Functions and their properties 1 0.1 Triple integrals, volumes and centroids Let f : R3 →Rhave for its domain some solid region R in R3. We divide R into m subregions of volume,∆V1,...,∆Vm in any fashion what- soever. Figure 1 Next, as in the definition of the double integral, we pick a point (xi,yi,zi)in each subregion of volume δVi,r=1, 2,...,m, and form the sum: m f(xi,yi,zi)∆Vi = f(x1,y1,z1)∆V1 + ...f(xm,ym,zm)δVm i=1 If this sum approaches a finite limit as m →∞and this limit is independent of the way in which ∆Vi → 0, we call the limit the triple integral of f over R, fdV ≡ Triple integral of f over R R dV = volume element in coordinate system which describes R. (If w = f(x, y, z) ≥ 0overR thetripleintegraloff over R actually describes the volume under the hypersurface w = f(x, y, z)inR4 whose projection on R3 is R!). By definition, we see that (if f =1): dV =VolumeofR R The triple integral has the same properties as the double integral. Its evaluation is by means of iterated integrals as before. thus if: R = {(x, y, z):g1(x, y) ≤ z ≤ g2(x, y),f1(x) ≤ y ≤ f2(x),a≤ x ≤ b} Then: b f2(x) g2(x,y) fdV = f(x, y, z)dzdydx R 1 f(x) g1(x,y) Basically, to describe R one first describes its projection on one of the coordi- nate planes, (say,z =0). Then, by cutting R by means of a fibre, one measures the height of z-variation along this fibre as (x, y) varies over the projection region. Example 1 Evaluate: 1 x xy I = dzdydx 0 x2 0 2 The ABC’s of Calculus 1 x xy I = { dz }dydx 0 x2 0 evaluate innermost first 1 x |z=xy = [z z=0 ]dydx 2 0 x 1 x = (xy)dydx 2 0 x 1 x = x( ydy)dx 2 0 x 1 x2 x4 = x( − )dx 0 2 2 4 6 1 x x 1 = ( − )|0 2 4 6 1 1 1 = ( − ) 2 4 6 1 = 24 Example 2 Evaluate the preceeding integral by changing the order of integra- tion from: dzdydx to dzdxdy Idea R as given is: R = {(x, y, z):0≤ z ≤ xy, x2 ≤ y ≤ x, 0 ≤ x ≤ 1} So, the projection of R onto the xy-plane is given by: R = {(x, y):x2 ≤ y ≤ x, 0 ≤ x ≤ 1} Sketch R and change the direction of the fibre used to describe R. √ The coordinates of the ends of a horizontal fibre are: (y,y) and ( y,y),Thus: √ R = {(x, y):y ≤ x ≤ y, 0 ≤ y ≤ 1} and, √ 1 x xy 1 y xy dzdydx = dzdxdy 0 x2 0 0 y 0 √ 1 y = xydxdy 0 y √ 1 y = y{ xdx}dy 0 y 1 1 = y(y − y2)dy 2 0 3 4 1 y y 1 = [ − ]|0 2 3 4 1 4 − 3 = ( ) 2 12 1 = 24 (as before) Functions and their properties 3 Example 3 Find the volume inside x2 + y2 =9,abovez =0,andbelow x + z =4 Figure 2 1) Here z =4− x. First, we determine the projection of this solid on, say, z =0. This gives the region x2 + y2 ≤ 9 whose description is, say: R = {(x, y):− 9 − x2 ≤ y ≤ 9 − x2, 0 ≤ x ≤ 3} A fibre extending through the solid starts at (x, y, 0) and ends at (x, y, z) where z =4− x.Hence0 ≤ z ≤ 4 − x,forthez-variation along this fibre. R = {(x, y, z):0≤ z ≤ 4 − x, − 9 − x2 ≤ y ≤ 9 − x2, 0 ≤ x ≤ 3} √ 3 + 9−x2 4−x Volume= dV = √ dzdydx R 0 − 9−x2 0 is one expression for the volume. 2) In cylindrical coordinates, (as we are dealing with a solid carved out of a cylinder), R = {(r, Θ) : 0 ≤ r ≤ 3, 0 ≤ θ<2π} and: R = {(r, Θ,z):0≤ z ≤ 4 − r cos θ, 0 ≤ r ≤ 3, 0 ≤ θ<2π} 2π 3 4−r cos Θ Volume= dV = rdzdrdΘ R 0 0 0 which is easier to evaluate. 