**Unformatted text preview: **MATH 1080, Calculus II
Homework #2 Name: ________SOLUTIONS_________
Group: __________ Table: __________
Due: Monday, January 27th (beginning of class) Topics Covered:
• Section 6.3: Volume by Slicing & Disk/Washer Method
• Section 6.4: Volume by Shell Method
Directions: Answer the following questions. Do all your work on your own paper. When you are finished,
staple all your work to this sheet. To receive full credit you must show all of your work and provide
1) Proper and complete notation!!
2) Legible, organized, and logical (relevant) justification for your answer!!
3) Each question should be concluded with a complete sentence, either mathematical or in words as
appropriate!!
Conceptual Questions:
1. True/False: Determine whether the following statements are true or false. If true, give an explanation; if
false, give a counterexample.
a. When using the shell method, the line segment used to represent a typical shell is parallel to the axis of
revolution.
Solution: True, if the line segment were not parallel, then rotating it around the axis of revolution would not result in a shell. b. If a region is revolved about the y-axis, then the disk method must be used.
Solution: False, in principle either disk/washer or shell methods can be used for revolution around the y-axis. c. If a region is revolved about the x-axis, then in principle it is possible to use the disk/washer method and
integrate with respect to x or the shell method and integrate with respect to y.
Solution: True, either method can be used for revolution around the x-axis, and revolution around the x-axis would require x
for the disk/washer method and y for the shell method. 2. Explanation:
a. Suppose a cut is made through a solid object perpendicular to the x-axis at a particular value of x.
Explain the meaning of A(x).
Solution: A(x) is the area of a typical cross section that is cut through the solid at the particular value of x. b. Explain why the disk method is a special case of the general slicing method.
Solution: For the disk method, the cross sections are disks or circles, and the area of the cross section A(x) is the area of a
circle, ߨሾݎሺݔሻሿଶ . c. A solid has a circular base and cross sections perpendicular to the base are squares. What method should
be used to find the volume of the solid that is generated? Explain why the method you have chosen must
be used and why another method cannot be used.
Solution: The general slicing method must be used because the cross-sections are square. Because of the square crosssections, the solid would not be able to be formed by rotation about an axis thereby ruling out the disk/washer and shell
methods. Problem Solving Questions:
3. Use the method of slicing to compute the volume, in cubic feet, of the Great Pyramid of Egypt, whose base
is a square 755 feet by 755 feet and whose height is 410 feet. Note: your final answer will be large;
therefore, you can use a calculator to find your final answer, and you can round your final answer to two
significant digits. However, you must show all of your work up until the last step or arithmetic.
Solution: We slice the pyramid horizontally. Each slice is a square
with area ݏଶ . To find a formula for ,ݏa side of the square, we will
draw a triangle in the xy-plane to represent the silhouette of the
pyramid with a horizontal slice in the middle (see below). We will
place the x-axis on the bottom of the triangle and the y-axis in the
middle of the triangle. The left side of the triangle is a line with
଼ଶ
ଵହଵ
ହହ
equation ݕൌ
ݔ 410 ⇒ ݔൌ
ݕെ , and the right side is a
ହହ line with equation ݕൌ 410 െ ଼ଶ
ହହ ଵସ ݔ⇒ݔൌ ଶ
ହହ
ଶ െ ଵହଵ
ଵସ .ݕ ݕ ݔ Then ݏൌ
݀ ݑൌ െ ହହ ଶ
ଵହଵ െ ଵହଵ
ଵସ ݕെቀ ݕെ ଵହଵ ݀ ⇒ ݕെ ଼ଶ ଵସ ହହ
ଶ ቁ ൌ 755 െ ଵହଵ
଼ଶ .ݕTherefore, ܸ ൌ
ସଵ ቀ755 െ ଵହଵ
଼ଶ ݕቁ ݀ ݑ ݃݊݅ݐݐ݈݁ ݓܰ .ݕൌ 755 െ
ଶ ݀ ݑൌ ݀ ݕ ݄݊݁ݓ ݏ݈ܣ .ݕൌ 0, ݑൌ 755; ݕ ݄݊݁ݓൌ 410, ݑൌ 0. ݄ܶ݁,݁ݎ݂݁ݎ ଵହଵ
଼ଶ ⇒ ,ݕ
151 ଶ
82 ଶ
82 ହହ ଶ
82 ଷ ହହ
82
ሺ755ሻଷ ൎ 78 ݈݈݉݅݅ ݐ݂݁݁ ݊ଷ .
ܸ ൌ න ൬755 െ
ݕ൰ ݀ ݕൌ න െ
ݑ݀ ݑൌ
න ݑ݀ ݑൌ
ݑ൨
ൌ
82
151
151
453
453
ହହ
Consequently, the volume of the great pyramid is approximately ܸ ൎ 78 ݈݈݉݅݅ ݐ݂݁݁ ݊ଶ .
