12 Week 12 amazonaws com
12 12.1
Week 12 Volumes of Revolution You should read section 6.2 of the textbook and do exercises 3, 9, 11, 13, 39, 41
We will use integration to define and calculate volumes of certain solids of revolution. First we will recall that the volume of a cylinder whose base has radius r and whose height is h is given by V = Area of base ⇥ height = (⇡r2 )h. Now suppose we have a solid extending from x = a to x = b, (a < b).
If we divide [a, b] into n slices of equal width x = xi xi 1 and choose sample point x⇤i 2 [xi 1 , xi ], then each slice has volume A(x⇤i ) x, where A(x⇤i ) is the cross-sectional area of the slice. If we sum the volumes of all the slices together from x = a to x = b, and let the number of slices increase to infinity, we get Z b n X ⇤ V = lim A(xi ) x = A(x)dx n!1
a
i=1
Now suppose we have a curve y = f (x) for x 2 [a, b] such that f (x) > 0. We can create a solid of revolution simply by revolving this curve about the x-axis.
Note that the cross-section of our solid is a circle of radius f (x) and area A(x) = ⇡r2 = ⇡[f (x)]2 . Thus, Z b Z b V = A(x)dx = ⇡[f (x)]2 dx a
a
is the volume of the solid obtained by revolving the curve y = f (x) from x = a to x = b about the x-axis. 112
If instead we revolve a curve x = g(y) about the y-axis from y = c to y = d, our formula becomes V =
Z
d
⇡[g(y)]2 dy c
since this time the radius is in the x direction. 12.1.1
Example
Find the volume of a right circular cone of radius R and height h.
Now suppose the region we want to revolve about the x-axis is bounded between two curves y = f (x) and y = g(x) from x = a to x = b. Then the cross section of the solid is a washer whose area is the area of the outer circle minus the area of the inner circle, or A(x) = ⇡[f (x)]2 ⇡[g(x)]2 . Thus, V =⇡
Z
b
[f (x)]2
[g(x)]2 dx
a
We could also choose to revolve the region between two curves about a line parallel to the x-axis, say y = k. Then since the outer radius has height f (x) k and inner radius has height g(x) l, we have V =⇡
Z
b
[f (x)
k]2
[g(x)
k]2 dx
a
We have analogous formulas if we rotate the region between curves x = f (y) and x = g(y) from y = c to y = d about the y-axis (the line x = 0) V =⇡
Z
d
[f (y)]2
[g(y)]2 dy
c
or about a line parallel to the y-axis, say x = l. Z d V =⇡ [f (y) l]2 c
113
[g(y)
l]2 dy
12.1.2
Example
Let R be the region between the curves y = x1/3 and y = x2 from x = 0 to x = 1. Find the volume of the region obtained when R is revolved: (a) about the line y = 2.
(b) about the line x =
3.
114
12.2
Work You should read section 6.4 of the textbook and do exercises 7, 15
In physics, work done on an object is equal to the force (push/pull) on the object times the distance that the object moves, or W = F d. If F is measured in newtons (N) and d in metres, then the unit for W is a newton-metre called a joule (J). If F is measured in pounds and d in feet, then the unit for W is a foot-pound (ft-lb). Suppose an object moves along the x-axis in the positive direction from x = a to x = b and at each point x, a force f (x) acts on the object, where f is a continuous function. We can divide the interval [a, b] into n subintervals of equal width x. We choose a sample point x⇤i 2 [xi 1 , xi ]. Then the force at that point is f (x⇤i ). We will assume the force is constant over [xi 1 , xi ] so that the work W done in moving the particle from xi 1 to xi is W ⇡ f (xi ) x. n X The total work done over [a, b] is then W ⇡ f (x⇤i ) x and if we let n ! 1, we get that the actual i=1
work done in moving the object from a to b is W = lim
n!1
n X
f (x⇤i )
x=
i=1
Z
b
f (x)dx a
Hooke’s Law states that the force required to maintain a spring stretch x units beyond its natural length is proportional to x, so f (x) = kx, where k is a positive constant called the spring constant. 12.2.1
Example
A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm?
115
12.2.2
Example
A 200-lb cable is 100 ft long and hangs vertically from the top of a tall building. How much work is required to lift the cable to the top of the building?
116
12.3
Average Value of a Function You should read section 6.5 of the textbook and do exercises 1, 3, 5
Recall that the average value of a finite set of numbers y1 , y2 , . . . , yn is yavg =
y1 + y2 + . . . + yn n
Suppose we want to calculate the average value of a continuous function y = f (x), a x b. We b a divide the interval [a, b] into n equal subintervals of width x = . Then we select a sample n point x⇤1 , . . . x⇤n from each subinterval and calculate the average of the numbers f (x⇤1 ), . . . , f (x⇤n ): f (x⇤1 ) + . . . + f (x⇤n ) n Since
x=
b
a n
, we can write n =
b
a and the average value becomes x
f (x⇤1 ) + . . . + f (x⇤n ) f (x⇤1 ) + . . . + f (x⇤n ) = b a n x n ⇤ ⇤ X f (x1 ) x + . . . + f (xn ) x 1 = = f (x⇤i ) x b a b a i=1 If we let n ! 1, then the average value of a function f (x) on the interval [a, b] is given by favg = lim
n!1
12.3.1
1 b
a
n X
f (x⇤i )
i=1
x=
1 b
a
Z
b
f (x)dx a
Example
Find the average value of the function f (x) = 1 + x2 on the interval [ 1, 2].
