Appendix A Answers to Selected Exercises
I Math 1560: Calculus I
1 Limits
1.1 An Introduction To Limits
Exercises
1.3 Finding Limits Analytically
Exercises
Problems
1.4 One-Sided Limits
Exercises
Problems
1.4.5.
1.4.5.a
1.4.5.b
1.4.5.c
1.4.5.d
1.4.5.e
1.4.5.f
1.4.6.
1.4.6.a
1.4.6.b
1.4.6.c
1.4.6.d
1.4.6.e
1.4.6.f
1.4.7.
1.4.7.a
1.4.7.b
1.4.7.c
1.4.7.d
1.4.7.e
1.4.7.f
1.4.8.
1.4.8.a
1.4.8.b
1.4.8.c
1.4.8.d
1.4.9.
1.4.9.a
1.4.9.b
1.4.9.c
1.4.9.d
1.4.10.
1.4.10.a
1.4.10.b
1.4.10.c
1.4.10.d
1.4.11.
1.4.11.a
1.4.11.b
1.4.11.c
1.4.11.d
1.4.11.e
1.4.11.f
1.4.11.g
1.4.11.h
1.4.12.
1.4.12.a
1.4.12.b
1.4.12.c
1.4.12.d
1.4.13.
1.4.13.a
1.4.13.b
1.4.13.c
1.4.13.d
1.4.14.
1.4.14.a
1.4.14.b
1.4.14.c
1.4.14.d
1.4.15.
1.4.15.a
1.4.15.b
1.4.15.c
1.4.15.d
1.4.15.e
1.4.15.f
1.4.15.g
1.4.15.h
1.4.16.
1.4.16.a
1.4.16.b
1.4.16.c
1.4.16.d
1.4.17.
1.4.17.a
1.4.17.b
1.4.17.c
1.4.17.d
1.4.18.
1.4.18.a
1.4.18.b
1.4.18.c
1.4.18.d
1.4.19.
1.4.19.a
1.4.19.b
1.4.19.c
1.4.19.d
1.4.20.
1.4.20.a
1.4.20.b
1.4.20.c
1.4.20.d
1.4.21.
1.4.21.a
1.4.21.b
1.4.21.c
1.4.21.d
1.5 Continuity
Exercises
Problems
1.6 Limits Involving Infinity
1.6.5 Exercises
Problems
2 Derivatives
2.1 Instantaneous Rates of Change: The Derivative
2.1.3 Exercises
Problems
2.1.3.27.
2.1.3.27.a
2.1.3.27.b
2.1.3.27.c
2.1.3.28.
2.1.3.28.a
2.1.3.28.b
2.1.3.28.c
2.2 Interpretations of the Derivative
2.2.5 Exercises
Problems
2.2.5.4.
2.2.5.5.
2.2.5.6.
2.2.5.7.
2.2.5.8.
2.2.5.9.
2.2.5.10.
2.2.5.11.
2.2.5.12.
2.3 Basic Differentiation Rules
Exercises
Terms and Concepts
2.3.2.
2.3.3.
2.3.4.
2.3.7.
2.3.9.
2.3.10.
Problems
2.4 The Product and Quotient Rules
Exercises
Problems
2.4.15.
2.4.16.
2.4.17.
2.4.18.
2.4.19.
2.4.20.
2.4.21.
2.4.22.
2.4.23.
2.4.24.
2.4.25.
2.4.26.
2.4.27.
2.4.28.
2.4.29.
2.4.30.
2.4.31.
2.4.32.
2.4.33.
2.4.34.
Answer.
\(\frac{\left(3r^{2}\cos\mathopen{}\left(r\right)-r^{3}\sin\mathopen{}\left(r\right)\right)\mathopen{}\left(r^{3}\sin\mathopen{}\left(r\right)-9\right)-\left(r^{3}\cos\mathopen{}\left(r\right)+8\right)\mathopen{}\left(3r^{2}\sin\mathopen{}\left(r\right)+r^{3}\cos\mathopen{}\left(r\right)\right)}{\left(r^{3}\sin\mathopen{}\left(r\right)-9\right)^{2}}\)
2.4.35.
2.4.36.
2.5 The Chain Rule
Exercises
Problems
2.5.41.
2.5.42.
2.6 Implicit Differentiation
2.6.4 Exercises
Problems
2.7 Derivatives of Inverse Functions
Exercises
Problems
3 The Graphical Behavior of Functions
3.1 Extreme Values
Exercises
Terms and Concepts
3.1.6.
Problems
3.2 The Mean Value Theorem
Exercises
Problems
3.3 Increasing and Decreasing Functions
Exercises
Terms and Concepts
3.3.3.
Problems
3.4 Concavity and the Second Derivative
3.4.3 Exercises
Problems
4 Applications of the Derivative
4.1 Newton’s Method
Exercises
Problems
4.2 Related Rates
Exercises
Problems
4.2.3.
4.2.3.a
4.2.3.b
4.2.3.c
4.2.4.
4.2.4.a
4.2.4.b
4.2.4.c
4.2.5.
4.2.6.
4.2.6.a
4.2.6.b
4.2.7.
4.2.7.a
4.2.7.b
4.2.7.c
4.2.8.
4.2.8.a
4.2.8.b
4.2.8.c
4.2.9.
4.2.9.a
4.2.9.b
4.2.9.c
4.2.9.d
4.2.10.
4.2.10.a
4.2.10.b
4.2.10.c
4.2.11.
4.2.11.a
4.2.11.b
4.2.11.c
4.2.11.d
4.2.12.
4.2.12.a
4.2.12.b
4.2.12.c
4.2.13.
4.2.13.a
4.2.13.b
4.2.13.c
4.2.13.d
4.2.14.
4.2.14.a
4.2.14.b
4.2.15.
4.3 Optimization
Exercises
Problems
4.3.3.
4.3.4.
4.3.5.
4.3.6.
4.3.7.
4.3.8.
4.3.9.
4.3.10.
4.3.11.
4.3.12.
4.3.13.
4.3.14.
4.3.15.
4.3.16.
4.3.17.
4.3.18.
4.4 Differentials
Exercises
Problems
4.4.31.
4.4.32.
4.4.32.a
4.4.32.b
4.4.33.
4.4.34.
4.5 Taylor Polynomials
Exercises
Terms and Concepts
4.5.3.
4.5.4.
Problems
4.5.13.
4.5.14.
4.5.15.
Answer.
\(0.707107-0.707107\mathopen{}\left(x-\frac{\pi }{4}\right)-0.353553\mathopen{}\left(x-\frac{\pi }{4}\right)^{2}+0.117851\mathopen{}\left(x-\frac{\pi }{4}\right)^{3}+0.0294628\mathopen{}\left(x-\frac{\pi }{4}\right)^{4}-0.00589256\mathopen{}\left(x-\frac{\pi }{4}\right)^{5}-0.000982093\mathopen{}\left(x-\frac{\pi }{4}\right)^{6}\)
4.5.16.
Answer.
4.5.17.
4.5.18.
4.5.19.
4.5.20.
5 Integration
5.1 Antiderivatives and Indefinite Integration
Exercises
Terms and Concepts
5.1.8.
Problems
5.2 The Definite Integral
Exercises
Terms and Concepts
5.2.4.
Problems
5.2.5.
5.2.5.a
5.2.5.b
5.2.5.c
5.2.5.d
5.2.5.e
5.2.5.f
5.2.6.
5.2.6.a
5.2.6.b
5.2.6.c
5.2.6.d
5.2.6.e
5.2.6.f
5.2.7.
5.2.7.a
5.2.7.b
5.2.7.c
5.2.7.d
5.2.7.e
5.2.7.f
5.2.8.
5.2.8.a
5.2.8.b
5.2.8.c
5.2.8.d
5.2.8.e
5.2.8.f
5.2.9.
5.2.9.a
5.2.9.b
5.2.9.c
5.2.9.d
5.2.10.
5.2.10.a
5.2.10.b
5.2.10.c
5.2.10.d
5.2.17.
5.2.17.a
5.2.17.b
5.2.17.c
5.2.17.d
5.2.18.
5.2.18.a
5.2.18.b
5.2.18.c
5.2.18.d
5.3 Riemann Sums
5.3.5 Exercises
Terms and Concepts
5.3.5.2.
5.3.5.3.
Problems
5.4 The Fundamental Theorem of Calculus
5.4.6 Exercises
Terms and Concepts
5.4.6.2.
Problems
5.5 Numerical Integration
5.5.6 Exercises
Problems
5.5.6.5.
5.5.6.5.a
5.5.6.5.b
5.5.6.5.c
5.5.6.6.
5.5.6.6.a
5.5.6.6.b
5.5.6.6.c
5.5.6.7.
5.5.6.7.a
5.5.6.7.b
5.5.6.7.c
5.5.6.8.
5.5.6.8.a
5.5.6.8.b
5.5.6.8.c
5.5.6.9.
5.5.6.9.a
5.5.6.9.b
5.5.6.9.c
5.5.6.10.
5.5.6.10.a
5.5.6.10.b
5.5.6.10.c
5.5.6.11.
5.5.6.11.a
5.5.6.11.b
5.5.6.11.c
5.5.6.12.
