big(1+e^{x^2}\big)^4}}\, dy\, dx$

Problems

$dz\, dx\, dy\text{:}$ $\ds\int_0^2\int_1^{3}\int_0^{(3-x)/2}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_1^3\int_0^{(3-x)/2}\int_0^{2}\, dy\, dz\, dx$

$dy\, dx\, dz\text{:}$ $\ds\int_0^1\int_1^{3-2z}\int_0^{2}\, dy\, dx\, dz$

$dx\, dz\, dy\text{:}$ $\ds\int_0^2\int_0^{1}\int_1^{3-2z}\, dx\, dz\, dy$

$dx\, dy\, dz\text{:}$ $\ds\int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz$

$\ds V = \int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz =2\text{.}$

15.6.4.11.

Answer.

$dz\, dy\, dx\text{:}$ $\ds\int_0^2\int_{-2}^{0}\int_{y^2/2}^{-y}\, dz\, dy\, dx$

$dz\, dx\, dy\text{:}$ $\ds\int_{-2}^0\int_0^{2}\int_{y^2/2}^{-y}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx$

$dy\, dx\, dz\text{:}$ $\ds\int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dx\, dz$

$dx\, dz\, dy\text{:}$ $\ds\int_{-2}^0\int_{y^2/2}^{-y}\int_0^{2}\, dx\, dz\, dy$

$dx\, dy\, dz\text{:}$ $\ds\int_0^2\int_{-\sqrt{2z}}^{-z}\int_0^{2}\, dx\, dy\, dz$ $\ds V = \int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx =4/3\text{.}$

15.6.4.12.

Answer.

$dz\, dy\, dx\text{:}$ $\ds\int_0^3\int_{3x}^{9}\int_{0}^{\sqrt{y^2-9x^2}}\, dz\, dy\, dx$

$dz\, dx\, dy\text{:}$ $\ds\int_{0}^9\int_0^{y/3}\int_{0}^{\sqrt{y^2-9x^2}}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_0^3\int_0^{\sqrt{81-9x^2}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dz\, dx$

$dy\, dx\, dz\text{:}$ $\ds\int_0^9\int_0^{\sqrt{9-z^2/9}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dx\, dz$

$dx\, dz\, dy\text{:}$ $\ds\int_{0}^9\int_{0}^{y}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dz\, dy$

$dx\, dy\, dz\text{:}$ $\ds\int_0^9\int_{z}^{9}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dy\, dz$

15.6.4.13.

Answer.

$dz\, dy\, dx\text{:}$ $\ds\int_0^2\int_{1-x/2}^{1}\int_{0}^{2x+4y-4}\, dz\, dy\, dx$

$dz\, dx\, dy\text{:}$ $\ds\int_{0}^1\int_{2-2y}^{2}\int_{0}^{2x+4y-4}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_0^2\int_0^{2x}\int_{z/4-x/2+1}^{1}\, dy\, dz\, dx$

$dy\, dx\, dz\text{:}$ $\ds\int_0^4\int_{z/2}^{2}\int_{z/4-x/2+1}^{1}\, dy\, dx\, dz$

$dx\, dz\, dy\text{:}$ $\ds\int_{0}^1\int_{0}^{4y}\int_{z/2-2y+2}^2\, dx\, dz\, dy$

$dx\, dy\, dz\text{:}$ $\ds\int_0^4\int_{z/4}^{1}\int_{z/2-2y+2}^2\, dx\, dy\, dz$ $\ds V = \int_0^4\int_{z/4}^{1}\int_{z/2-2y-2}^2\, dx\, dy\, dz = 4/3\text{.}$

15.6.4.14.

Answer.

