Sketch the cross-section, (disk, shell, washer) and determine the appropriate formula. The endpoints of the slice in the xy-plane are y = ± √ a2 − x2, so h = 2 √ a2 − x2. Volume and Area from Integration a) Since the region is rotated around the x-axis, we'll use 'vertical partitions'. 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. a) Set up the integral for surface area using integration dx b) Set up the integral for surface area using integration dy Set up the definite integral, and integrate. Figure 14.2:4 shows the area accumulated from ato x: 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. (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. Set up the definite integral, and integrate. APPLICATIONS OF INTEGRATION 4G-5 Find the area of y = x2, 0 ≤ x ≤ 4 revolved around the y-axis. Determine the boundaries of the solid, 4. 1. 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. For example, the accumu-lated area used in the second half of the Fundamental Theorem of Integral Calculus is additive. Sketch the cross-section, (disk, shell, washer) and determine the appropriate formula. 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. We are familiar with calculating the area of regions that have basic geometrical shapes such as rectangles, squares, triangles, circles and trapezoids. Determine the boundaries of the solid, 4. Khan Academy Solids of Revolution (10:04) . Finding volume of a solid of revolution using a disc method. Applications of Integrals Brief Review The application of Integrals we will focus on this week is area and volume. 1. 4. Sometimes the same volume problem can be solved in two different ways (14.0)(16.0). A … One very useful application of Integration is finding the area and volume of “curved” figures, that we couldn’t typically get without using Calculus. y-axis orientation is RIGHT-LEFT and your limits come from y. Revolution Applet. In the following video the narrator walks trough the steps of setting up a volume integration (14.0)(16.0). 4G-6 Find the area of the astroid x2/3 +y2/3 = a2/3 revolved around the x-axis. 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. In all the volume is a a (h2/4)dx = (a 2 − x 2 )dx = 4a 3 /3 −a −a Sketch the area and determine the axis of revolution, (this determines the variable of integration) 2. 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. 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') Therefore the area of a slice is x2. 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. 3. In this section we shall consider the 17. This means that we can apply Duhamel’s Principle to finding integral formulas of many geometric quantities. 3 0 4 CHAPTER 6 APPLICATIONS OF THE DEFINITE INTEGRAL 6.1 AREA FIGURE 6.1 Y a \. 1. In these two videos, the This means that we can apply Duhamel’s Principle to finding integral formulas of many geometric quantities. 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 APPLICATION OF INTEGRATION Measure of Area Area is a measure of the surface of a two-dimensional region.