5.5.6.12.a
5.5.6.12.b
5.5.6.12.c
II Math 2560: Calculus II
6 Techniques of Antidifferentiation
6.1 Substitution
6.1.6 Exercises
Problems
6.2 Integration by Parts
Exercises
Problems
6.3 Trigonometric Integrals
6.3.4 Exercises
Problems
6.4 Trigonometric Substitution
Exercises
Terms and Concepts
6.4.2.
6.4.3.
Problems
6.5 Partial Fraction Decomposition
Exercises
Terms and Concepts
Problems
6.6 Hyperbolic Functions
6.6.3 Exercises
Problems
6.7 L’Hospital’s Rule
6.7.4 Exercises
Problems
6.7.4.9.
6.7.4.10.
6.7.4.11.
6.7.4.12.
6.7.4.13.
6.7.4.14.
6.7.4.15.
6.7.4.16.
6.7.4.17.
6.7.4.18.
6.7.4.19.
6.7.4.20.
6.7.4.21.
6.7.4.23.
6.7.4.24.
6.7.4.25.
6.7.4.26.
6.7.4.27.
6.7.4.28.
6.7.4.29.
6.7.4.30.
6.7.4.31.
6.7.4.32.
6.7.4.33.
6.7.4.34.
6.7.4.35.
6.7.4.36.
6.7.4.37.
6.7.4.38.
6.7.4.39.
6.7.4.40.
6.7.4.41.
6.7.4.42.
6.7.4.43.
6.7.4.44.
6.7.4.45.
6.7.4.46.
6.7.4.47.
6.7.4.48.
6.7.4.49.
6.7.4.50.
6.7.4.51.
6.7.4.52.
6.7.4.53.
6.7.4.54.
6.8 Improper Integration
6.8.4 Exercises
Problems
7 Applications of Integration
7.1 Area Between Curves
Exercises
Problems
7.1.19.
7.1.20.
7.1.31.
7.1.32.
7.2 Volume by Cross-Sectional Area; Disk and Washer Methods
Exercises
Problems
7.3 The Shell Method
Exercises
Problems
7.4 Arc Length and Surface Area
7.4.3 Exercises
Problems
7.5 Work
7.5.4 Exercises
Terms and Concepts
7.5.4.1.
Problems
7.5.4.5.
7.5.4.5.a
7.5.4.5.b
7.5.4.6.
7.5.4.6.a
7.5.4.6.b
7.5.4.7.
7.5.4.7.a
7.5.4.7.b
7.5.4.7.c
7.5.4.8.
7.5.4.9.
7.5.4.9.a
7.5.4.9.b
7.5.4.9.c
7.5.4.10.
7.5.4.11.
7.5.4.12.
7.5.4.13.
7.5.4.14.
7.5.4.15.
7.5.4.16.
7.5.4.17.
7.5.4.18.
7.5.4.19.
7.5.4.20.
7.5.4.21.
7.5.4.21.a
7.5.4.21.b
7.5.4.21.c
7.5.4.22.
Answer.
192,767 ft–lb. Note that the tank is oriented horizontally. Let the origin be the center of one of the circular ends of the tank. Since the radius is 3.75 ft, the fluid is being pumped to \(y=4.75\text{;}\) thus the distance the gas travels is \(h(y)=4.75-y\text{.}\) A differential element of water is a rectangle, with length 20 and width \(2\sqrt{3.75^2-y^2}\text{.}\) Thus the force required to move that slab of gas is \(F(y) = 40\cdot45.93\cdot\sqrt{3.75^2-y^2}dy\text{.}\) Total work is \(\int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75-y)\sqrt{3.75^2-y^2}\, dy\text{.}\) This can be evaluated without actual integration; split the integral into \(\int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75)\sqrt{3.75^2-y^2}\, dy + \int_{-3.75}^{3.75} 40\cdot45.93\cdot(-y)\sqrt{3.75^2-y^2}\, dy\text{.}\) The first integral can be evaluated as measuring half the area of a circle; the latter integral can be shown to be 0 without much difficulty. (Use substitution and realize the bounds are both 0.)
7.5.4.23.
7.5.4.24.
7.5.4.24.a
7.5.4.24.b
7.5.4.24.c
7.5.4.25.
7.5.4.26.
7.5.4.27.
7.6 Fluid Forces
Exercises
Problems
7.6.19.
7.6.20.
8 Differential Equations
8.1 Graphical and Numerical Solutions to Differential Equations
8.1.4 Exercises
Terms and Concepts
8.1.4.1.
8.1.4.3.
8.1.4.4.
8.1.4.5.
8.1.4.6.
Problems
8.1.4.13.
Answer.
The \(x\) and \(y\) axes are uncalibrated.In the first quadrant in the top left, the field lines are north-east facing and in the bottom right they are southeast facing. In the second quadrant the field lines are all north-east facing. In the third quadrant like in the first quadrant in the top left the field lines are northeast facing and in the bottom right they are southeast facing. In the fourth quadrant all lines are southeast facing.
