1 / 73

740 likes | 865 Views

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.

Download Presentation
## APPLICATIONS OF INTEGRATION

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**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**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.**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.**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**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.**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.**IRREGULAR SOLIDS**We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of S.**IRREGULAR SOLIDS**Let A(x) be the area of the cross-section of S in a plane Pxperpendicular to the x-axis and passing through the point x, where a≤x≤ b. • Think of slicing Swith a knife through x and computing the area of this slice.**IRREGULAR SOLIDS**The cross-sectional area A(x) will vary as x increases from a to b.**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.**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.**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:**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.**IRREGULAR SOLIDS**Therefore, we definethe 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.**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:**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.**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.**SPHERES**Example 1 Show that the volume of a sphere of radius r is**SPHERES**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:**SPHERES**Example 1 So, the cross-sectional area is:**SPHERES**Example 1 Using the definition of volume with a = -r and b = r, we have: (The integrand is even.)**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.**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 .**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**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. VOLUMES Example 2**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 .**VOLUMES**Example 2 The area of the cross-section is: The volume of the approximating cylinder (a disk with thickness ∆x) is:**VOLUMES**Example 2 The solid lies between x =0 and x = 1. So, its volume is:**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.**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**VOLUMES**Example 3 So, the area of a cross-section through y is: The volume of the approximating cylinder is:**VOLUMES**Example 3 Since the solid lies between y =0 and y =8, its volume is:**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.**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.**VOLUMES**Example 4 A cross-section in the plane Pxhas the shape of a washer(an annular ring) with inner radius x2 and outer radius x.**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:**VOLUMES**Example 5 Find the volume of the solid obtained by rotating the region in Example 4 about the line y =2.**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.**VOLUMES**Example 5 The cross-sectional area is:**VOLUMES**Example 5 So, the volume is:**SOLIDS OF REVOLUTION**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**In general, we calculate the volume of a solid of revolution by using the basic defining formula**SOLIDS OF REVOLUTION**We find the cross-sectional area A(x) or A(y) in one of the following two ways.**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**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

More Related