27.4. Changing Coordinates. Introduction. Prerequisites. Learning Outcomes

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1 Changing Coordinates 27. Introduction We have seen how changing the variable of integration of a single integral or changing the coordinate sstem for multiple integrals can make integrals easier to evaluate. In this section we introduce the Jacobian. The Jacobian gives a general method for transforming the coordinates of an multiple integral. Prerequisites Before starting this Section ou should... Learning Outcomes After completing this Section ou should be able to... 1 have a thorough understanding of the various techniques of integration 2 be familiar with the concept of a function of several variables 3 be able to evaluate the determinant of a matri have completed the previous block decide which coordinate transformation simplif an integral determine the Jacobian for a coordinate transformation evaluate multiple integrals using a transformation of coordinates

2 1. You have seen the Jacobian before When the method of substitution is used to solve an integral of the form b f() d a three parts of the integral are changed, the limits, the function and the infinitesimal d. So if the substitution is of the form = (u) the u limits, c and d, are found b solving a = (c) and b = (d) and the function is epressed in terms of u as f ( (u)) (u) δ δ δu δu u The above diagram shows wh the d needs to be changed. While the δu is the same length for all u, the δ change as u changes. The rate at which the change is precisel d (u). This du gives the relation δ = d du δu Hence the transformed integral can be written as b a f() d = d c f ( (u)) d du du Here the d is plaing the part of the Jacobian that we will define. du Another change of coordinates that ou have seen is the transformations from cartesian coordinates (, ) topolar coordinates (r, θ). Recall that a double integral in polar coordinates is epressed as f(, ) dd = g(r, θ)rdrdθ δr δr δθ A B r 1 r 1 + δr r 2 r 2 + δr HELM (VERSION 1: March 18, 2): Workbook Level 1 2

3 We can see from the diagram that the area elements change in size as r increases. The circumference of a circle of radius r is 2πr, sothe length of an arc spanned b an angle θ is 2πr θ = rθ. 2π Hence the area elements in polar coordinates are approimated b rectangles of width δr and length rδθ. Thus under the transformation from cartesian to polar coordinates we have the relation δδ rδrδθ that is, rδrδθ plas the same role as δδ. This is wh the r term appears in the integrand. Here r is plaing the part of the Jacobian. 2. The Jacobian Given an integral of the form A f(, ) dd Assume we have a change of variables of the form = (u, v) and = (u, v) then the Jacobian of the transformation is defined as u v J(u, v) = u v Ke Point Notice the pattern occurring in the,, u and v. Across a row of the determinant the numerators are the same and down a column the denominators are the same. Notation Different tetbooks use different notation for the Jacobian. The following are equivalent. ( ), J(u, v) =J(, ; u, v) =J = (, ) u, v (u, v) 3 HELM (VERSION 1: March 18, 2): Workbook Level 1

4 The Jacobian correctl describes how area elements change under such a transformation. The required relationship is dd J(u, v) dudv that is, J(u, v) dudv plas the role of dd. Ke Point When transforming area elements emploing the Jacobian it is the modulus of the Jacobian that must be used. Eample Find the area of the circle of radius R. θ = π/2 θ = π θ θ = r = θ =2π r = R Solution Let A be the region bounded b a circle of radius R centred at the origin. Then the area of this region is da. Wewill calculate this area b changing to polar coordinates, so consider A the usual transformation = r cos θ, = r sin θ from cartesian to polar coordinates. First we require all the partial derivatives Thus r = cos θ J(r, θ) = r r r = sin θ θ θ = θ cos θ sin θ = r sin θ r sin θ rcos θ θ = r cos θ = cos θ r cos θ ( r sin θ) sin θ = r ( cos 2 θ + sin 2 θ ) = r HELM (VERSION 1: March 18, 2): Workbook Level 1

5 Solution (contd.) This confirms the previous result for polar coordinates, dd rdrdθ. The limits on r are r =(centre) to r = R (edge). The limits on θ are θ =toθ =2π, i.e. starting to the right and going once round anticlockwise. The required area is A da = 2π R J(r, θ) drdθ = Note that here r>so J(r, θ) = J(r, θ) =r. 2π R r drdθ =2π R2 2 = πr2 Eample The diamond shaped region A in the diagram is bounded b the lines +2 = 2, 2 =2, +2 = 2 and 2 = 2. We wish to evaluate the integral I = (3 +6) 2 da A over this region. Since the region A is neither verticall nor horizontall simple, evaluating I without changing coordinates would require separating the region into two simple triangular regions. So we use a change of coordinates to transform A to a square region and evaluate I. 1 v v =2 2 = 2 +2 =2 2 A 2 +2 = 2 2 =2 u = 2 A u =2 u 1 v = 2 Solution B considering the equations of the boundar lines of region A it is eas to see that the change of coordinates u = +2 (1) v = 2 (2) will transform the boundar lines to u =2,u = 2, v =2and v = 2. These values of u and v are the new limits of integration. The region A will be transformed to the square region A shown above. We require the inverse transformations so that we can substitute for and in terms of u and v. Badding (1) and (2) we obtain u + v =2 and b subtracting (1) and (2) we obtain u v =, thus the required change of coordinates is = 1 2 (u + v) = 1 (u v) 5 HELM (VERSION 1: March 18, 2): Workbook Level 1

6 Solution (contd.) Substituting for and in the integrand (3 +6) 2 of I gives ( ) (u + v)+6 (u v) =9u 2 We have the new limits of integration and the new form of the integrand, we now require the Jacobian. The required partial derivatives are u = 1 2 v = 1 2 u = 1 Then the Jacobian is J(u, v) = = v = 1 Then da = J(u, v) da = 1 da. Using the new limits, integrand and the Jacobian, the integral can be written I = u2 dudv. You should evaluate this integral and check that I = Given the transformations u = +, v = epress and in terms of u and v to find the inverse transformations. 2. Find the Jacobian J(u, v) for the transformation in part 1. ( 3. Epress the integral I = 2 + 2) dd in terms of u and v, using the transformations in part 1.. Find the limits on u and v for the rectangle with vertices (, ) =(, ), (2, 2), ( 1, 5), ( 3, 3) and hence evaluate I. Your solution 1.) = 1 2 (u + v), = 1 2 (u v) HELM (VERSION 1: March 18, 2): Workbook Level 1 6

7 Your solution 2. J(u, v) = 1 2 Your solution 3. 1 (u 2 + v 2) dudv Your solution. 1 v= 6 v= (u2 + v 2 )dudv = 1 3. The Jacobian in 3 dimensions When changing the coordinate sstem of a triple integral I = f(,, z)dv V we need to etend the above definition of the Jacobian to 3 dimensions. For given transformations = (u, v, w), = (u, v, w) and z = z(u, v, w) the Jacobian is u v w J(u, v, w) = u v w z z z u v w Ke Point The same pattern persists as in the 2 dimensional case. Across a row of the determinant the numerators are the same and down a column the denominators are the same. 7 HELM (VERSION 1: March 18, 2): Workbook Level 1

8 The volume element dv = dddz becomes dv = J(u, v, w) dudvdw. As before the limits and integrand must also be transformed. Eample Use spherical coordinates to find the volume of a sphere of radius R. z (,, z) φ r θ Solution The change of coordinates from Cartesian to spherical polar coordinates is given b the transformation equations = r cos θ sin φ = r sin θ sin φ z = r cos φ We now need the nine partial derivatives = cos θ sin φ r = sin θ sin φ r z r = cos φ θ = r sin θ sin φ = r cos θ sin φ θ z θ = φ = r cos θ cos φ = r sin θ cos φ φ z φ = r sin φ Hence we have J(r, θ, φ) = cos θ sin φ r sin θ sin φ rcos θ cos φ sin θ sin φ rcos θ sin φ rsin θ cos φ cos φ r sin φ It is easiest to epand this determinant along the bottom row. J(r, θ, φ) =cos φ r sin θ sin φ rcos θ cos φ r cos θ sin φ rsin θ cos φ + rsin φ cos θ sin φ sin θ sin φ r sin θ sin φ rcos θ sin φ Check that this gives J(r, θ, φ) = r 2 sin φ. Notice that J(r, θ, φ) for φ π, so J(r, θ, φ) = r 2 sin φ. The limits are found as follows. The variable φ is related to latitude with φ =representing the North Pole with φ = π/2 representing the equator and φ = π representing the South Pole. HELM (VERSION 1: March 18, 2): Workbook Level 1 8