8.1.4.14.
Answer.
The \(x\) and \(y\) axes are uncalibrated. The field lines form concentric ovals facing away from the origin on both positive and negative \(x\) and \(y\) axes. The concentric shorter arcs are on either end of the \(x\) axis. On the two ends of the \(y\) axis concentric wider arcs are drawn. The field lines intermix to form an ’X’ with centre at the origin.
8.1.4.15.
Answer.
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.
8.1.4.18.
Answer.
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.
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.
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.
The \(x\) and \(y\) axes are uncalibrated, the field lines in the first quadrant are shown. In the top right and the centre the field lines are southeast facing, very close to the \(x\) and \(y\) axis the field lines are almost parallel to the \(x\) axis. A curve is drawn that starts from a \(y\) intercept and decreases along the slope lines coming close to the \(x\) axis.
8.1.4.22.
8.1.4.23.
Answer.
\begin{align*}
x_i \amp \quad \amp \quad \amp y_i \\
0.0 \amp \quad \amp \quad \amp 1.0000 \\
0.1 \amp \quad \amp \quad \amp 1.0000 \\
0.2 \amp \quad \amp \quad \amp 1.0037 \\
0.3 \amp \quad \amp \quad \amp 1.0110 \\
0.4 \amp \quad \amp \quad \amp 1.0219 \\
0.5 \amp \quad \amp \quad \amp 1.0363
\end{align*}
8.1.4.24.
Answer.
\begin{align*}
x_i \amp \quad \amp \quad \amp y_i \\
0.0 \amp \quad \amp \quad \amp 2.0000 \\
0.2 \amp \quad \amp \quad \amp 2.4000 \\
0.4 \amp \quad \amp \quad \amp 2.9197 \\
0.6 \amp \quad \amp \quad \amp 3.5816 \\
0.8 \amp \quad \amp \quad \amp 4.4108 \\
1.0 \amp \quad \amp \quad \amp 5.4364
\end{align*}
8.1.4.25.
8.1.4.26.
Answer.
| \(x\) | \(0.0\) | \(0.2\) | \(0.4\) | \(0.6\) | \(0.8\) | \(1.0\) |
| \(y(x)\) | 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 |
8.1.4.27.
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 |
8.2 Separable Differential Equations
8.2.2 Exercises
Problems
8.3 First Order Linear Differential Equations
8.3.2 Exercises
Problems
8.3.2.21.
Answer.
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.
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 \(y=0\text{.}\)
\(\displaystyle y = \frac{2 + \ln(x+1)}{x+1}\)
8.4 Modeling with Differential Equations
8.4.3 Exercises
Problems
9 Curves in the Plane
9.1 Conic Sections
9.1.4 Exercises
Problems
9.1.4.45.
9.2 Parametric Equations
9.2.4 Exercises
Problems
9.2.4.5.
Answer.
The sketch for this exercise is a curve that lies mostly in the fourth quadrant. It resembles part of a slingshot orbit for a comet passing around the sun: the curve passes through the origin from below, turns quickly in the second quadrant, crossing the \(y\) axis at \((0,1)\text{,}\) and then the \(x\) axis at \((2,0)\text{,}\) where it returns to the fourth quadrant.
9.2.4.6.
9.2.4.7.
9.2.4.8.
Answer.
The curve begins to the left of the \(y\) axis, and crosses near \((0,4)\text{.}\) It passes through the first quadrant to the point \((3,2)\text{;}\) it then bends downward and makes a teardrop-shaped loop before passing through \((3,2)\) a second time, and then continuing up and to the right, through the first quadrant.
9.3 Calculus and Parametric Equations
9.3.5 Exercises
Problems
9.4 Introduction to Polar Coordinates
9.4.4 Exercises
Problems
9.4.4.5.
Answer.
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.
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.
9.4.4.8.
9.4.4.9.
9.4.4.10.
9.4.4.11.
9.4.4.12.
9.4.4.13.
9.4.4.14.
9.4.4.15.
9.4.4.16.
9.4.4.17.
9.4.4.18.
9.4.4.19.
9.4.4.20.
Answer.
The curve is a limaçon with an inner loop, symmetric about the \(x\) axis, but it is shifted horizontally, to the left, relative to the example in the gallery of polar curves.
The point of self-intersection is at \((-1/2, 0)\text{.}\) The other end of the inner loop is at the origin, and the far end of the outer loop is at the point \((1,0)\text{.}\)
9.4.4.21.
Answer.