$dz\, dy\, dx\text{:}$ $\ds\int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx$

$dz\, dx\, dy\text{:}$ $\ds\int_{0}^4\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\int_{0}^{2x+4y-4}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_{-2}^2\int_0^{8-2x^2}\int_{z/2}^{4-x^2}\, dy\, dz\, dx$

$dy\, dx\, dz\text{:}$ $\ds\int_0^8\int_{-\sqrt{4-z/2}}^{\sqrt{4-z/2}}\int_{z/2}^{4-x^2}\, dy\, dx\, dz$

$dx\, dz\, dy\text{:}$ $\ds\int_{0}^4\int_{0}^{2y}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dz\, dy$

$dx\, dy\, dz\text{:}$ $\ds\int_0^8\int_{z/2}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dy\, dz$ $\ds V = \int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx = 512/15\text{.}$

15.6.4.15.

Answer.

$dz\, dy\, dx\text{:}$ $\ds\int_{0}^1\int_{0}^{1-x^2}\int_{0}^{\sqrt{1-y}}\, dz\, dy\, dx$

$dz\, dx\, dy\text{:}$ $\ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_{0}^1\int_0^{x}\int_{0}^{1-x^2}\, dy\, dz\, dx + \int_{0}^1\int_x^{1}\int_{0}^{1-z^2}\, dy\, dz\, dx$

$dy\, dx\, dz\text{:}$ $\ds\int_0^1\int_{0}^{z}\int_{0}^{1-z^2}\, dy\, dx\, dz+\int_0^1\int_{z}^{1}\int_{0}^{1-x^2}\, dy\, dx\, dz$

$dx\, dz\, dy\text{:}$ $\ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dx\, dz\, dy$

$dx\, dy\, dz\text{:}$ $\ds\int_0^1\int_{0}^{1-z^2}\int_{0}^{\sqrt{1-y}}\, dx\, dy\, dz$ Answers will vary. Neither order is particularly “hard.” The order $dz\, dy\, dx$ requires integrating a square root, so powers can be messy; the order $dy\, dz\, dx$ requires two triple integrals, but each uses only polynomials.

15.6.4.16.

Answer.

$dz\, dy\, dx\text{:}$ $\ds\int_{0}^1\int_{0}^{3x}\int_{0}^{1-x}\, dz\, dy\, dx+\int_{0}^1\int_{3x}^{3}\int_{0}^{1-y/3}\, dz\, dy\, dx$

$dz\, dx\, dy\text{:}$ $\ds\int_{0}^3\int_{0}^{y/3}\int_{0}^{1-y/3}\, dz\, dy\, dx+\int_{0}^3\int_{y/3}^{1}\int_{0}^{1-x}\, dz\, dx\, dy$

$dy\, dz\, dx\text{:}$ $\ds\int_{0}^1\int_0^{1-x}\int_{0}^{3-3z}\, dy\, dz\, dx$

Answer.

$M\approx 0.237\text{;}$ center of mass is approximately $(0.173, 0.099,0.065)\text{.}$

Terms and Concepts

16.2.3.1.

Answer.

Answers will vary. Appropriate answers include velocities of moving particles (air, water, etc.); gravitational or electromagnetic forces.

16.2.3.2.

Answer.

Specific answers will vary, though should relate to the idea that “more of the vector field is moving into that point than out of that point.”

16.2.3.3.

Answer.

Specific answers will vary, though should relate to the idea that the vector field is spinning clockwise at that point.

16.2.3.4.

Link to full-sized image

A two-dimensional vector field is plotted relative to $x$ and $y$ coordinate axes, with the origin at the center.

Along the lines $y=2$ and $y=-2\text{,}$ vectors point up and to the right, with a relatively small slope. These vectors have the largest magnitude.

Answer.

We can conclude that $\vec F$ is conservative.

16.3.4.6.

Answer.

Answer.

$24$ ft-lbs.

16.3.4.17.

Answer.

$\displaystyle f(x,y) = xy+x$
$\curl \vec F = 0\text{.}$
$1\text{.}$ (One parametrization for $C$ is $\vec r(t) = \langle t,t-1\rangle$ on $0\leq t\leq 1\text{.}$)
$1$ (with $A = (0,1)$ and $B = (1,0)\text{,}$ $f(B) - f(A) = 1\text{.}$)

16.3.4.18.