଼ଶ ସଵ ଵହଵ 4. A solid has a base in the -ݕݔplane that is bounded by the curves ݕൌ ݔଶ and the line ݕൌ 1. Cross sections
through the solid perpendicular to the -ݕaxis are triangles that are twice as tall as their base. Find the volume
of the solid.
Solution: The cross sections are triangles therefore, ܣൌ ܾ݄ ൌ ܾሺ2ܾሻ ൌ ܾ ଶ .
ଶ
ଶ
Now the base of the triangle is perpendicular to the ݕaxis running from right
side of the parabola to the left side (see figure to right) – therefore we need the
parabola to be in terms of ݕ :ݕൌ ݔଶ ⇒ ݔൌ േඥ .ݕThen ܾሺݕሻ ൌ ඥ ݕെ
ଵ ଵ ൫െඥݕ൯ ൌ 2ඥ .ݕTherefore, ܣൌ ൫2ඥݕ൯ ൌ 4 .ݕAlso, ݕranges from 0 to 1 so
ଶ ܸ ൌ න 4 ݕ݀ݕൌ ሾ2 ݕଶ ሿଵ ൌ 2 ݏݐ݅݊ݑଶ .
ଵ Consequently, the volume of the solid is ܸ ൌ 2 ݏݐ݅݊ݑଶ .
ଵ ௫ 5. Washers vs. Shells: Let ܴ be the region bounded by ݕൌ
and ݕൌ 1 െ . Let ܵ be the solid generated
௫ାଵ
ଷ
when ܴ is rotated about the -ݔaxis. Find the volume of ܵ by both the disk/washer and shell methods. Check
that your results agree. State which method is easiest to use and why.
Solution: First, we must find the points of intersection:
ݔ
1
3െݔ
1
ൌ 1െ ⇒
ൌ
⇒ 3 ൌ ሺ3 െ ݔሻሺ ݔ 1ሻ ⇒ 0 ൌ 2 ݔെ ݔଶ ⇒ 0 ൌ ሺ2 െ ݔሻ ݔ ⇒ ݔൌ 0,2.
ݔ1
3 ݔ1
3
ଵ
Then the two points of intersection are ሺ0,1ሻ and ቀ2, ቁ.
ଷ
For the disk/washer method, rotation about the -ݔaxis implies integration with
respect to .ݔGraphing the region described shows that there is a gap between
the region and the axis of revolution; therefore, we use the washer method with
ଵ
௫
ݔ
ݎሺݔሻ ൌ
and ܴሺݔሻ ൌ 1 െ . Then
ݕൌ 1െ
ݔଶ
1 ଶ
ܸ ൌ න ߨ ቈቀ1 െ ቁ െ ൬
൰ ݀ݔ
3
ݔ1
ଶ
ଶ
2
1
1
ൌ ߨ න ൬1 െ ݔ ݔଶ ൰ ݀ ݔെ ߨ න
݀ݔ
ሺ ݔ 1ሻଶ
3
9
ᇣᇧᇤᇧᇥ ௫ାଵ ଶ ଷ 3 ݕൌ 1
ݔ1 ଷ
1 ଷ ଶ
4 8
1
ଷ
ݔ൨ െ ߨ න ିݑଶ ݀ ݑൌ ߨ 2 െ ൨ െ ߨሾെିݑଵ ሿଵ
ൌ ߨ ݔെ ݔଶ
3
27
3 27
ଵ
26ߨ
1
8ߨ
ൌ
െ ߨ ൬െ 1൰ ൌ ݐ݅݊ݑଶ .
27
3
27
௧ ௨ୀ௫ାଵ For the shell method, rotation about the -ݕaxis implies integration with respect to .ݕTherefore, solving both functions for ݕ
gives:
1
1
ݔ
⇒ ݔൌ െ 1 and ݕൌ 1 െ ⇒ 3 െ 3 ݕൌ .ݔ
ݕൌ
ݔ1
ݕ
3
Now as the axis of revolution is the -ݔaxis, the shell radius is ;ݔalso from the graph of the region, it is clear that the shell height
ଵ
is 3 െ 3 ݕെ ቀ െ 1ቁ. Consequently,
௬ ଵ
ଵ
1
8ߨ
ܸ ൌ න 2ߨ ݕ൭3 െ 3 ݕെ ൬ 1൰൱ ݀ ݕൌ 2ߨ න ሺ4 ݕെ 3 ݕଶ െ 1ሻ݀ ݕൌ 2ߨሾ2 ݕଶ െ ݕଷ െ ݕሿଵ ൌ ݏݐ݅݊ݑ .