This is a more complicated curve. It passes several times through the origin, and has eight other points of self-intersection. The largest loops in the curve are similar to cardioids; there are four of these passing through the origin, with a second intercept at one of the four points \((\pm 1, 0)\text{,}\) \((0,\pm 1)\text{.}\) As these loops intersect each other, they create four other loops of intermediate size, and four smaller loops in the center.
9.4.4.22.
9.4.4.23.
9.4.4.24.
9.4.4.25.
9.4.4.26.
Answer.
The curve for this exercise is a fairly strange shape. It is symmetric about the \(x\) axis. There is a large, outer loop that looks like a leaf or a raindrop. It has a cusp at \((-7,0)\text{,}\) and also passes through the origin. There is a smaller, inner loop that looks almost like a circle. It passes through the origin and the point \((-3,0)\text{.}\)
9.4.4.27.
9.4.4.28.
9.4.4.29.
9.4.4.30.
9.5 Calculus and Polar Functions
9.5.5 Exercises
Problems
III Math 2570: Calculus III
10 Sequences and Series
10.1 Sequences
Exercises
Problems
10.2 Infinite Series
10.2.4 Exercises
Terms and Concepts
10.2.4.1.
10.3 Integral and Comparison Tests
10.3.4 Exercises
Problems
10.4 Ratio and Root Tests
10.4.3 Exercises
Terms and Concepts
10.4.3.3.
Problems
10.5 Alternating Series and Absolute Convergence
Exercises
Terms and Concepts
10.5.3.
11 Vectors
11.1 Introduction to Cartesian Coordinates in Space
11.1.7 Exercises
Terms and Concepts
11.1.7.5.
11.1.7.6.
Problems
11.1.7.7.
11.1.7.9.
11.1.7.10.
11.1.7.15.
Answer.
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.
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.
A “classic” cylinder, in the shape of a tube. Cross-sections parallel to the \(xy\) plane are ellipses.
11.1.7.18.
Answer.
Viewed along the \(z\) axis, the surface appears to be the hyperbola \(y=1/x\text{.}\) From other angles, it is a pair of curved sheets, bent in the shape of a hyperbola.
11.1.7.27.
Answer.
The surface is a standard hyperbolic paraboloid, opening upward along the \(y\) axis, and downward along the \(x\) axis.
11.1.7.28.
Answer.
The surface is a cone, opening along the \(z\) axis. Cross-sections parallel to the \(xy\) plane are ellipses.
11.1.7.29.
Answer.
The surface is a circular paraboloid, opening along the negative \(x\) axis.
11.1.7.30.
Answer.
A hyperboloid of two sheets with circular cross sections, opening along the \(x\) axis.
11.1.7.31.
Answer.
A hyperboloid of one sheet. It opens along the \(y\) axis, and cross sections parallel to the \(xz\) plane are ellipses.
11.1.7.32.
Answer.
An ellpsoid, somewhat in the shape of a squashed olive.
11.2 An Introduction to Vectors
Exercises
Terms and Concepts
11.2.2.
11.2.3.
11.2.4.
Problems
11.2.11.
11.2.11.a
11.2.11.c
11.2.12.
11.2.12.a
11.2.12.c
11.2.26.
11.2.27.
11.3 The Dot Product
11.3.2 Exercises
Problems
11.3.2.11.
11.3.2.12.
11.3.2.33.
11.3.2.34.
11.3.2.35.
11.3.2.36.
11.3.2.37.
11.3.2.39.
11.4 The Cross Product
11.4.3 Exercises
Problems
11.4.3.17.
11.4.3.18.
11.4.3.39.
11.4.3.40.
11.4.3.41.
11.4.3.42.
11.5 Lines
11.5.4 Exercises
Terms and Concepts
11.5.4.1.
Problems
11.6 Planes
11.6.2 Exercises
Terms and Concepts
11.6.2.1.
11.6.2.2.
Problems
12 Vector Valued Functions
12.1 Vector-Valued Functions
12.1.4 Exercises
Problems
12.1.4.14.
Answer.
Graph of the function \(\vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra\) on \([0,2\pi]\text{.}\) The graph of the function is an oval-shaped spiral centered about the \(z\)-axis. Ignoring the \(z\)-axis, the curve is simply an oval having a horizontal width of \(6\) and a height of \(2\) in the \(xy\)-plane. Incorporating the \(z\) coordinate then creates the linearly increasing oval spiral, which begins at the point \((3,0,0)\text{,}\) completes precisely one full revolution and ends at the point \((3,0,2)\text{.}\)
12.1.4.15.
Answer.
Graph of the function \(\vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra\) on \([0,2\pi]\text{.}\) The graph of the function is an oval lying in the plane coming from rotating the \(xy\) plane \(45\) degrees towards the \(z\)-axis. The oval lying in this plane has a horizontal width of \(\sqrt{2}\) and a height of \(1\text{.}\) Ignoring the \(z\) coordinate, the curve is a unit circle in the \(xy\) plane. Similarly ignoring the \(y\) coordinate, the curve is a unit circle in the \(xz\) plane. If we now ignore the \(x\) coordinate, the resulting curve is a diagonal line given by \(z=y\) in the \(yz\) plane. This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes.
12.1.4.16.
Answer.
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 \((1,0,0)\text{.}\)
12.2 Calculus and Vector-Valued Functions
12.2.5 Exercises
Terms and Concepts
12.2.5.4.
Problems
12.3 The Calculus of Motion
12.3.3 Exercises
Problems
12.4 Unit Tangent and Normal Vectors
12.4.4 Exercises
Terms and Concepts
12.4.4.3.
Problems
12.5 The Arc Length Parameter and Curvature
12.5.4 Exercises
Terms and Concepts
12.5.4.3.
12.5.4.4.
12.5.4.6.
Problems
13 Introduction to Functions of Several Variables
13.2 Limits and Continuity of Multivariable Functions
13.2.5 Exercises
Terms and Concepts
13.2.5.2.
13.2.5.3.
13.2.5.4.
13.2.5.5.
13.2.5.6.
Problems
13.3 Partial Derivatives
13.3.7 Exercises
Problems
IV Math 2580: Calculus IV
14 Functions of Several Variables, Continued
14.2 The Multivariable Chain Rule
14.2.3 Exercises
Problems
14.3 Directional Derivatives
14.3.3 Exercises
Problems
14.3.3.19.
14.3.3.19.a
14.3.3.19.b
14.3.3.19.c
14.3.3.19.d
14.3.3.20.
14.3.3.20.a
14.3.3.20.b
14.3.3.20.c
14.3.3.20.d
14.3.3.21.
14.3.3.21.a
14.3.3.21.b
14.3.3.21.c
14.3.3.21.d
14.3.3.22.
14.3.3.22.a
14.3.3.22.b
14.3.3.22.c
14.3.3.22.d
14.3.3.23.
14.3.3.23.a
14.3.3.23.b
14.3.3.23.c
14.3.3.23.d
14.3.3.24.
14.3.3.24.a
14.3.3.24.b
14.3.3.24.c
14.3.3.24.d
14.4 Tangent Lines, Normal Lines, and Tangent Planes
14.4.5 Exercises
Problems
14.5 Extreme Values
14.5.3 Exercises
Problems
15 Multiple Integration
15.1 Iterated Integrals and Area
15.1.4 Exercises
Problems
15.3 Double Integration with Polar Coordinates
Exercises
Problems
15.5 Surface Area
Exercises
Problems
15.6 Volume Between Surfaces and Triple Integration
15.6.4 Exercises
Problems
15.6.4.9.
Answer.
\(\ds V = \int_0^3\int_0^{1-x/3}\int_0^{2-2x/3-2y}\, dz\, dy\, dx =1\text{.}\)
15.6.4.10.
Answer.
\(\ds V = \int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz =2\text{.}\)
15.6.4.11.
Answer.
\(dx\, dy\, dz\text{:}\) \(\ds\int_0^2\int_{-\sqrt{2z}}^{-z}\int_0^{2}\, dx\, dy\, dz\) \(\ds V = \int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx =4/3\text{.}\)
15.6.4.12.
Answer.
\(dy\, dz\, dx\text{:}\) \(\ds\int_0^3\int_0^{\sqrt{81-9x^2}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dz\, dx\)
\(dy\, dx\, dz\text{:}\) \(\ds\int_0^9\int_0^{\sqrt{9-z^2/9}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dx\, dz\)
\(dx\, dz\, dy\text{:}\) \(\ds\int_{0}^9\int_{0}^{y}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dz\, dy\)
15.6.4.13.
Answer.
\(dx\, dy\, dz\text{:}\) \(\ds\int_0^4\int_{z/4}^{1}\int_{z/2-2y+2}^2\, dx\, dy\, dz\) \(\ds V = \int_0^4\int_{z/4}^{1}\int_{z/2-2y-2}^2\, dx\, dy\, dz = 4/3\text{.}\)
15.6.4.14.
Answer.
\(dz\, dx\, dy\text{:}\) \(\ds\int_{0}^4\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\int_{0}^{2x+4y-4}\, dz\, dx\, dy\)
\(dy\, dx\, dz\text{:}\) \(\ds\int_0^8\int_{-\sqrt{4-z/2}}^{\sqrt{4-z/2}}\int_{z/2}^{4-x^2}\, dy\, dx\, dz\)
\(dx\, dz\, dy\text{:}\) \(\ds\int_{0}^4\int_{0}^{2y}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dz\, dy\)
\(dx\, dy\, dz\text{:}\) \(\ds\int_0^8\int_{z/2}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dy\, dz\) \(\ds V = \int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx = 512/15\text{.}\)
15.6.4.15.
