APPLICATIONS OF INTEGRATION 6 APPLICATIONS OF INTEGRATION

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

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.

We start with a simple type of solid VOLUMES We start with a simple type of solid called a cylinder or, more precisely, a right cylinder.

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.)

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: