applications of integration
DESCRIPTION
6. APPLICATIONS OF INTEGRATION. APPLICATIONS OF INTEGRATION. 6.2 Volumes. In this section, we will learn about: Using integration to find out the volume of a solid. VOLUMES. In trying to find the volume of a solid, we face the same type of problem as in finding areas. VOLUMES. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/1.jpg)
APPLICATIONS OF INTEGRATIONAPPLICATIONS OF INTEGRATION
6
![Page 2: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/2.jpg)
6.2
Volumes
APPLICATIONS OF INTEGRATION
In this section, we will learn about:
Using integration to find out
the volume of a solid.
![Page 3: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/3.jpg)
In trying to find the volume of a solid,
we face the same type of problem as
in finding areas.
VOLUMES
![Page 4: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/4.jpg)
We have an intuitive idea of what volume
means.
However, we must make this idea precise
by using calculus to give an exact definition
of volume.
VOLUMES
![Page 5: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/5.jpg)
We start with a simple type of solid
called a cylinder or, more precisely,
a right cylinder.
VOLUMES
![Page 6: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/6.jpg)
As illustrated, a cylinder is bounded by
a plane region B1, called the base, and
a congruent region B2 in a parallel plane.
The cylinder consists of all points on line segments perpendicular to the base and join B1 to B2.
CYLINDERS
![Page 7: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/7.jpg)
If the area of the base is A and the height of
the cylinder (the distance from B1 to B2) is h,
then the volume V of the cylinder is defined
as:
V = Ah
CYLINDERS
![Page 8: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/8.jpg)
In particular, if the base is a circle with
radius r, then the cylinder is a circular
cylinder with volume V = πr2h.
CYLINDERS
![Page 9: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/9.jpg)
If the base is a rectangle with length l and
width w, then the cylinder is a rectangular box
(also called a rectangular parallelepiped) with
volume V = lwh.
RECTANGULAR PARALLELEPIPEDS
![Page 10: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/10.jpg)
For a solid S that isn’t a cylinder, we first
‘cut’ S into pieces and approximate each
piece by a cylinder.
We estimate the volume of S by adding the volumes of the cylinders.
We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large.
IRREGULAR SOLIDS
![Page 11: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/11.jpg)
We start by intersecting S with a plane
and obtaining a plane region that is called
a cross-section of S.
IRREGULAR SOLIDS
![Page 12: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/12.jpg)
Let A(x) be the area of the cross-section of S
in a plane Px perpendicular to the x-axis and
passing through the point x, where a ≤ x ≤ b.
Think of slicing S with a knife through x and computing the area of this slice.
IRREGULAR SOLIDS
![Page 13: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/13.jpg)
The cross-sectional area A(x) will vary
as x increases from a to b.
IRREGULAR SOLIDS
![Page 14: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/14.jpg)
We divide S into n ‘slabs’ of equal width ∆x
using the planes Px1, Px2, . . . to slice the solid.
Think of slicing a loaf of bread.
IRREGULAR SOLIDS
![Page 15: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/15.jpg)
If we choose sample points xi* in [xi - 1, xi], we
can approximate the i th slab Si (the part of S
that lies between the planes and ) by a
cylinder with base area A(xi*) and ‘height’ ∆x.
1ixP ix
P
IRREGULAR SOLIDS
![Page 16: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/16.jpg)
The volume of this cylinder is A(xi*).
So, an approximation to our intuitive
conception of the volume of the i th slab Si
is: ( ) ( *) iV S A xi x
IRREGULAR SOLIDS
![Page 17: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/17.jpg)
Adding the volumes of these slabs, we get an
approximation to the total volume (that is,
what we think of intuitively as the volume):
This approximation appears to become better and better as n → ∞.
Think of the slices as becoming thinner and thinner.
1
( *)n
ii
V A x x
IRREGULAR SOLIDS
![Page 18: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/18.jpg)
Therefore, we define the volume as the limit
of these sums as n → ∞).
