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APPLICATIONS OF INTEGRATION 6

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6.2 Volumes APPLICATIONS OF INTEGRATION 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, we face the same type of problem as in finding areas. VOLUMES

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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

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

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As illustrated, a cylinder is bounded by a plane region B 1, called the base, and a congruent region B 2 in a parallel plane. The cylinder consists of all points on line segments perpendicular to the base and join B 1 to B 2. CYLINDERS

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If the area of the base is A and the height of the cylinder (the distance from B 1 to B 2 ) is h, then the volume V of the cylinder is defined as: V = Ah CYLINDERS

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In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with volume V = πr 2 h. CYLINDERS

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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 = l wh. RECTANGULAR PARALLELEPIPEDS

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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

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

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Let A(x) be the area of the cross-section of S in a plane P x 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

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

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We divide S into n ‘slabs’ of equal width ∆x using the planes P x1, P x2,... to slice the solid. Think of slicing a loaf of bread. IRREGULAR SOLIDS

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If we choose sample points x i * in [x i - 1, x i ], we can approximate the i th slab S i (the part of S that lies between the planes and ) by a cylinder with base area A(x i *) and ‘height’ ∆x. IRREGULAR SOLIDS

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The volume of this cylinder is A(x i *). So, an approximation to our intuitive conception of the volume of the i th slab S i is: IRREGULAR SOLIDS

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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

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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

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Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plane P x, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is: DEFINITION OF VOLUME

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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

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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. VOLUMES

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

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If we place the sphere so that its center is at the origin, then the plane P x intersects the sphere in a circle whose radius, from the Pythagorean Theorem, is: Example 1 SPHERES

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

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

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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

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The three parts show the geometric interpretations of the Riemann sums when n = 5, 10, and 20 if we choose the sample points x i * to be the midpoints. SPHERES

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

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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. Example 2 VOLUMES

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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

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

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

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

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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

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

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

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The region R enclosed by the curves y = x and y = x 2 is rotated about the x-axis. Find the volume of the resulting solid. Example 4 VOLUMES

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The curves y = x and y = x 2 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

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A cross-section in the plane P x has the shape of a washer (an annular ring) with inner radius x 2 and outer radius x. Example 4 VOLUMES

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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

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Thus, we have: Example 4 VOLUMES

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

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

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The cross-sectional area is: Example 5 VOLUMES

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So, the volume is: VOLUMES Example 5

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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. SOLIDS OF REVOLUTION

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

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

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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

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If the cross-section is a washer, we first find the inner radius r in and outer radius r out 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

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

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

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

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The volume is: Example 6 SOLIDS OF REVOLUTION

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

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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 7 VOLUMES

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Let’s take the circle to be x 2 + y 2 = 1. The solid, its base, and a typical cross-section at a distance x from the origin are shown. Example 7 VOLUMES

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

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

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

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The volume of the solid is: Example 7 VOLUMES

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

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We place the origin O at the vertex of the pyramid and the x-axis along its central axis. Any plane P x 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

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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) Example 8 VOLUMES

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

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

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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

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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: NOTE

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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 9 VOLUMES

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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 VOLUMES Example 9

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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 ° = Example 9 VOLUMES

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

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The volume is: Example 9 VOLUMES

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