This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: What is Integration Good For? Or how I learned to stop worrying and love the integral, Or why the heck are we learning all of this? Marty Weissman April 19, 2000 Math 21a This handout is to explain what an integral “means.” This handout does not explain how to evaluate integrals. This handout will hopefully help you convert concepts into mathematical language. This handout will not help you use that mathematical language to find answers. Part I: Finding the size of a region Single variable calculus : Let ) ( x f y = be a function of one variable. You should be familiar with integrals of the form: ∫ b a dx x f ) ( . If the function ) ( x f is positive, then this (by definition) computes the area between the xaxis and the graph of ) ( x f over the domain of integration b x a ≤ ≤ . Remember, every integral involves a domain of integration, and an “integrand,” i.e. the thing being integrated. We can also use integrals (inefficiently in this case) to find the size of the domain of integration. Here in the onedimensional case, if the domain of integration is given by b x a ≤ ≤ , then the size of the domain of integration is just the length of the interval [ ] b a , , which is just a b . We could alternatively compute the size of the domain of integration by evaluating the integral: a b x dx b a b a = = ⋅ ∫ 1 . Twovariable calculus : Let ) , ( y x f z = be a function of two variables, and let R be a “filledin” region in the plane. You should be familiar with integrals of the form: ∫∫ R dA y x f ) , ( . We will call integrals of this kind abstract area integrals , to emphasize the fact that we don’t specify whether dxdy dA = or whether q rdrd dA = , and we don’t give a parametrization of the region R in terms of coordinates. If the function ) , ( y x f is positive, then this integral evaluates the volume of the solid contained above the region R in the xyplane, and below the graph of ) , ( y x f z = . We can also use integrals (sometimes efficiently in this case) to find the size of the domain of integration, i.e. the area of the region R . Just like the onevariable case, we have the formula: ). ( 1 R Area dA R = ⋅ ∫∫ To evaluate an abstract double integral, you have to convert it into an iterated integral by choosing coordinates (Cartesian or polar), expressing the region R in terms of these coordinates, substituting ) sin( ), cos( q q r y r x = = in the case of polar coordinates, and expressing dA in the proper coordinates. Threevariable calculus : Let ) , , ( z y x f be a function of three variables, and let R be a solid region in space. You should be somewhat familiar with integrals of the form: ∫∫∫ R dV z y x f ) , , ( . Such an expression is called an abstract volume integral , to emphasize the fact that we don’t specify whether dxdydz dV = or dz rdrd dV q = or r j q j r d d d dV ) sin( 2 = , i.e. we don’t yet make a choice of coordinates, nor do we choose a parametrization of the solid region...
View
Full
Document
This note was uploaded on 02/18/2012 for the course MATH 310 taught by Professor Gregoryj.galloway during the Fall '11 term at University of Miami.
 Fall '11
 GregoryJ.Galloway
 Math, Multivariable Calculus

Click to edit the document details