2π 3 = r(4 − r cos θ)drdΘ 0 0 2π 3 2 r r=3 = (2r − cos Θ| =0)dΘ 3 r 0 2π = (18 − 9cosθ)dΘ 0 =18· 2π − 0 = 36π Example 4 Find the centroid of the volume under z2 = xy above the triangle y = x, y =0,x=4in the plane z =0. It is clear that the projection of this solid under the surface z2 = xy on z =0is given by the triangle R: R = {(x, y):0≤ y ≤ x, 0 ≤ x ≤ 4} 4 The ABC’s of Calculus Figure 3 R 2 Furthermore√ the ”height” of a vertical fibre from to the surface z = xy is z = ± xy (since x ≥ 0,y ≥ 0), we take z ≥ 0. So: Volume of solid region (above z =0) = dV R √ 4 x xy = dzdydx 0 0 0 4 x 1 1 = x 2 y 2 dydx 0 0 4 3 2 1 2y x = x 2 ( )|0 dx 0 3 4 1 2 3 = x 2 ( x 2 − 0)dx 3 0 4 2 = x2dx 0 3 3 2 x 4 = · |0 3 3 128 = 9 128 256 [or Volume above and below z =0, =2× ( 9 )= 9 ] the centroid (x, y, z) is given by (in Cartesian coordinates): xdV ydV zdV x = R , y = R , z = R R dV R dV R dV Now √ 4 x xy xdV = xdzdydx R 0 0 0 4 x √ = x xydydx 0 0 4 x 3 = x 2 { y 1 dy}dx 0 0 2 4 3 2 3 = x 2 · x 2 dx 0 3 4 2 = x3dx 3 0 2 x4 = · |4 3 4 0 2 · 43 = 3 Functions and their properties 5 128 = 3 Therefore: 128 128 3 3 x = R = 128 =3 vol 9 Similarly: √ 4 x xy ydV = ydzdydx R 0 0 0 4 x 3 1 = y 2 x 2 dydx 0 0 4 2 5 1 = x 2 · x 2 dx 5 0 2 4 = x3dx 5 0 2 44 = · 5 4 2 = · 43 5 128 = 5 Therefore 128 5 9 y = 128 = 9 5 Finally, √ 4 x xy zdV = zdzdydx R 0 0 0 4 x 1 = (xy)dydx 2 0 0 1 4 x = x( ydy)dx 2 0 0 4 1 x2 = x · dx 2 2 0 1 4 = x3dx 4 0 1 44 = · 4 4 =16 Therefore: 16 144 9 z = 128 = = 9 128 8 Example 5 Find the volume of the four -dimensional sphere x2 +y2+z2 +w2 = a2, of ”radius” a>0. 6 The ABC’s of Calculus 2 2 2 2 Here,W = ± a − x − y − z = f(x, y, z). consider the hemi-hypersphere w = a2 − x2 − y2 − z2.Ifx2 + y2 + z2 ≤ a2 ⇒ w is defined and f(x, y, z) ≥ 0 So by definition, the volume of this hemi-hypersphere is: = f(x, y, z)dV R where R is the ”ordinary” spherical region: R = {(x, y, z):x2 + y2 + z2 ≤ a2} =7{(ρ, Θ,φ):0≤ ρ ≤ a, 0 ≤ θ<2π, 0 ≤ φ ≤ π} Therefore the volume of this sphere: Volume =2 fdV R 2π π a =2 a2 − ρ2 · ρ2 sin φdρdφdΘ 0 0 0 2π π a =2 sin φ( ρ2 a2 − ρ2dρ)dφdΘ 0 0 0 constant a 2π π =2· ( ρ2 a2 − ρ2dρ) sin φdφdΘ 0 0 0 a =2· 4π ρ2 a2 − ρ2dρ 0 ρ = a sin u, dρ = a cos udu π 2 =2· 4π a2 sin2 u · a cos u · a cos udu 0 π 2 =2· 4πa4 sin2 u cos2 udu 0 4 π 4πa 2 =2· sin2(2u)du 4 0 (sin(2u)=2sinu cos u) πa4 π =2· sin2(v)dv (2u = v) 2 0 πa4 1 1 =2· [ v − sin(2v)]|v=π 2 2 4 v=0 πa4 π =2· [ ] 2 2 π2a4 =2· 4 π2a4 = is the volume of a 4-D sphere with radius a>0 2 Functions and their properties 7 0.2 Leibnitz’s Rule for Differentiating an inte- gral: If f : R2 →Ris continuous and differentiable in both variables, and: b(α) I(α)= f(x, α)dx a(α) where a(α),b(α) are differentiable functions of x,thenI(α) exists, and: dI b(α) ∂f db da I(α)= = (a, α)dx + f(b(α),α) − f(a(α),α) dα a(α) ∂α dα dα Example 6 Find I(α) if: cos α I(α)= (x2 sin α − x3)dx sin α By Leibnitz’s rule: cos(α) ∂ 2 3 2 3 d 2 3 d I (α)= (x sin α − x )dx +(x sin α − x )|x=cos α · (cos α) − (x sin α − x )|x=sin α · (sin α) sin(α) ∂α dα dα cos(α) = (x2 cos α)dx +(cos2 α sin α − cos3 α) · (− sin α) − (sin3 α − sin3 α) · (cosα) sin(α) cos(α) =cosα x2dx +cos3 α sin α − cos2 α sin2 α sin(α) cos3 α sin3 α =cosα · − cos α +cos3 α sin α − cos2 α sin2 α 3 3 Example 7 2) Let 1 x+ct u(x, t)= g(s)ds 2c x−ct where g is a differentiable function on R,Then: ∂u 1 x+ct g(x + ct) · c g(x − ct)(−c) = g(s)ds + − ∂t 2c x−ct 2c 2c by Leibnitz cg(x + ct)+cg(x − ct) =0+ 2c g(x + ct + g(x − ct) =( ) 2 ∂2u 1 = (g(x + ct) · c + g(x − ct)(−c)) ∂t2 2 c = (g(x + ct) − g(x − ct)) 2 1 ∂2u g(x + ct) − g(x − ct) or = c ∂t2 2 8 The ABC’s of Calculus Now, ∂u 1 x+ct ∂ g(x + ct) · (1) g(x − ct) · (1) = g(s)ds + − ∂x 2c x−ct ∂x 2c 2c (g(x + ct) − g(x − ct)) = 2c ∂2u (g(x + ct)(1) + g(x − ct)(1)) = ∂x2 2c (g((x + ct) − g(x − ct)) = 2c ∂2u (g((x + ct) − g(x − ct)) or c = ∂x2 2 1 ∂2u = c ∂t2 1 ∂2u ∂2u ⇒ − =0 c2 ∂t2 ∂x2 a one dimensional wave equation: Also ∂u u(x, 0) = 0, (x, 0) = g(s) ∂t.