ଵ/ଷ
ଵ
ଵ
ݕ
27
ଷ Consequently, ܸ ൌ ଷ ଼గ
ଶ ݏݐ݅݊ݑଶ from both methods. Since both methods resulted in an integral that could be evaluated using our current integration techniques, an argument could
be made for either method being the easier option. The washer method may be considered the easier option since it didn’t require
solving for the other variable. However, the integration in terms of y was much simpler, so an argument could also be made for
the shell method being the easier option in this case. 6. Let ܴ be the region bounded by the following curves. Let ܵ be the solid generated when ܴ is revolved about
the given axis. Find the volume of ܵ by the method of your choice; in your solution, state the method that
you use.
ଵ
a. ݕൌ cosሺ ݔଶ ሻ , ݔൌ 0, ݔൌ √ߨ, ݕൌ 0, about the -ݕaxis.
ଶ
Solution: Notice, rotation around the -ݕaxis requires disk/washer
method to be in terms of .ݕHowever, if we solve for ݕin the first
equation, we obtain ݔൌ ඥܿି ݏଵ ݕwhich we cannot integrate.
Therefore, we must integrate with respect to .ݔConsequently, we must
use the shell method. Then
ଵ
ଶ√గ ܸൌන గ
ସ 1
ݔൌ √ݔ
2 ݕൌ0 2ߨݏܿݔሺ ݔଶ ሻ
ᇣᇧᇧᇤᇧᇧᇥ ݀ ݔൌ ߨ න ܿ ݑ݀ ݑ ݏൌ ߨሾݑ ݊݅ݏሿ
ᇧ ᇧ గ/ସ ߨ
√2
ൌ ߨ ቂ ݊݅ݏെ 0 ݊݅ݏቃ ൌ
ߨ.
4
2
√ଶ
Therefore, the volume of the solid is ܸ ൌ ߨ ݐ݅݊ݑଷ .
௧ ௨ୀ௫ మ ݔൌ0 ଶ b. ݔൌ ݔ , 2ݕൌ െ ݕ ,ݕൌ 2, ݕ 0, about the -ݔaxis.
ݕൌ2 Solution (Shell Method):
Shell radius is ݕand shell height is ݕଶ െ ሺെݕሻ ൌ ݕଶ .ݕThen
ଶ
ଶ
ଶ
ݕସ ݕଷ
ଶ
ଷ
ଶ ሻ݀ݕ
ൌ 2ߨ ቈ
ܸ ൌ න 2ߨݕሺ ݕ ݕሻ݀ ݕൌ න 2ߨሺ ݕ ݕ
4
3
8
40ߨ
ൌ 2ߨ ൬4 ൰ ൌ ݐ݅݊ݑଷ .
3
3
ସగ
Therefore, the volume of the solid is ܸ ൌ
ݏݐ݅݊ݑଷ . ݔൌ ݕଶ ଷ ݔൌ െݕ Solution (Washer Method):
First, put the equations in terms of ݔ :ݔൌ ݕଶ ⇒ ݕൌ േ ,ݔand ݔൌ െ ݕ ⇒ ݕൌ െ .ݔThen ܸ ൌ ߨ න ሺ2ଶ െ ሺെݔሻଶ ሻ݀ ݔ ߨ න ቀ2ଶ െ ൫√ݔ൯ ቁ ݀ ݔൌ ߨ න ሺ4 െ ݔଶ ሻ݀ ݔ ߨ න ሺ4 െ ݔሻ݀. ݔ
ସ ଶ ସ
ସ
1
1
8
40ߨ
ൌ ߨ 4 ݔെ ݔଷ ൨ ߨ 4 ݔെ ݔଶ ൨ ൌ ߨ 0 െ ൬െ8 ൰൨ ߨሾሺ16 െ 8ሻ െ 0ሿ ൌ ݏݐ݅݊ݑଷ .
3
2
3
3
ିଶ
ସగ
Therefore, the volume of the solid is ܸ ൌ ݏݐ݅݊ݑଷ .
ିଶ ିଶ ଷ c. ݕൌ sec ݕ , ݔൌ tan ݔ , ݔൌ 0, ݔൌ 1, about the -ݔaxis.
Solution: If we use the shell method, then we must solve for ݔ
which gives us two inverse trig functions which we don’t know
how to integrate right now. Therefore, we must integrate with
respect to .ݔThen
ܸ ൌ න ߨሾሺݔ ܿ݁ݏሻଶ െ ሺݔ ݊ܽݐሻଶ ሿ݀ݔ
ଵ ൌ න ߨሾ ܿ݁ݏଶ ݔെ ሺ ܿ݁ݏଶ ݔെ 1ሻሿ݀ݔ
ଵ ݕൌ sec ݔ ଵ ൌ න ߨሺ1ሻ݀ ݔൌ ߨሾݔሿଵ
ൌ ߨ . ݏݐ݅݊ݑ
Therefore, the volume of the solid is ܸ ൌ ߨ ݏݐ݅݊ݑଷ .
ݕൌ tan ݔ ଷ ݔൌ0 ݔൌ1 ...

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