# APPLICATIONS OF INTEGRATION

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APPLICATIONS OF INTEGRATION
6 APPLICATIONS OF INTEGRATION

6.2 Volumes In this section, we will learn about:
APPLICATIONS OF INTEGRATION 6.2 Volumes In this section, we will learn about: Using integration to find out the volume of a solid.

In trying to find 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.

We have an intuitive idea of what volume means.
VOLUMES 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 We start with a simple type of solid called a cylinder or, more precisely, a right cylinder.

As illustrated, a cylinder is bounded by
CYLINDERS 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.

If the area of the base is A and the height of
CYLINDERS 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

In particular, if the base is a circle with
CYLINDERS In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with volume V = πr2h.

RECTANGULAR PARALLELEPIPEDS
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.

For a solid S that isn’t a cylinder, we first
IRREGULAR SOLIDS 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.

We start by intersecting S with a plane
IRREGULAR SOLIDS We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of S.

Let A(x) be the area of the cross-section of S
IRREGULAR SOLIDS 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.

The cross-sectional area A(x) will vary as x increases from a to b.
IRREGULAR SOLIDS The cross-sectional area A(x) will vary as x increases from a to b.

We divide S into n ‘slabs’ of equal width ∆x
IRREGULAR SOLIDS 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.

If we choose sample points xi* in [xi - 1, xi], we
IRREGULAR SOLIDS 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.

The volume of this cylinder is A(xi*).
IRREGULAR SOLIDS 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:

Adding the volumes of these slabs, we get an
IRREGULAR SOLIDS 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.

Therefore, we define the volume as the limit of these sums as n → ∞).
IRREGULAR SOLIDS 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.

Let S be a solid that lies between x = a and x = b.
DEFINITION OF VOLUME 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:

When we use the volume formula , it is important to remember
VOLUMES 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.

Notice that, for a cylinder, the cross-sectional
VOLUMES 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.

Show that the volume of a sphere of radius r is
SPHERES Example 1 Show that the volume of a sphere of radius r is

If we place the sphere so that its center is
Example 1 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:

So, the cross-sectional area is:
SPHERES Example 1 So, the cross-sectional area is:

Using the definition of volume with a = -r and b = r, we have:
SPHERES Example 1 Using the definition of volume with a = -r and b = r, we have: (The integrand is even.)

The figure illustrates the definition of volume
SPHERES 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.

The three parts show the geometric interpretations of the Riemann sums
SPHERES 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 .

Notice that as we increase the number
SPHERES Notice that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume.

Find the volume of the solid obtained by
VOLUMES Example 2 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.

The region is shown in the first figure.
VOLUMES Example 2 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

The area of the cross-section is:
VOLUMES Example 2 The area of the cross-section is: The volume of the approximating cylinder (a disk with thickness ∆x) is:

The solid lies between x = 0 and x = 1. So, its volume is:
VOLUMES Example 2 The solid lies between x = 0 and x = 1. So, its volume is:

Find the volume of the solid obtained
VOLUMES Example 3 Find the volume of the solid obtained by rotating the region bounded by y = x3, y = 8, and x = 0 about the y-axis.

As the region is rotated about the y-axis, it
VOLUMES Example 3 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

So, the area of a cross-section through y is:
VOLUMES Example 3 So, the area of a cross-section through y is: The volume of the approximating cylinder is:

Since the solid lies between y = 0 and y = 8, its volume is:
VOLUMES Example 3 Since the solid lies between y = 0 and y = 8, its volume is:

The region R enclosed by the curves y = x
VOLUMES Example 4 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.

The curves y = x and y = x2 intersect at the points (0, 0) and (1, 1).
VOLUMES Example 4 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.

A cross-section in the plane Px has the shape
VOLUMES Example 4 A cross-section in the plane Px has the shape of a washer (an annular ring) with inner radius x2 and outer radius x.

Thus, we find the cross-sectional area by
VOLUMES Example 4 Thus, we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle:

VOLUMES Example 4 Thus, we have:

Find the volume of the solid obtained
VOLUMES Example 5 Find the volume of the solid obtained by rotating the region in Example 4 about the line y = 2.

Again, the cross-section is a washer.
VOLUMES Example 5 Again, the cross-section is a washer. This time, though, the inner radius is 2 – x and the outer radius is 2 – x2.

The cross-sectional area is:
VOLUMES Example 5 The cross-sectional area is:

VOLUMES Example 5 So, the volume is:

The solids in Examples 1–5 are all called solids of revolution because
they are obtained by revolving a region about a line.

In general, we calculate the volume of
SOLIDS OF REVOLUTION In general, we calculate the volume of a solid of revolution by using the basic defining formula

We find the cross-sectional area A(x) or A(y) in one of the following
SOLIDS OF REVOLUTION We find the cross-sectional area A(x) or A(y) in one of the following two ways.

If the cross-section is a disk, we find
WAY 1 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

If the cross-section is a washer, we first find
WAY 2 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

Find the volume of the solid obtained
SOLIDS OF REVOLUTION Example 6 Find the volume of the solid obtained by rotating the region in Example 4 about the line x = -1.

The figure shows the horizontal cross-section.
SOLIDS OF REVOLUTION Example 6 The figure shows the horizontal cross-section. It is a washer with inner radius 1 + y and outer radius

So, the cross-sectional area is:
SOLIDS OF REVOLUTION Example 6 So, the cross-sectional area is:

SOLIDS OF REVOLUTION Example 6 The volume is:

In the following examples, we find
VOLUMES In the following examples, we find the volumes of three solids that are not solids of revolution.

The figure shows a solid with a circular base
VOLUMES Example 7 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.

Let’s take the circle to be x2 + y2 = 1.
VOLUMES Example 7 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.

As B lies on the circle, we have So, the base of the triangle ABC is
VOLUMES Example 7 As B lies on the circle, we have So, the base of the triangle ABC is |AB| =

Since the triangle is equilateral, we see that its height is
VOLUMES Example 7 Since the triangle is equilateral, we see that its height is

Thus, the cross-sectional area is :
VOLUMES Example 7 Thus, the cross-sectional area is :

The volume of the solid is:
VOLUMES Example 7 The volume of the solid is:

Find the volume of a pyramid whose base is a square with side L
VOLUMES Example 8 Find the volume of a pyramid whose base is a square with side L and whose height is h.

We place the origin O at the vertex
VOLUMES Example 8 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.

We can express s in terms of x by observing
VOLUMES Example 8 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)

Thus, the cross-sectional area is:
VOLUMES Example 8 Thus, the cross-sectional area is:

The pyramid lies between x = 0 and x = h. So, its volume is:
VOLUMES Example 8 The pyramid lies between x = 0 and x = h. So, its volume is:

In the example, we didn’t need to place
NOTE 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 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:

A wedge is cut out of a circular cylinder of
VOLUMES Example 9 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.

If we place the x-axis along the diameter
VOLUMES Example 9 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

A cross-section perpendicular to the x-axis at
VOLUMES Example 9 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° =

Thus, the cross-sectional area is:
VOLUMES Example 9 Thus, the cross-sectional area is:

VOLUMES Example 9 The volume is: