The exam will have 5 questions covering multiple

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1 Math Notes of 11/14/2018 Math 2210 Exam 3 ummary The exam will have 5 questions covering multiple integrals. Multiple Integrals are integrals of functions of n variables over regions in IR n. For our purposes n =2,orn = 3. For n =2we obtain a double integral, forn =3,atriple integral. To obtain a multiple integral, think in terms of what happens in a part of the integration region that corresponds to small changes in the integration variables. Add up those contributions, take the limit as the number of contributions goes to infinity and their size goes to zero, and replace the sum by an integral. ouble integrals. Let be a region in IR 2. f(x, y)dxdy = f(x, y)da = where x =max i x i and y =max i y i lim f(xi,y j ) x i y i x, y 0 (1) Math Notes of 11/14/2018 page 1

2 Examples area = mass = average value of f = da δ(x, y)da f(x, y)da da (2) For example, f might be temperature, depth, or density. imilar formulas apply for triple integrals. uppose is a region in IR 3.Conceptually f(x, y, z)dxdydz = For example: volume of = f(x, y, z)dv = lim x, y, z 0 where δ is density f(xi,y i,z i ) x i y i z i. (3) dv. (4) The formulas for the center of mass familiar from Calculus 1 actually simplify, and become more general. Let δ(x, y) denotedensity. Then the center of mass of a subset Math Notes of 11/14/2018 page 2

3 of IR 2 is ( x, ȳ) where x = xδ(x, y)da δ(x, y)da and ȳ = yδ(x, y)da δ(x, y)da (5) The formulas for the center of mass of a region in IR 3 are straightforward generalizations of the 2 case: xδ(x, y, z)dv x =, δ(x, y, z)dv yδ(x, y, z)dv ȳ = and (6) δ(x, y, z)dv zδ(x, y, z)dv z = δ(x, y, z)dv The practical way of computing multiple integrals is to convert them to iterated integrals which are nested one variable integrals. In general the limits of integration of the inner integrals depend on the variables of the outer integrals. Math Notes of 11/14/2018 page 3

4 For example, to integrate the function f over the unit sphere W (defined by x 2 +y 2 +z 2 1) we can write 1 z 2 1 z 2 y 2 W f(x, y, z)dv = z 2 For example, to obtain the volume of the unit sphere we set f(x, y, z) =1andobtain V = = = 1 1 z z z z 2 y 2 1 z 2 y 2 dxdydz 1 z z 2 y 2 dydz 1 z 2 y 2 f(x, y, z)dxdydz. (7) (and, since this is the area of a circle of radius 1 z 2 ) π(1 z 2 )dz ) = π (z z3 1 3 = 4π 3 1 which of course is no surprise. (8) For iterated integrals, the sequence of the integration variables matters! To get the limits of integration, and the sequence of the integration variables, figure them out geometrically. There is no set rule or simple recipe. Math Notes of 11/14/2018 page 4

5 In many applications, integrals get much simplified by using appropriate coordinates. To figure out an integration formula in a certain coordinate system, consider a part of the region in which the coordinates change by some small value. on t try to memorize formulas. However, in polar coordinates, letting be aregioninir 2,and the same region expressed in terms of polar coordinates, we obtain f(x, y)dxdy = rf ( r cos(θ),rsin(θ) ) drdθ. (9) Note the additional factor r! In cylindrical coordinates we use polar coordinates in xy-plane and keep z as a variable. We get f(r, θ, z)da = rf(r, θ, z)drdθdz (10) with appropriate limits of integration. Again, note the additional factor r. In spherical coordinates, ρ = x 2 + y 2 + z 2, θ is as in cylindrical coordinates, and φ is the angle that <x,y,z>makes with the z-axis. The triple integral is given by f(ρ, θ, φ)da = ρ 2 sin θf(r, θ, z)drdθdz. (11) g at's of Math Notes of 11/14/2018 page 5

6 Note the additional factor ρ 2 sin θ. The surface area of the graph of f(x, y) defined on a domain in R 2 is Area = 1+fx 2 + f y 2 dxdy. (12) This is the natural analog of the one variable formula for arc length. The individual integrals in an iterated integral can of course be computed using any single variable integration technique. The concept of improper integrals can be generalized to the integration of functions of several variables. We saw one example: IR 2e x 2 y 2 da = π f which implies that e x2 dx = π. Math Notes of 11/14/2018 page 6

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