Figure 14.2:4 shows the area accumulated from ato x: (z/h)M = (z/h)2LM = (z/h)2b Therefore, the volume is h (z/h)2bdz = bz3/3h2 h= bh/3 0 0 4B-4 The slice perpendicular to the xz-plane are right triangles with base of length x and height z = 2x. 1. g(.l) h x If a function I is continuous and f(x) 0 on [a, h], then, by Theo- rem (5.19), the area of the region under the graph of f from a to b is given by the definite integral f(x) dx. Applications of Integration 9.1 Area between ves cur We have seen how integration can be used to find an area between a curve and the x-axis. APPLICATION OF INTEGRATION Measure of Area Area is a measure of the surface of a two-dimensional region. We are familiar with calculating the area of regions that have basic geometrical shapes such as rectangles, squares, triangles, circles and trapezoids. A … APPLICATIONS OF INTEGRATION I YEAR B.Tech . E. Solutions to 18.01 Exercises 4. lated area, length, volume, and surface area, have. Sol: We know that the volume of the solid generated by the revolution of the area bounded by the curve , the and the lines is given by Now, given curve Required volume is given by This means that we can apply Duhamel’s Principle to finding integral formulas of many geometric quantities. Set up the definite integral, and integrate. In this section we shall consider the 3 0 4 CHAPTER 6 APPLICATIONS OF THE DEFINITE INTEGRAL 6.1 AREA FIGURE 6.1 Y a \. Therefore the area of a slice is x2. Sketch the area and determine the axis of revolution, (this determines the variable of integration) 2. Revolution Applet. 1. a) Set up the integral for surface area using integration dx b) Set up the integral for surface area using integration dy With very little change we can find some areas between curves; indeed, the area between a curve and the x-axis may be interpreted as the area between the curve and a second “curve” with equation y = 0. Problems on Volume of solid of Revolution 1) Find the volume of the solid that result when the region enclosed by the curve is revolved about the . In all the volume is a a (h2/4)dx = (a 2 − x 2 )dx = 4a 3 /3 −a −a Sketch the cross-section, (disk, shell, washer) and determine the appropriate formula. Applications of Integrals Brief Review The application of Integrals we will focus on this week is area and volume. One very useful application of Integration is finding the area and volume of “curved” figures, that we couldn’t typically get without using Calculus. Title: APPLICATIONS OF INTEGRATION Author: Y.P.REDDY Subject: I YEAR B.Tech Created Date: 4/18/2011 8:01:01 AM For example, the accumu-lated area used in the second half of the Fundamental Theorem of Integral Calculus is additive. APPLICATIONS OF INTEGRATION 4G-5 Find the area of y = x2, 0 ≤ x ≤ 4 revolved around the y-axis. Set up the definite integral, and integrate. Sketch the cross-section, (disk, shell, washer) and determine the appropriate formula. 4G-6 Find the area of the astroid x2/3 +y2/3 = a2/3 revolved around the x-axis. Finding volume of a solid of revolution using a disc method. UNIT-4 APPLICATIONS OF INTEGRATION ... Find the volume of the solid that result when the region enclosed by the curve ... Sol: We know that the volume of the solid generated by the revolution of the area bounded by the curve , the and the lines is given by Now, given curve Required volume is given by 3. 1. lated area, length, volume, and surface area, have. Determine the boundaries of the solid, 4. Sketch the area and determine the axis of revolution, (this determines the variable of integration) 2. 1. Determine the boundaries of the solid, 4. 4G-7 Conside the torus of Problem 4C-1. Finding volume of a solid of revolution using a disc method. In these two videos, the 17. The left boundary will be x = O and the fight boundary will be x = 4 The upper boundary will be y 2 = 4x The 2-dimensional area of the region would be the integral Area of circle Volume (radius) (ftnction) dx sum of vertical discs') Volume and Area from Integration a) Since the region is rotated around the x-axis, we'll use 'vertical partitions'. y-axis orientation is RIGHT-LEFT and your limits come from y. Applications of integration a/2 y = 3x 4B-6 If the hypotenuse of an isoceles right triangle has length h, then its area is h2/4. Figure 14.2:4 shows the area accumulated from ato x: For example, the accumu-lated area used in the second half of the Fundamental Theorem of Integral Calculus is additive. In the following video the narrator walks trough the steps of setting up a volume integration (14.0)(16.0). The endpoints of the slice in the xy-plane are y = ± √ a2 − x2, so h = 2 √ a2 − x2. 4. APPLICATION OF INTEGRALS 361 Example 1 Find the area enclosed by the circle x2 + y2 = a2. Sometimes the same volume problem can be solved in two different ways (14.0)(16.0). AREA BETWEEN TWO CURVES: You can do this with an x-axis orientation in which case your limits of integration come from the x-axis and you do TOP-BOTTOM. 3. Khan Academy Solids of Revolution (10:04) . This means that we can apply Duhamel’s Principle to finding integral formulas of many geometric quantities.