However, we recognize the limit of Riemann
sums as a definite integral and so we have
the following definition.
IRREGULAR SOLIDS
![Page 19: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/19.jpg)
Let S be a solid that lies between x = a
and x = b.
If the cross-sectional area of S in the plane Px,
through x and perpendicular to the x-axis,
is A(x), where A is a continuous function, then
the volume of S is:
1
lim ( *) ( )n b
i axi
V A x x A x dx
DEFINITION OF VOLUME
![Page 20: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/20.jpg)
When we use the volume formula
, it is important to remember
that A(x) is the area of a moving
cross-section obtained by slicing through
x perpendicular to the x-axis.
VOLUMES
( )b
aV A x dx
![Page 21: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/21.jpg)
Notice that, for a cylinder, the cross-sectional
area is constant: A(x) = A for all x.
So, our definition of volume gives:
This agrees with the formula V = Ah.
b
aV Adx A b a
VOLUMES
![Page 22: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/22.jpg)
Show that the volume of a sphere
of radius r is 34
3 .V r
Example 1SPHERES
![Page 23: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/23.jpg)
If we place the sphere so that its center is
at the origin, then the plane Px intersects
the sphere in a circle whose radius, from the
Pythagorean Theorem,
is:2 2y r x
Example 1SPHERES
![Page 24: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/24.jpg)
So, the cross-sectional area is:2 2 2( ) ( )A x y r x
Example 1SPHERES
![Page 25: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/25.jpg)
Using the definition of volume with a = -r and
b = r, we have:
(The integrand is
even.)
2 2
2 2
0
3 32 3
0
343
( )
2 ( )
2 23 3
r r
r r
r
r
V A x dx r x dx
r x dx
x rr x r
r
Example 1SPHERES
![Page 26: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/26.jpg)
The figure illustrates the definition of volume
when the solid is a sphere with radius r = 1.
From the example, we know that the volume of the sphere is
The slabs are circular cylinders, or disks.
43 4.18879
SPHERES
![Page 27: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/27.jpg)
The three parts show the geometric
interpretations of the Riemann sums
when n = 5, 10,
and 20 if we choose the sample points xi*
to be the midpoints .
2 2
1 1
( ) (1 )n n
i ii i
A x x x x
ix
SPHERES
![Page 28: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/28.jpg)
Notice that as we increase the number
of approximating cylinders, the corresponding
Riemann sums become closer to the true
volume.
SPHERES
![Page 29: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/29.jpg)
Find the volume of the solid obtained by
rotating about the x-axis the region under
the curve from 0 to 1.
Illustrate the definition of volume by sketching
a typical approximating cylinder.
y x
Example 2VOLUMES
![Page 30: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/30.jpg)
The region is shown in the first figure.
If we rotate about the x-axis, we get the solid
shown in the next figure. When we slice through the point x, we get a disk
with radius .
VOLUMES
x
Example 2
![Page 31: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/31.jpg)
The area of the cross-section is:
The volume of the approximating cylinder
(a disk with thickness ∆x) is:
2( ) ( )A x x x
( )A x x x x
Example 2VOLUMES
![Page 32: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/32.jpg)
The solid lies between x = 0 and x = 1.
So, its volume is:1
0
1
0
12
0
( )
2 2
V A x dx
xdx
x
VOLUMES Example 2
![Page 33: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/33.jpg)
Find the volume of the solid obtained
by rotating the region bounded by y = x3,
Y = 8, and x = 0 about the y-axis.
Example 3VOLUMES
![Page 34: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/34.jpg)
As the region is rotated about the y-axis, it
makes sense to slice the solid perpendicular
to the y-axis and thus to integrate with
respect to y.
Slicing at height y, we get a circular disk with radius x, where
VOLUMES Example 3
3x y
![Page 35: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/35.jpg)
So, the area of a cross-section through y is:
The volume of the approximating
cylinder is:
2 2 2/33( ) ( )A y x y y
2/3( )A y y y y
Example 3VOLUMES
![Page 36: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/36.jpg)
Since the solid lies between y = 0 and
y = 8, its volume is:
8
0
8 2 3
0
853 35
0
( )
96
5
V A y dy
y dy
y
Example 3VOLUMES
![Page 37: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/37.jpg)
The region R enclosed by the curves y = x
and y = x2 is rotated about the x-axis.
Find the volume of the resulting solid.
Example 4VOLUMES
![Page 38: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/38.jpg)
The curves y = x and y = x2 intersect at
the points (0, 0) and (1, 1).
The region between them, the solid of rotation, and cross-section perpendicular to the x-axis are shown.
VOLUMES Example 4
![Page 39: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/39.jpg)
A cross-section in the plane Px has the shape
of a washer (an annular ring) with inner
radius x2 and outer radius x.
Example 4VOLUMES
![Page 40: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/40.jpg)
Thus, we find the cross-sectional area by
subtracting the area of the inner circle from
the area of the outer circle:
2 2 2
2 4
( ) ( )
( )
A x x x
x x
VOLUMES Example 4
![Page 41: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/41.jpg)
Thus, we have:1
0
1 2 4
0
13 5
0
( )
( )
3 5
2
15
V A x dx
x x dx
x x
Example 4VOLUMES
![Page 42: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/42.jpg)
Find the volume of the solid obtained
by rotating the region in Example 4
about the line y = 2.
Example 5VOLUMES
![Page 43: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/43.jpg)
Again, the cross-section is a washer.
This time, though, the inner radius is 2 – x
and the outer radius is 2 – x2.
VOLUMES Example 5
![Page 44: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/44.jpg)
The cross-sectional area is:
2 2 2( ) (2 ) (2 )A x x x
Example 5VOLUMES
![Page 45: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/45.jpg)
So, the volume is:
1
0
1 22 2
0
1 4 2
0
15 3 2
0
( )
2 (2 )
5 4
85 4
5 3 2 5
V A x dx
x x dx
x x x dx
x x x
VOLUMES Example 5
![Page 46: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/46.jpg)
The solids in Examples 1–5 are all
called solids of revolution because
they are obtained by revolving a region
about a line.
SOLIDS OF REVOLUTION
![Page 47: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/47.jpg)
In general, we calculate the volume of
a solid of revolution by using the basic
defining formula
( ) orb d
a cV A x dx V A y dy
SOLIDS OF REVOLUTION
![Page 48: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/48.jpg)
We find the cross-sectional area
A(x) or A(y) in one of the following
two ways.
SOLIDS OF REVOLUTION
![Page 49: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/49.jpg)
If the cross-section is a disk, we find
the radius of the disk (in terms of x or y)
and use:
A = π(radius)2
WAY 1
![Page 50: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/50.jpg)
If the cross-section is a washer, we first find
the inner radius rin and outer radius rout from
a sketch. Then, we subtract the area of the inner disk from
the area of the outer disk to obtain: A = π(outer radius)2 – π(outer radius)2
WAY 2
![Page 51: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/51.jpg)
Find the volume of the solid obtained
by rotating the region in Example 4
about the line x = -1.
Example 6SOLIDS OF REVOLUTION
![Page 52: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/52.jpg)
The figure shows the horizontal cross-section.
It is a washer with inner radius 1 + y and
outer radius
Example 6
1 .y
SOLIDS OF REVOLUTION
![Page 53: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/53.jpg)
So, the cross-sectional area is:
2 2
2 2
( ) (outer radius) (inner radius)
1 1
A y
y y
Example 6SOLIDS OF REVOLUTION
![Page 54: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/54.jpg)
The volume is:
1
0
21 2
0
1 2
0
13 2 32
0
( )
1 1
2
4
3 2 3 2
V A y dy
y y dy
y y y dy
y y y
Example 6SOLIDS OF REVOLUTION
![Page 55: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/55.jpg)
In the following examples, we find
the volumes of three solids that are
not solids of revolution.
