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**APPLICATIONS OF INTEGRATION**

6 APPLICATIONS OF INTEGRATION

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**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.

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**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.

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**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.

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**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.

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**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.

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**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

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**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.

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**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.

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**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.

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**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.

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**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.

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**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.

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**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.

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**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.

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**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:

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**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.

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**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.

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**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:

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**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.

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**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.

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**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

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**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:

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**So, the cross-sectional area is:**

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

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**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.)

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**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.

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**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 .

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**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.

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**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.

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**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

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**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:

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**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:

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**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.

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**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

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**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:

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**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:

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**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.

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**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.

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**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.

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**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:

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VOLUMES Example 4 Thus, we have:

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**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.

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**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.

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**The cross-sectional area is:**

VOLUMES Example 5 The cross-sectional area is:

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VOLUMES Example 5 So, the volume is:

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**The solids in Examples 1–5 are all called solids of revolution because **

they are obtained by revolving a region about a line.

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**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

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**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.

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**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

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**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

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**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.

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**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

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**So, the cross-sectional area is:**

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

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SOLIDS OF REVOLUTION Example 6 The volume is:

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**In the following examples, we find **

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

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**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.

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**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.

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**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| =

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**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

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**Thus, the cross-sectional area is :**

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

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**The volume of the solid is:**

VOLUMES Example 7 The volume of the solid is:

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**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.

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**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.

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**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)

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**Thus, the cross-sectional area is:**

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

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**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:

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**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.

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**Instead, if we had placed the center of **

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:

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**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.

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**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

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**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° =

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**Thus, the cross-sectional area is:**

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

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VOLUMES Example 9 The volume is:

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