Answer.
\(dz\, dx\, dy\text{:}\) \(\ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dz\, dx\, dy\)
\(dy\, dz\, dx\text{:}\) \(\ds\int_{0}^1\int_0^{x}\int_{0}^{1-x^2}\, dy\, dz\, dx + \int_{0}^1\int_x^{1}\int_{0}^{1-z^2}\, dy\, dz\, dx\)
\(dy\, dx\, dz\text{:}\) \(\ds\int_0^1\int_{0}^{z}\int_{0}^{1-z^2}\, dy\, dx\, dz+\int_0^1\int_{z}^{1}\int_{0}^{1-x^2}\, dy\, dx\, dz\)
\(dx\, dz\, dy\text{:}\) \(\ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dx\, dz\, dy\)
\(dx\, dy\, dz\text{:}\) \(\ds\int_0^1\int_{0}^{1-z^2}\int_{0}^{\sqrt{1-y}}\, dx\, dy\, dz\) 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.
\(dz\, dy\, dx\text{:}\) \(\ds\int_{0}^1\int_{0}^{3x}\int_{0}^{1-x}\, dz\, dy\, dx+\int_{0}^1\int_{3x}^{3}\int_{0}^{1-y/3}\, dz\, dy\, dx\)
\(dz\, dx\, dy\text{:}\) \(\ds\int_{0}^3\int_{0}^{y/3}\int_{0}^{1-y/3}\, dz\, dy\, dx+\int_{0}^3\int_{y/3}^{1}\int_{0}^{1-x}\, dz\, dx\, dy\)
\(dx\, dy\, dz\text{:}\) \(\ds\int_0^1\int_{0}^{3-3z}\int_{0}^{1-z}\, dx\, dy\, dz\) \(\ds V = \int_0^1\int_{0}^{3-3z}\int_{0}^{1-z}\, dx\, dy\, dz = 1\text{.}\)
15.7 Triple Integration with Cylindrical and Spherical Coordinates
15.7.3 Exercises
Problems
16 Vector Analysis
16.1 Introduction to Line Integrals
16.1.4 Exercises
Terms and Concepts
16.1.4.1.
16.1.4.2.
16.1.4.3.
Answer.
The variable \(s\) denotes the arc-length parameter, which is generally difficult to use. Theorem 16.1.4 allows one to parametrize a curve using another, ideally easier-to-use, parameter.
Problems
16.2 Vector Fields
16.2.3 Exercises
Terms and Concepts
16.2.3.1.
16.2.3.2.
16.2.3.3.
16.2.3.4.
Problems
16.2.3.5.
Answer.
Correct answers should look similar to
A two-dimmensional vector field is plotted relative to \(x\) and \(y\) coordinate axes, with the origin at the center. All of the vectors are horizontal.
The vectors all point away from the \(y\) axis: those with \(x\gt 0\) point to the right, while those with \(x\lt 0\) point to the left. Vectors close to the \(y\) axis are small, and those near the left and right edges of the image are largest.
16.2.3.6.
Answer.
Correct answers should look similar to
A two-dimmensioanl vector field is plotted relative to \(x\) and \(y\) coordinate axes, with the origin at the center.
All of the vectors are vertical. Those with \(x\gt 0\) point up, while those with \(x\lt 0\) point down. Vectors close to the \(y\) axis are small, and those near the left and right edges of the image are largest.
16.2.3.7.
Answer.
Correct answers should look similar to
A two-dimensional vector field is plotted relative to \(x\) and \(y\) coordinate axes, with the origin at the center of the image.
The vector field in this image is constant: many vectors are plotted, but they all have the same magnitude and direction. In particular, each vector points down and to the right.
16.2.3.8.
Answer.
Correct answers should look similar to
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.
Along the lines \(y=1\) and \(y=-1\) vectors point up and to the right, but with a steeper slope. The magnitudes of these vectors are less than the ones at the top and bottom of the image.
Along the \(x\) axis, the vectors are vertical, and have the smallest magnitude.
It looks like the vectors could lie tangent to a family of cubic curves of the form \(y = a\sqrt[3]{x}+b\text{.}\)
16.3 Line Integrals over Vector Fields
16.3.4 Exercises
Terms and Concepts
16.3.4.2.
16.3.4.5.
16.3.4.6.
Answer.