VOLUMES
![Page 56: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/56.jpg)
The figure shows a solid with a circular base
of radius 1. Parallel cross-sections
perpendicular to the base are equilateral
triangles.
Find the volume of the solid.
Example 7VOLUMES
![Page 57: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/57.jpg)
Let’s take the circle to be x2 + y2 = 1.
The solid, its base, and a typical cross-section
at a distance x from the origin are shown.
Example 7VOLUMES
![Page 58: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/58.jpg)
As B lies on the circle, we have
So, the base of the triangle ABC is |AB| =
21y x
22 1 x
Example 7VOLUMES
![Page 59: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/59.jpg)
Since the triangle is equilateral, we see
that its height is 23 3 1y x
VOLUMES Example 7
![Page 60: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/60.jpg)
Thus, the cross-sectional area is :
VOLUMES Example 7
2 212
2
( ) 2 1 3 1
3(1 )
A x x x
x
![Page 61: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/61.jpg)
The volume of the solid is:
1
1
1 12 2
1 0
13
0
( )
3(1 ) 2 3(1 )
4 32 3
3 3
V A x dx
x dx x dx
xx
Example 7VOLUMES
![Page 62: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/62.jpg)
Find the volume of a pyramid
whose base is a square with side L
and whose height is h.
Example 8VOLUMES
![Page 63: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/63.jpg)
We place the origin O at the vertex
of the pyramid and the x-axis along its
central axis.
Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square with side of length s.
VOLUMES Example 8
![Page 64: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/64.jpg)
We can express s in terms of x by observing
from the similar triangles that
Therefore, s = Lx/h
Another method is to observe that the line OP has slope L/(2h)
So, its equation is y = Lx/(2h)
2
2
x s s
h L L
Example 8VOLUMES
![Page 65: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/65.jpg)
Thus, the cross-sectional area is:VOLUMES Example 8
22 2
2( )
LA x s x
h
![Page 66: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/66.jpg)
The pyramid lies between x = 0 and x = h.
So, its volume is:0
22
20
2 3 2
20
( )
3 3
h
h
h
V A x dx
Lx dx
h
L x L h
h
Example 8VOLUMES
![Page 67: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/67.jpg)
In the example, we didn’t need to place
the vertex of the pyramid at the origin.
We did so merely to make the equations simple.
NOTE
![Page 68: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/68.jpg)
Instead, if we had placed the center of
the base at the origin and the vertex on
the positive y-axis, as in the figure, you can
verify that we would have
obtained the integral:2
220
2
( )
3
h LV h y dy
h
L h
NOTE
![Page 69: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/69.jpg)
A wedge is cut out of a circular cylinder of
radius 4 by two planes. One plane is
perpendicular to the axis of the cylinder.
The other intersects the first at an angle of 30°
along a diameter of the cylinder.
Find the volume of the wedge.
Example 9VOLUMES
![Page 70: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/70.jpg)
If we place the x-axis along the diameter
where the planes meet, then the base of
the solid is a semicircle
with equation
-4 ≤ x ≤ 4 216 ,y x
VOLUMES Example 9
![Page 71: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/71.jpg)
A cross-section perpendicular to the x-axis at
a distance x from the origin is a triangle ABC,
whose base is and whose height
is |BC| = y tan 30° =
216y x 216 3.x
Example 9VOLUMES
![Page 72: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/72.jpg)
Thus, the cross-sectional area is:
2 212
2
1( ) 16 16
3
16
2 3
A x x x
x
VOLUMES Example 9
![Page 73: APPLICATIONS OF INTEGRATION](https://reader036.vdocument.in/reader036/viewer/2022081515/56812bc0550346895d9002a3/html5/thumbnails/73.jpg)
The volume is:
4
4
24 4 2
4 0
43
0
( )
16 116
2 3 3
1 12816
33 3 3
V A x dx
xdx x dx
xx
Example 9VOLUMES