1
The area between two curves
The Volume of the Solid of revolution (by slicing)
1. AREA BETWEEN the CURVES
𝑑𝐴 = {(
𝑏
𝑜𝑢𝑡𝑒𝑟
𝑖𝑛𝑛𝑒𝑟
)−(
)} dx
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑏
𝑑
𝐴 = ∫ 𝑑𝐴 = ∫[𝑦1 (𝑥) − 𝑦2 (𝑥)]𝑑𝑥
𝑎
𝑑
𝐴 = ∫ 𝑑𝐴 = ∫[𝑥1 (𝑦) − 𝑥2 (𝑦)]𝑑𝑦
𝑎
𝑐
2
𝑐
EX: Determine the area of the region bounded by y = 2x +10 and y = 4x +16 between x = – 2 and x = 5
𝑜𝑢𝑡𝑒𝑟
𝑖𝑛𝑛𝑒𝑟
𝑏
𝑏
𝐴 = ∫𝑎 𝑑𝐴 = ∫𝑎 [(𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) − (
)] 𝑑𝑥
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
2
EX: Determine the area of the region enclosed by y = sin x and y = cos x and the y -axis for 0 ≤ 𝑥 ≤
1
EX: Determine the area of the enclosed area by 𝑥 = 2 𝑦 2 − 3 and 𝑦 = 𝑥 − 1
Intersection: (-1,-2) and (5,4).
1
THE SAME: Determine the area of the enclosed area by 𝑥 = 2 𝑦 2 − 3 and 𝑦 = 𝑥 − 1
So, in this last example we’ve seen a case where we could
use either method to find the area.
However, the second was definitely easier.
𝜋
2
3
EX: Find area
Intersection points are:
y=-1
y=3
3
𝐴 = ∫[(−𝑦 2 + 10) − (𝑦 − 2)2 ]𝑑𝑦
−1
3
= ∫[−2𝑦 2 + 4𝑦 + 6]𝑑𝑦
−1
3
2
= [− 𝑦 3 + 2𝑦 2 + 6𝑦]
3
−1
64
𝐴=
3
Volume of REVOLUTION
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▪
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Find the Volume of revolution using the disk method
Find the volume of revolution using the washer method
Find the volume of revolution using the shell method
Find the volume of a solid with known cross sections
Area is only one of the applications of integration. We can add up representative volumes in the same
way we add up representative rectangles. When we are measuring volumes of revolution, we can slice
representative disks or washers.
DISK METHOD
𝑏
𝑉 = ∫ 𝑑𝑉
𝑑𝑉 = 𝜋𝑟2 𝑑𝑥
𝑎
𝑑𝑉 = 𝜋[𝑓(𝑥)]2 𝑑𝑥
𝑑𝑉 = 𝜋[𝑓(𝑥) + 𝑘]2 𝑑𝑥
𝑑𝑉 = 𝜋[𝑓(𝑥) − 𝑘]2 𝑑𝑥
𝑑𝑉 = 𝜋[𝑘 − 𝑓(𝑥)]2 𝑑𝑥
4
𝑑
𝑉 = ∫ 𝑑𝑉
𝑐
WASHER METHOD
A solid obtained by revolving a region around a line.