Answer.

$\displaystyle f(x,y) = x^2+xy+y^2$
$\curl \vec F = 0\text{.}$
$0\text{.}$
$0$ (with $A = (0,0)$ and $B = (0,0)\text{,}$ $f(B) - f(A) = 0\text{.}$)

16.3.4.19.

Answer.

$\displaystyle f(x,y) = x^2yz$
$\curl \vec F = \vec 0\text{.}$
$250\text{.}$
$250$ (with $A = (1,-1,0)$ and $B = (5,5,2)\text{,}$ $f(B) - f(A) = 250\text{.}$)

16.3.4.20.

Answer.

$\displaystyle f(x,y) = x^2+y^2+z^2$
$\curl \vec F = \vec 0\text{.}$
$0\text{.}$
$0$ (with $A = (1,0,0)$ and $B = (1,0,0)\text{,}$ $f(B) - f(A) = 0\text{.}$)

16.4.4.15.

Answer.

Three line integrals need to be computed to compute $\oint_C \vec F\cdot d\vec r\text{.}$ It does not matter which corner one starts from first, but be sure to proceed around the triangle in a counterclockwise fashion.

From $(0,0)$ to $(2,0)\text{,}$ the line integral has a value of 0. From $(2,0)$ to $(1,1)$ the integral has a value of $7/3\text{.}$ From $(1,1)$ to $(0,0)$ the line integral has a value of $-1/3\text{.}$ Total value is 2.

The double integral of $\curl\vec F$ over $R$ also has value 2.

16.4.4.16.

Answer.

The double integral of $\divv\vec F$ over $R$ also has value $2/3\text{.}$

16.4.4.24.

Answer.

Two line integrals need to be computed to compute $\oint_C \vec F\cdot \vec n\, ds\text{.}$ Along the parabola, the line integral has value $159/20\text{.}$ Along the line, the line integral has value $6\text{.}$ Together, the total value is $279/20\text{.}$

The double integral of $\divv\vec F$ over $R$ also has value $279/20\text{.}$

Terms and Concepts

16.5.3.1.

Answer.

Answers will vary, though generally should meaningfully include terms like “two sided”.

16.5.3.2.

Answer.

Many possible answers exist; the one given by the book is the Möbius band.

Problems

16.5.3.3.

Answer.

$\vec r(u,v) = \langle u, v, 3u^2v\rangle$ on $-1\leq u\leq 1\text{,}$ $0\leq v\leq 2\text{.}$
$\vec r(u,v) = \langle 3v\cos(u)+1, 3v\sin(u)+2, 3(3v\cos(u)+1)^2(3v\sin(u)+2)\rangle\text{,}$ on $0\leq u\leq 2\pi\text{,}$ $0\leq v\leq 1\text{.}$
$\vec r(u,v) = \langle u, v(2-2u), 3u^2v(2-2u)\rangle$ on $0\leq u, v\leq 1\text{.}$
$\vec r(u,v) = \langle u, v(1-u^2), 3u^2v(1-u^2)\rangle$ on $-1\leq u\leq 1\text{,}$ $0\leq v\leq 1\text{.}$

16.5.3.4.

Answer.

$\vec r(u,v) = \langle u, v, 4u+2u^2\rangle$ on $1\leq u\leq 4\text{,}$ $5\leq v\leq 7\text{.}$
$\vec r(u,v) = \langle 4v\cos(u), 3v\sin(u), 16v\cos(u)+2(3v\sin(u))^2\rangle\text{,}$ on $0\leq u\leq 2\pi\text{,}$ $0\leq v\leq 1\text{.}$
$\vec r(u,v) = \langle u, u+v(4-2u), 4u+2\big(u+v(4-2u)\big)^2\rangle$ on $0\leq u\leq 2\text{,}$ $0\leq v\leq 1\text{.}$
$\vec r(u,v) = \langle v\cos(u), v\sin(u), 4v\cos(u) + 2(v\sin(u))^2\rangle$ on $0\leq u\leq 2\pi\text{,}$ $2\leq v\leq 5\text{.}$

