10.7 Triple Integrals. The Divergence Theorem of Gauss

Transcription

1 10.7 riple Integrals. he Divergence heorem of Gauss We begin by recalling the definition of the triple integral f (x, y, z) dv, (1) where is a bounded, solid region in R 3 (for example the solid ball {(x, y, z) R 3 : x 2 + y 2 + z 2 1}). 1 / 14

2 Definition of a triple integral First partition into finitely many small solid subregions ijk where the indices i, j, k vary over some finite indexing sets. Let V ijk denote the volume of ijk. For each triple of indexes (i, j, k) let (xi, yj, z k ) be any point in ijk. Form the Riemann sum f ( xi, yj, zk ) Vijk i j k Now take the limit of all possible such Riemann sums, where the limit is taken as the partition becomes finer and finer in the sense that the diameter of all of the sets ijk goes to 0. If the limit exists, then we call that limit the triple integral written above in (1). It can be shown that the limit exists provided f is a function defined on an open set which contains the solid such that f is continuous throughout its domain. 2 / 14

3 riple integrals in rectangular coordinates as iterated integrals Given a triple integral f (x, y, z) dv, we would like to be able to integrate it as an iterated integral. Our ability to do that is made easier if is of a special type. he following describes the sorts of regions for which it is natural to use rectangular coordinates. riple integrals in rectangular coordinates Say is a solid which is bounded below by a bottom surface z = g(x, y) and above by a top surface z = h(x, y). Suppose also that the projection of onto the xy-plane is a region R. hen f (x, y, z) dv = where the outer double integral is done over R. R h(x,y) g(x,y) f (x, y, z) dz da, here are similar formulas in case it is natural to instead project onto either the xz-plane or onto the yz-plane. 3 / 14

4 riple integrals in rectangular coordinates as iterated integrals Exercise. Calculate 2z dv, where is the solid tetrahedron in the first octant with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1). Clearly identify what are the upper and lower surfaces and what is the projection R over which we do the outer double integral. 4 / 14

5 riple integrals in cylindrical coordinates as iterated integrals he regions for which it is natural to use cylindrical coordinates are described next. riple integrals in cylindrical coordinates is similar to the one we described on the rectangular coordinates slide, namely it is a solid which is bounded below by a bottom surface z = g(x, y) and above by a top surface z = h(x, y). Again let R be the projection of onto the xy-plane. Relative to the 2-dimensional polar coordinates (r, θ), we want R to be a region over the form {(r, θ) : α θ β, h 1(θ) r h 2(θ)}. hen β h2 (θ) h(r cos θ,r sin θ) f (x, y, z) dv = f (r cos θ, r sin θ, z) r dz dr dθ, α h 1 (θ) g(r cos θ,r sin θ) where the outer double integral is done over R. Note that region R is bounded by an inner curve and an outer curve. 5 / 14

6 riple integrals in cylindrical coordinates as iterated integrals Exercise. Calculate x 2 + y 2 + z 2 : dv, where is the solid cylinder x 2 + y 2 9, 1 z 1, x and y 0. Clearly identify what are the upper and lower surfaces, the projection R, and the inner and outer curves. 6 / 14

7 riple integrals in spherical coordinates as iterated integrals he regions for which it is natural to use spherical coordinates are described next. riple integrals in spherical coordinates is the region between two surfaces of the form ρ = g(θ, φ) and ρ = h(θ, φ) where θ and φ vary independly between constant values θ 1 θ θ 2, φ 1 φ φ 2. Since ρ measures distance back to the origin, we have an inner surface and an outer surface (rather than a top and bottom surface as for rectangular and cylindrical coordinates). hen f (x, y, z) dv = θ2 φ2 h(θ,φ) θ 1 φ 1 g(θ,φ) f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ 2 sin φ dρ dφ dθ, but before just blindly substituting the above expression into f, notice that any terms of the form x 2 + y 2 + z 2 should be replaced with ρ 2. 7 / 14

8 riple integrals in spherical coordinates as iterated integrals Exercise. Calculate x 2 + y 2 + z 2 dv, where is the region inside the cone z = x 2 + y 2, 0 z 1. Be sure to identify the inner and outer surfaces 8 / 14

9 he Divergence heorem of Gauss We ve already seen the two-dimensional version of the theorem at the end of 10.4 on Green s heorem: heorem(wo-dimensional Divergence heorem) Let R be a bounded open subset of R 2 whose boundary C consists of finitely many simple closed curves. Let F = [F 1 (x, y), F 2 (x, y)] be a vector field such that F 1, F 2, and all of their first order partial derivatives are continuous on a neighborhood of R C. hen div( F ) da = F n ds, R where n (x, y) is the unit outward point normal at (x, y) in C, and we orient C in such a way that R is on our left. In words it says that the flux of F out of R equals the integral of the divergence of F over R. It s useful to sketch R and C with the direction of n and the orientation of C indicated. C 9 / 14

10 he Divergence heorem of Gauss If we generalize each piece of the wo-dimensional Divergence heorem to three dimensions, we get the statement of the Divergence heorem (we just call it the Divergence heorem, not the three-dimensional Divergence heorem). he region R with a bounding curve is replaced by a solid with C replaced by S, the entire boundary of. F = [F1, F 2 ] is replaced by a three-dimensional vector field. heorem. Let S be a closed bounded surface which encloses a solid. Let F = [F 1, F 2, F 3 ] be a vector field such that the components of F and all of their first order partial derivatives are continuous on a neighborhood of. hen F n da = div( F ) dv. S 10 / 14

11 he Divergence heorem of Gauss Exercise. a) In the Divergence heorem, if we take the solid to be the solid ball = {(x, y, z) : x 2 + y 2 + z 2 1}, what is the surface S? b) In the Divergence heorem, if we take the solid to be the solid upper hemisphere: = {(x, y, z) : x 2 + y 2 + z 2 1 and z 0}, what is the surface S? c) In the Divergence heorem, if we take the solid to be the inside of a truncated cone: = {(x, y, z) : 0 z x 2 + y 2 and z 1}, what is the surface S? 11 / 14

12 he Divergence heorem of Gauss In order to get a sense of why the Divergence heorem holds, we confirm it for a special case in the following exercise. Exercise. Suppose is the region between an upper surface z = S 2 (x, y) and a lower surface z = S 1 (x, y), where x and y vary over a region D in the xy-plane. a) Calculate F 3 z dv by writing it as an iterated integral and just evaluating the inner integral with respect to z. b) Calculate S [0, 0, F 3] n da by first integrating over the top surface z = S 2(x, y) and then integrating over the bottom surface z = S 3(x, y). Be sure you use the outward pointing normal when integrating. c) Do you see how to confirm that the Divergence heorem holds for the vector field [0, 0, F 3]? 12 / 14

13 he Divergence heorem of Gauss Exercise. Verify the Divergence heorem holds for F = [x 2, z 2, y 2 ] where is the unit cube: = {(x, y, z) : 0 x, y, z 1}. his means you should independently calculate both sides of the Divergence heorem for this specific example, and verify that they come out the same. 13 / 14

14 he Divergence heorem of Gauss Exercise. Verify the Divergence heorem holds for F = [x, y, z] where is the region inside the cone z = x 2 + y 2 below the plane z = / 14