5
6
Volumes by Cylindrical Shells
7
8
o
o
o
o
Procedure: volume by slicing sketch the solid and a typical
cross section find a formula for the area, A(x), of the
cross section find limits of integration integrate
A(x) to get volume
Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to
the x-axis are
a) squares
𝑦 = √4 − 𝑥 2
x2 + y 2 = 4
length of a side is : 2√4 − 𝑥2
𝐴 = 𝑎2
𝑑𝑉 = 𝐴 𝑑𝑥
2
𝑉 = 4 ∫(4 − 𝑥 2 )𝑑𝑥 =
−2
128
3
b) Equilateral triangles
𝑦 = √4 − 𝑥 2
x2 + y 2 = 4
𝐴=
1
𝑎 2 √3 2
𝑎 √𝑎2 − ( ) =
𝑎 = √3(4 − 𝑥 2 )
2
2
4
2
𝑉 = ∫ √3(4 − 𝑥 2 )𝑑𝑥 =
−2
c) semicircles
√3
≈ 18.475
𝑦 = √4 − 𝑥 2
x2 + y 2 = 4
𝐴=
32
1
𝑎 2 1
4 − 𝑥2
𝜋 ( ) = 𝜋 𝑎2 = 𝜋
2
2
8
2
𝑑𝑉 = 𝐴 𝑑𝑥
2
𝑉 = ∫𝜋
−2
4 − 𝑥2
16𝜋
𝑑𝑥 =
≈ 16.755
2
3
9
d) Isosceles right triangles
x2 + y 2 = 4
𝑦 = √4 − 𝑥 2
𝑎
1
𝑎2
2
𝐴= 𝑎
=
= 4 − 𝑥2
2 tan 𝜋⁄4
4
𝑑𝑉 = 𝐴 𝑑𝑥
2
𝑉 = ∫(4 − 𝑥 2 ) 𝑑𝑥 =
−2
32
≈ 10.667
3
PRACTICE:
1. Find the volume of the solid generated by revolving about the x-axis the region bounded by the graph
of 𝑓(𝑥) = √𝑥 − 1
the x-axis, and the line x = 5. Draw a sketch. 1. ANS: 8π
2. Find the volume of the solid generated by revolving about the x -axis the region bounded by the graph
of 𝑦 = √cos 𝑥
where 0 ≤ 𝑥 ≤
𝜋
2
the x-axis, and the y-axis. Draw a sketch. 2. ANS: π
3. Find the volume of the solid generated by revolving about the y-axis the region in the first quadrant
bounded by the graph of y = x2, the y-axis, and the line y = 6. Draw a sketch. 3. ANS: 18 π
4. Using a calculator, find the volume of the solid generated by revolving about the line y = 8 the region
bounded by the graph of y = x 2 + 4, the line y = 8. Draw a sketch. ANS: 512/15 π
5. Using a calculator, find the volume of the solid generated by revolving about the line y = –3 the region
bounded by the graph of y = ex, the y-axis, the lines x = ln 2 and y = – 3. Sketch. 5. ANS: 13.7383 π
6. Using the Washer, find the volume of the solid generated by revolving the region bounded by y = x 3
and y = x in the first quadrant about the x-axis. Draw a sketch. Method (just a fancy name – use sketch
and common sense!!! instead of given boundaries, you have to find it as intersection of two curves and
then use sketch to subtract one volume from another ) 6. ANS: 4π/21
7. Using the Washer Method and a calculator, find the volume of the solid generated by revolving the
region bounded by y = x 3 and y = x about the line y = 2. Draw a sketch. 7. ANS: 17π/21
8. Using the Washer Method and a calculator, find the volume of the solid generated by revolving the
region bounded by y = x2 and x = y2 about the y-axis. Draw a sketch. 8. ANS: 3π/10
10
AGAIN PRACTICE:
1. The base of a solid is the region enclosed by the ellipse
𝑥2
4
+
𝑦2
25
=1
. The cross sections are perpendicular to the x-axis and are isosceles
right triangles whose hypotenuses are on the ellipse. Find the volume
of the solid. 1. ANS: V = 200/3
2. The base of a solid is the region enclosed by a triangle whose vertices are (0, 0), (4, 0) and (0, 2). The
cross sections are semicircles perpendicular to the x-axis. Using a calculator, find the volume of the solid.
2. ANS: V = 2.094
3. Find the volume of the solid whose base is
the region bounded by the lines x + 4 y = 4, x
= 0, and y = 0, if the cross sections taken
perpendicular to the x-axis are semicircles.
3. ANS: V = π/6
4. The base of a solid is the region in the first quadrant bounded by the y-axis, the graph of y = arctanx,
the horizontal line y = 3, and the vertical line x = 1. For this solid, each cross section perpendicular to the
x-axis is a square. What is the volume of the solid? 4. ANS: V = ∫₀¹ (3 - arctan(x))² dx = 6.61233
5. A solid has its base is the region bounded by the lines x + 2y
= 6, x = 0 and y = 0 and the cross sections taken perpendicular
to x-axis are circles. Find the volume the solid.
5. ANS: 9/2 π
6. A solid has its base is the region bounded by the lines x + y = 4, x = 0 and y = 0 and the cross section is
perpendicular to the x-axis are equilateral triangles. Find its volume. 6. ANS: V = 16√3/3