By the Fundamental Theorem of Line Integrals, since \(\vec F\) is conservative, \(\oint_C \vec F\cdot d\vec r = f(B) - f(A)\text{,}\) where \(f\) is a potential function for \(\vec F\) and \(A\) and \(B\) are the initial and terminal points of \(C\text{,}\) respectively. Since \(C\) is a closed curve, \(A = B\text{,}\) and hence \(f(B) - f(A) = 0\text{.}\)
Problems
16.4 Flow, Flux, Green’s Theorem and the Divergence Theorem
16.4.4 Exercises
Terms and Concepts
16.4.4.2.
16.4.4.4.
Problems
16.4.4.13.
16.4.4.14.
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.
16.4.4.16.
Answer.
Two line integrals need to be computed to compute \(\oint_C \vec F\cdot d\vec r\text{.}\) Along the parabola, the line integral has value \(25.5\text{.}\) Along the line, the line integral has value \(-21\text{.}\) Together, the total value is \(4.5\)
16.4.4.21.
16.4.4.22.
16.4.4.23.
Answer.
Three line integrals need to be computed to compute \(\oint_C \vec F\cdot \vec n\, ds\text{.}\) 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 \(1/3\text{.}\) From \((1,1)\) to \((0,0)\) the line integral has a value of \(1/3\text{.}\) Total value is \(2/3\text{.}\)
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 \(279/20\text{.}\)
16.5 Parametrized Surfaces and Surface Area
16.5.3 Exercises
Terms and Concepts
16.5.3.1.
16.5.3.2.
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 \(0\leq u\leq 2\pi\text{,}\) \(0\leq v\leq 1\text{.}\)
-
\(\vec r(u,v) = \langle u, v(1-u^2), 3u^2v(1-u^2)\rangle\) 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 \(0\leq u\leq 2\pi\text{,}\) \(2\leq v\leq 5\text{.}\)
16.5.3.9.
Answer.
Answers may vary.
For \(z = \frac12(3-x)\text{:}\) \(\vec r(u,v) = \langle u, v , \frac12(3-u)\rangle\text{,}\) with \(1\leq u\leq 3\) and \(0\leq v\leq 2\text{.}\)
For \(x=1\text{:}\) \(\vec r(u,v) = \langle 1,u,v\rangle\text{,}\) with \(0\leq u\leq 2\text{,}\) \(0\leq v\leq 1\)
For \(y=0\text{:}\) \(\vec r(u,v) = \langle u,0,v/2(3-u)\rangle\text{,}\) with \(1\leq u\leq 3\text{,}\) \(0\leq v\leq 1\)
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.
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.
For \(z=4-x^2-4y^2\text{:}\) \(\vec r(u,v) = \langle 2v\cos(u),v\sin(u),4-(2v\cos(u))^2-4(v\sin(u))^2\rangle\) with \(0\leq u\leq 2\pi\) and \(0\leq v\leq 1\text{.}\)
For \(z=0\text{:}\) \(\vec r(u,v) = \langle 2v\cos(u),v\sin(u),0\rangle\) with \(0\leq u\leq 2\pi\) and \(0\leq v\leq 1\text{.}\)
16.6 Surface Integrals
16.6.3 Exercises
Terms and Concepts
16.6.3.2.
Problems
16.6.3.7.
16.6.3.8.
16.6.3.9.
16.6.3.10.
16.6.3.11.
16.6.3.12.
16.6.3.13.
16.6.3.14.
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 \(0\leq u\leq 2\pi\text{,}\) \(0\leq v\leq 1\text{.}\)
16.7 The Divergence Theorem and Stokes’ Theorem
16.7.4 Exercises
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.
16.7.4.3.
16.7.4.4.
Problems
16.7.4.5.
Answer.
Outward flux across the plane \(z=2-x/2-2y/3\) is 14; across the plane \(z=0\) the outward flux is \(-8\text{;}\) across the planes \(x=0\) and \(y=0\) the outward flux is 0.
Total outward flux: \(14\text{.}\)
\(\iint_D\divv\vec F\, dV = \int_0^{4}\int_0^{3-3x/4}\int_0^{2-x/2-2y/3}(2x+2y)\, dz\, dy\, dx = 14\text{.}\)
16.7.4.6.
Answer.
Outward flux across the cylinder \(x^2+y^2=1\) is 0; across the plane \(z=3\) the outward flux is \(3\pi\text{;}\) across the plane \(z=-3\) the outward flux is \(3\pi\text{.}\)
Total outward flux: \(6\pi\text{.}\)
\(\iint_D\divv\vec F\, dV = \int_0^{2\pi}\int_0^{1}\int_{-3}^{3}r\, dz\, dr\, d\theta = 6\pi\text{.}\)
16.7.4.7.
16.7.4.8.
16.7.4.9.
16.7.4.10.
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.
16.7.4.21.
16.7.4.22.
16.7.4.22.a
16.7.4.22.b
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.