For $y=2\text{:}$ $\vec r(u,v) = \langle u,2,v/2(3-u)\rangle\text{,}$ with $1\leq u\leq 3\text{,}$ $0\leq v\leq 1$

For $z=0\text{:}$ $\vec r(u,v) = \langle u,v,0\rangle\text{,}$ with $1\leq u\leq 3\text{,}$ $0\leq v\leq 2$

16.5.3.10.

Answer.

Answers may vary.

For $z=2x+4y-4\text{:}$ $\vec r(u,v) = \langle u, 1-u/2+uv/2, 2u+4(1-u/2+uv/2)-4\rangle\text{,}$ with $0\leq u\leq 2\text{,}$ $0\leq v\leq 1\text{.}$

For $x=2\text{:}$ $\vec r(u,v) = \langle 2,u,4uv\rangle\text{,}$ with $0\leq u\leq 1\text{,}$ $0\leq v\leq 1$

For $y=1\text{:}$ $\vec r(u,v) = \langle u,1,2uv\rangle\text{,}$ with $0\leq u\leq 2\text{,}$ $0\leq v\leq 1$

For $z=0\text{:}$ $\vec r(u,v) = \langle u, 1-u/2+uv/2,0\rangle\text{,}$ with $0\leq u\leq 2\text{,}$ $0\leq v\leq 1$

16.5.3.11.

Answer.

Answers may vary.

For $z=2y: \vec r(u,v) = \langle u, v(4-u^2), 2v(4-u^2)\rangle$ with $-2\leq u\leq 2$ and $0\leq v\leq 1\text{.}$

For $y=4-x^2: \vec r(u,v) = \langle u, 4-u^2, 2v(4-u^2)\rangle$ with $-2\leq u\leq 2$ and $0\leq v\leq 1\text{.}$

For $z=0\text{:}$ $\vec r(u,v) = \langle u, v(4-u^2), 0\rangle$ with $-2\leq u\leq 2$ and $0\leq v\leq 1\text{.}$

16.5.3.12.

Answer.

Answers may vary.

For $y=1-z^2\text{:}$ $\vec r(u,v) = \langle u, v(1-u^2), \sqrt{1-v(1-u^2)}\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

For $y=1-x^2\text{:}$ $\vec r(u,v) = \langle u, 1-u^2, uv\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

For $x=0\text{:}$ $\vec r(u,v) = \langle 0, v(1-u^2),u\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

For $y=0\text{:}$ $\vec r(u,v) = \langle u, 0,v\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

For $z=0\text{:}$ $\vec r(u,v) = \langle u, v(1-u^2), 0rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

16.5.3.13.

Answer.

Answers may vary.

For $x^2+y^2/9=1\text{:}$ $\vec r(u,v) = \langle \cos(u), 3\sin(u), v\rangle$ with $0\leq u\leq 2\pi$ and $1\leq v\leq 3\text{.}$

For $z=1\text{:}$ $\vec r(u,v) = \langle v\cos(u), 3v\sin(u), 1\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1\text{.}$

For $z=3\text{:}$ $\vec r(u,v) = \langle v\cos(u), 3v\sin(u), 3\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1\text{.}$

16.5.3.14.

Answer.

Answers may vary.

For $x^2+y^2=(z-1)^2\text{:}$ $\vec r(u,v) = \langle v\cos(u), v\sin(u), 1-v\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1\text{.}$

For $z=0\text{:}$ $\vec r(u,v) = \langle v\cos(u), v\sin(u), 0\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1\text{.}$

16.5.3.15.

Answer.

Answers may vary.