117
Week 12 Volumes of Revolution You should read section 6.2 of the textbook and do exercises 3, 9, 11, 13, 39, 41
We will use integration to define and calculate volumes of certain solids of revolution. First we will recall that the volume of a cylinder whose base has radius r and whose height is h is given by V = Area of base ⇥ height = (⇡r2 )h. Now suppose we have a solid extending from x = a to x = b, (a < b).
If we divide [a, b] into n slices of equal width x = xi xi 1 and choose sample point x⇤i 2 [xi 1 , xi ], then each slice has volume A(x⇤i ) x, where A(x⇤i ) is the cross-sectional area of the slice. If we sum the volumes of all the slices together from x = a to x = b, and let the number of slices increase to infinity, we get Z b n X ⇤ V = lim A(xi ) x = A(x)dx n!1
a
i=1
Now suppose we have a curve y = f (x) for x 2 [a, b] such that f (x) > 0. We can create a solid of revolution simply by revolving this curve about the x-axis.
Note that the cross-section of our solid is a circle of radius f (x) and area A(x) = ⇡r2 = ⇡[f (x)]2 . Thus, Z b Z b V = A(x)dx = ⇡[f (x)]2 dx a
a
is the volume of the solid obtained by revolving the curve y = f (x) from x = a to x = b about the x-axis. 112
If instead we revolve a curve x = g(y) about the y-axis from y = c to y = d, our formula becomes V =
Z
d
⇡[g(y)]2 dy c
since this time the radius is in the x direction. 12.1.1
Example
Find the volume of a right circular cone of radius R and height h.
Now suppose the region we want to revolve about the x-axis is bounded between two curves y = f (x) and y = g(x) from x = a to x = b. Then the cross section of the solid is a washer whose area is the area of the outer circle minus the area of the inner circle, or A(x) = ⇡[f (x)]2 ⇡[g(x)]2 . Thus, V =⇡
Z
b
[f (x)]2
[g(x)]2 dx
a
We could also choose to revolve the region between two curves about a line parallel to the x-axis, say y = k. Then since the outer radius has height f (x) k and inner radius has height g(x) l, we have V =⇡
Z
b
[f (x)
k]2
[g(x)
k]2 dx
a
We have analogous formulas if we rotate the region between curves x = f (y) and x = g(y) from y = c to y = d about the y-axis (the line x = 0) V =⇡
Z
d
[f (y)]2
[g(y)]2 dy
c
or about a line parallel to the y-axis, say x = l. Z d V =⇡ [f (y) l]2 c
113
[g(y)
l]2 dy
12.1.2
Example
Let R be the region between the curves y = x1/3 and y = x2 from x = 0 to x = 1. Find the volume of the region obtained when R is revolved: (a) about the line y = 2.
(b) about the line x =
3.
114
12.2
Work You should read section 6.4 of the textbook and do exercises 7, 15
In physics, work done on an object is equal to the force (push/pull) on the object times the distance that the object moves, or W = F d. If F is measured in newtons (N) and d in metres, then the unit for W is a newton-metre called a joule (J). If F is measured in pounds and d in feet, then the unit for W is a foot-pound (ft-lb). Suppose an object moves along the x-axis in the positive direction from x = a to x = b and at each point x, a force f (x) acts on the object, where f is a continuous function. We can divide the interval [a, b] into n subintervals of equal width x. We choose a sample point x⇤i 2 [xi 1 , xi ]. Then the force at that point is f (x⇤i ). We will assume the force is constant over [xi 1 , xi ] so that the work W done in moving the particle from xi 1 to xi is W ⇡ f (xi ) x. n X The total work done over [a, b] is then W ⇡ f (x⇤i ) x and if we let n ! 1, we get that the actual i=1
work done in moving the object from a to b is W = lim
n!1
n X
f (x⇤i )
x=
i=1
Z
b
f (x)dx a
Hooke’s Law states that the force required to maintain a spring stretch x units beyond its natural length is proportional to x, so f (x) = kx, where k is a positive constant called the spring constant. 12.2.1
Example
A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm?
115
12.2.2
Example
A 200-lb cable is 100 ft long and hangs vertically from the top of a tall building. How much work is required to lift the cable to the top of the building?
116
12.3
Average Value of a Function You should read section 6.5 of the textbook and do exercises 1, 3, 5
Recall that the average value of a finite set of numbers y1 , y2 , . . . , yn is yavg =
y1 + y2 + . . . + yn n
Suppose we want to calculate the average value of a continuous function y = f (x), a x b. We b a divide the interval [a, b] into n equal subintervals of width x = . Then we select a sample n point x⇤1 , . . . x⇤n from each subinterval and calculate the average of the numbers f (x⇤1 ), . . . , f (x⇤n ): f (x⇤1 ) + . . . + f (x⇤n ) n Since
x=
b
a n
, we can write n =
b
a and the average value becomes x
f (x⇤1 ) + . . . + f (x⇤n ) f (x⇤1 ) + . . . + f (x⇤n ) = b a n x n ⇤ ⇤ X f (x1 ) x + . . . + f (xn ) x 1 = = f (x⇤i ) x b a b a i=1 If we let n ! 1, then the average value of a function f (x) on the interval [a, b] is given by favg = lim
n!1
12.3.1
1 b
a
n X
f (x⇤i )
i=1
x=
1 b
a
Z
b
f (x)dx a
Example
Find the average value of the function f (x) = 1 + x2 on the interval [ 1, 2].
117
