CopyofCalculusIInotes - CalculusIInotes Chapter7. twoCurves . rectangleorcircle, easilydefinedshape?

CopyofCalculusIInotes - CalculusIInotes Chapter7. twoCurves...

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Unformatted text preview: Calculus II notes Chapter 7.1 Area of a Region between two Curves Calculus is the mathematics of change. We can apply basic algebra to find the area of a rectangle or circle, but what about a three dimensional object that does not have a constant or easily defined shape? It is easy to measure the circumference of a pie wedge using basic algebra but what about a curve that changes direction erratically? Find the area under a curve? How do we maximize or minimize spending and profit? How do we calculate instantaneous rates of acceleration and velocity? How do we find the centroid or the volume of an oddly shaped object? Limits are especially useful in general in calculus for finding the area under a function. We can also extend this idea to the area between two curves! ex) RECALL! Our study of limits and area. Draw a picture of area under a curve. How can we estimate the area under a curve? Maybe with rectangles! What happens as the width of the rectangles shrinks? The area becomes more accurate, less error. What happens as the limit of the width approaches zero? The area becomes exact. Now, what happens as we subtract the area of a function just underneath it, what are we left with? Area Between Two Curves​: If f and g are continuous on [a,b] and ​ for all x in [a,b] then the area of the region bounded by the graphs of fand g and the vertical lines x=a and x=b is ­ Recall that area would be negative if we had a single function, and it was under the x­axis. We took the negative of that value to get the area, since area was positive ­ When we subtract the two functions, we only get their height, which will be positive regardless of position on the y­axis, so the area is always positive! ­ Note! A vertical rectangle of width ​ implies integration with respect to x, and a horizontal rectangle of width ​ implies integration with respect to y ex) Find the area of the region bounded by the graphs of ex) Find the area of the region bounded by the graphs of ​ between x=1 and x=2 ex) Find the area of the region bounded by the graphs of ​, where do they intersect? Note that they switch positions! For the area equation to be correct our function but which is which at what intervals? ex) Find the area of the region bounded by the graphs of ­sinx and cosx, but just for a single region. Note that they cross over each other infinitely, creating an infinite area if we were to take all of them. Where do they intersect? Now we can also integrate with respect to the y­axis, taking equations and solving for x! Then integrating with a ​. Note that the greater function is the function that is higher up on the x­axis when deciding which function goes where in f(x)­g(x). For example, take the following two problems, we can integrate either with respect to y or x, but which is easier? Draw pictures. Find the areas, both ways! #1​ The area between #2​ The area between ​ and ​ and ​. Homework Chap 7.1 # 1,3,5,17­29 odd, 37­41 odd,71,78, Bonus 82 Chapter 7.2 Volume: The Disk Method Well, if we can find the area between two functions, why not spin that function and find it’s volume?! Well yeah, we can do that. I know if I spun a single point in a circle, that would give me an area equal to a circle of some radius spun about an axis. Now if I add all of the points of the area between two curves I’ll get the volume of that object of some radius spun about an axis. Volume of disk=(area of disk)(width of disk)= (Draw a picture) So looking at a function that changes in height, which would have a varying area, we can take the volume as a sum of ​ to get volume But this is along a ​horizontal axis of revolution​, or parallel to the x­axis. We can similarly write a formula for the vertical axis of revolution, because maybe that would be easier to integrate in certain situations, depending on the functions involved. Along a ​vertical axis of revolution​ Where c and d are y­values instead of a and b being x­values. The above equations fall under the ​Disk Method for finding Volume (Draw a picture of horizontal versus vertical) ex) Graph the following equation ​, find the volume when it is constrained between x=0 and x=3 and revolved about the x­axis and y­axis. ex) Revolve the following area about the y­axis of y=2, for the function constrained between and find the volume. Note that it is above the x­axis, what does it’s image look like? Note that R(x)=f(x)­2 now! The Washer Method​: In the disk method we are assuming the volume is a solid with no holes, but we can take the axis of revolution to be some other point (draw a picture) We get an image of a washer when we rotate about this new axis, since volume of a washer is then we can integrate with some function that isn’t constant to get when spinning about some x­axis, say y=c, where and ex) Find the volume of the solid formed by revolving the region bounded by the graphs of and ​ about the axis y=2 and the x­axis y=0. Draw a picture. Now lets try integrating around a vertical axis like x=c, here we will integrate with respect to y, have a ​ and solve the function for x. ex) Find the volume of the solid generated by revolving the region bounded by the equations about the given vertical axes. line x=2. ​ revolved about the y­axis and about the The disk method gave us volumes for objects with a circular cross section, but in general if we have an equation for the area that is known then we can find its volume also. It could be a square, a triangle, or some other constant or complex area. 1) For cross sections of A(x) that are perpendicular to the x­axis 2) For cross sections of A(x) that are perpendicular to the y­axis (Draw a picture) Ex #60) Lets find the volume of a sphere (similar to 75). If we take a cross section of a sphere of radius r we notice the slice, perpendicular to the x­axis will be ​, note that so ​, we will integrate with respect to x, and its bounds range from what to what? (Draw a picture, label your axes, draw your cross section, find its area, relate it to the x position) Ex #75) Let's find the volume of a wedge formed by cutting a right circular cylinder (draw a picture from pg 456) Find the volume for the wedge using an arbitrary angle theta. ​ where What is the area along the x­axis? What shape does it create? A triangle! So the triangles sides follow the trig identity ​, then we can create a function for the area in terms of y, since x is the base and z is the height then as y changes , but we know the shape is a circle so ­r to +r since ​ when x=0, so ​ and y goes from Now finish the integration… Homework Chap 7.2 #1­13 odd, 17,21,25,29,31,54,55,57,59 Chapter 7.3 Volume: The Shell Method We have another method for finding the volume of a solid of revolution, the shell method uses cylindrical shells to create the volume. Think of a circle, its length or outer circumference is if we want a cylinder we multiply it by its height, and if we multiply by the thickness of this shell we get a some object akin to a paint can. We can control these variables to get cans of differing shapes and sizes. where p is the average radius, h is the height, w is the width. If our axis of revolution is the y­axis​ ​ then p is the average radius measured from the y­axis, h is the height measured along the y­axis, and w is the thickness or dx. (draw a picture) ​If our axis of revolution is the x­axis​ ​ then p is the average radius measured from the x­axis, h is the height measured along the x­axis, and w is the thickness or dy. (draw a picture) ex) Use the shell method to find the volume of revolution about the x and y axis for the volume given by ex) Use the shell method to find the volume of the solid generated by rotating the plane region about the x­axis where ex) Use the shell method to find the volume of the solid generated by rotating the plane region about the x­axis where Why do we bother learning two separate methods? Well given a really ugly equation written in terms of x we can potentially integrate easily with the shell method, but the disk method could potentially be a couple integrals. ex) Find the volume of the solid of revolution formed by revolving the region bounded by the graph of ​ and rotated about the x­axis. Try it using the shell method first, how many integrals do you need? Now try it using the disk method… uh oh, what happens? How ugly is that integral!? Compare and contrast: Disk vs. Shell The representative rectangle ( ​) is always perpendicular to the axis of revolution for the disk method and is always parallel to the axis of revolution for the shell method. (Draw pictures of Disk method: vertical and horizontal, Shell method: vertical and horizontal) ­ ­ One method may be nicer or easier to use than another, or they could both be used interchangeably, you have to find what works For example, you might need to rotate something about a vertical axis, then you have to solve some function for y and it may be really hard or just not worth it, then you might consider the shell method so you can keep everything in terms of x! ­ Try writing the function ​ in terms of y... ex) Use the disk or shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations ​, y=0, x=1, x=5 about the a) x­axis b) y­axis and c) y=10 ex) Use the disk or shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations ​, x=0, y=0, about the y­axis ex) Find the volume of the ellipsoid bounded by the graph of Show that the volume is revolved about the x­axis. ​ where a>0, b>0. ​ when revolved about the y­axis, also find its volume when Homework Chap 7.3#1­29 odd,47,51 Chapter 7.4 Arc Length and Surfaces of Revolution If we go for a walk and it meanders about, how do we find the distance we have traveled? In a two dimensional coordinate system we have the distance equation for straight lines, but if we consider that any meandering path is just the sum of many small straight lines, we can find the distance traveled no matter how much we change direction! We can then algebraically manipulate this formula, the sum of these straight line distances we shall label s for arclength then Now if we take this sum as a limit, so that we can break our path up into as many tiny sections as we need to find its length we get where s is the arclength from a to b. Arc Length Let the function y=f(x) represent a smooth curve on the interval [a, b]. The arc length of f ​ If we solve the function for x then x=g(y) and between a and b is on the interval [c, d] ex) Find the arclength of the function ​ on the interval [1, 2] ex) Find the arclength of the function ​ on the interval [0, 8] in terms of x, and what would be the interval in terms of y? Find the arclength in terms of y too, are they the same? ex) Find the arclength of the function ex) Find the arclength of the function ​ on the interval ​ on the interval [ln2, ln3] ex) Find the arclength of the function ​ on the interval [1, 4] Now if we take this continuous arclength and rotate it around an axis we can create a surface of revolution or surface area! ​ (Draw a picture) In a manner similar to how we derived the arclength formula we can find the surface area S. Definition of the Area of a Surface of Revolution Let y=f(x) have a continuous derivative on the interval [a, b]. The area S of the surface of revolution formed by revolving the graph of f about a horizontal or vertical axis is where r(x) is the distance between the graph of f and the axis of revolution. If x=g(y) on the interval [c, d], then the surface area is Where r(y) is the distance between the graph of g and the axis of revolution. ex) Find the area of the surface of revolution by revolving interval [4, 9] ex) Find the area of the surface of revolution by revolving interval [0, 3] (x­values, convert to y­values) ​ about the x axis on the ​ about the x axis on the Homework Chap 7.4 #1­15 odd,21,23,31,37­45 odd,56, Chapter 7.5 Work work work work work In physics, work is defined as the force needed to move an object over a specific distance or W=FD From Physics I we know there are the gravitational (F=mg) and electromagnetic forces ( ) acting on objects and charges, and the strong and weak nuclear forces holding atoms together. In SI units we measure Work in N*m or newton*meters, in English units we have ft*lbs or foot*pounds, within the SI system we also have centimeter to gram units where work is measured in ergs or dyne*centimeters… I know, weird! Let’s just stick with the normal SI and English units. Ex) If I pick up a 100 lb weight and lift it three feet how much work have I done? What if I push it three feet? Will the work done be the same? Why or why not? Now work done by a variable force obviously will be different, that is where calculus steps in! We can think of every one of those changes in force as its own tiny little ​, if we sum all of them together we get the total work done on the object. Now if the force is always changing, but in a continuous manner, then we can apply limits and turn this sum into an integral. ­ So notice that the area under the graph of force is the total work done! (draw a picture) Since forces generally change when in a dynamic system, we will lots of opportunities to use our formula. The following formulas for different types of forces (push or pulls) are listed below. Hooke’s Law:​ the force required to compress a spring (sometimes written with a negative to denote it is a restorative force) ​ or ​ where k is the spring constant and x is the distance the spring was stretched or compressed from its natural state or length. ...
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