For $z=1-x^2\text{:}$ $\vec r(u,v) = \langle u,v,1-u^2\rangle$ with $-1\leq u\leq 1$ and $-1\leq v\leq 2\text{.}$

For $y=-1\text{:}$ $\vec r(u,v) = \langle u,-1,v(1-u^2)\rangle$ with $-1\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

For $y=2\text{:}$ $\vec r(u,v) = \langle u,2,v(1-u^2)\rangle$ with $-1\leq u\leq 1$ and $0\leq v\leq 1\text{.}$

For $z=0\text{:}$ $\vec r(u,v) = \langle u,v,0\rangle$ with $-1\leq u\leq 1$ and $-1\leq v\leq 2\text{.}$

16.5.3.16.

Answer.

Answers may vary.

Answer.

$9\pi/8\text{;}$ the flux over $\surfaceS_1$ is $3\pi/4$ (use $\vec r(u,v) = \langle \sin(u)\cos(v),\sin(u)\sin(v),\cos(u)\rangle$ on $\pi/3\leq u\leq \pi\text{,}$ $0\leq v\leq 2\pi$) and the flux over $\surfaceS_2$ is $3\pi/8$ (use $\vec r(u,v) = \langle v\sqrt{3}\cos (u)/2, v\sqrt{3}\sin(u)/2,1/2\rangle$ for $0\leq u\leq 2\pi\text{,}$ $0\leq v\leq 1\text{.}$

Terms and Concepts

16.7.4.1.

Answer.

Answers will vary; in Section 16.4, the Divergence Theorem connects outward flux over a closed curve in the plane to the divergence of the vector field, whereas in this section the Divergence Theorem connects outward flux over a closed surface in space to the divergence of the vector field.

16.7.4.2.

Answer.

Divergence.

16.7.4.3.

Answer.

Curl.

16.7.4.4.

$\iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = \pi\text{.}$

16.7.4.11.

Answer.

Circulation on $C\text{:}$ The flow along the line from $(0,0,2)$ to $(4,0,0)$ is 0; from $(4,0,0)$ to $(0,3,0)$ it is $-6\text{,}$ and from $(0,3,0)$ to $(0,0,2)$ it is 6. The total circulation is $0+(-6)+6=0\text{.}$

$\iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = \iint_\surfaceS 0 \, dS = 0\text{.}$

16.7.4.12.

Answer.

16.7.4.22.a

Answer.

$\curl\vec F = 1\text{.}$

16.7.4.22.b

Answer.

$\curl\vec F\cdot \vec n = 1\text{,}$ where $\vec n$ is a unit vector normal to $\surfaceS\text{.}$

16.7.4.23.

Answer.

Answers will vary. Often the closed surface $\surfaceS$ is composed of several smooth surfaces. To measure total outward flux, this may require evaluating multiple double integrals. Each double integral requires the parametrization of a surface and the computation of the cross product of partial derivatives. One triple integral may require less work, especially as the divergence of a vector field is generally easy to compute.

16.7.4.24.

Answer.

Answers will vary. Often the closed curve $C$ is composed of several smooth curves. To measure the total circulation, one may have to evaluate line integrals along each curve. Each line integral requires the parametrization of its curve. It may be less work to evaluate one single double (i.e., surface) integral.

\(x\)	\(0.0\)	\(0.2\)	\(0.4\)	\(0.6\)	\(0.8\)	\(1.0\)
\(y(x)\)	1.0000	1.0204	1.0870	1.2195	1.4706	2.0000
\(h = 0.2\)	1.0000	1.0000	1.0400	1.1265	1.2788	1.5405
\(h = 0.1\)	1.0000	1.0100	1.0623	1.1687	1.3601	1.7129

Appendix A Answers to Selected Exercises

I Math 1560: Calculus I1 Limits1.1 An Introduction To Limits

Exercises

Problems

1.3 Finding Limits Analytically

Exercises

Problems

1.4 One-Sided Limits

Exercises

Problems

I Math 1560: Calculus I
1 Limits
1.1 An Introduction To Limits