9 Solution (contd.) The variable θ is related to longitude with values of to 2π covering ever point for each value of φ. Thus limits on φ are toπ and limits on θ are to2π. The limits on r are r = (centre) to r = R (surface). To find the volume of the sphere we then integrate the volume element dv = r 2 sin φ drdθdφ between these limits. π 2π R π 2π Volume = r 2 1 π 2π sin φ drdθdφ = 3 R3 sin φ dθdφ = 3 R3 sin φ dφ = 3 πr3 Eample Find the volume integral of the function f(,, z) = over the parallelepipid with the vertices of the base at (,, z) =(,, ), (2,, ), (3, 1, ) and (1, 1, ) and the vertices of the upper face at (,, z) =(, 1, 2), (2, 1, 2), (3, 2, 2) and (1, 2, 2) z Solution This will be a difficult integral to derive limits for in terms of, and z. However, it can be noted that the base is described b z =while the upper face is described b z =2. Similarl, the front face is described b 2 z =with the back face being described b 2 z =2. Finall the left face satisfies z =while the right face satisfies z =. The above suggests a change of variable with the new variables satisfing u =2 2 + z, v =2 z and w = z and the limits on u being to, the limits on v being to 2 and the limits on w being to 2. Inverting the relationship between u, v, w and, and z, gives = 1 2 (u + v) = 1 (v + w) z = w 2 9 HELM (VERSION 1: March 18, 2): Workbook Level 1

10 Solution (contd.) The Jacobian is given b J(u, v, w) = u u z u v v z v w w z w = = 1 Note that the function f(,, z) = equals 1(u + v) 1(v + w) = 1 (u w). Thus the integral is = = = 2 2 w= v= u= 2 2 w= v= 2 2 w= v= 1 2 (u w)1 du dv dw 1 (u w) du dv dw = u= 8 (1 12 ) 2 w dv dw = [ 2w 1 2 w2 ] 2 = 2 w= 2 w= 2 [ v vw 2 v= ] 2 [ 1 16 u2 1 8 uw dw = 2 w= ] dv dw (2 w) dw = 2 Find the Jacobian for the following transformations. 1. =2u +3v w, = v 5w, z = u +w 2. = u 2 + vw, =2v + u 2 w, z = uvw 3. Clindrical polar coordinates. = ρ cos θ, = ρ sin θ, z = z z ρ (,, z) θ HELM (VERSION 1: March 18, 2): Workbook Level 1 1

11 Your solution 1. 6 Your solution 2. u 2 v 2u w + u 2 vw 2 2v 2 w Your solution 3. ρ 11 HELM (VERSION 1: March 18, 2): Workbook Level 1

12 Eercises 1. The function f = is to be integrated over an elliptical cone with base being the ellipse, 2 /+ 2 =1,z =and ape (point) at the point (,, 5). This integral has relevance in topics such as moments of inertia. The integral can be made simpler b means of the change of variables = 2(1 w)τ cos θ, =(1 w )τ sin θ, z = w. 5 5 z (,, 5) ( 2,, ) (, 1, ) (2,, ) (a) Find the limits on the variables τ, θ and w. (b) Find the Jacobian J(τ,θ,w) for this transformation. (c) Epress the integral ( ) dz d dz in terms of τ, θ and w. (d) Evaluate this integral. [Hint :- it ma be worth noting that cos 2 θ = 1 (1 + cos 2θ)] Using clindrical polar coordinates, integrate the function f = z over the volume between the surfaces z =and z = for A torus (doughnut) has major radius R and minor radius r. Using the transformation =(R + τ cos α) cos θ, =(R + τ cos α) sin θ,z = τ sin α, find the volume of the torus. [Hints :- limits on α and θ are to 2π, limits on τ are to r. Show that Jacobian is τ(r + τ cos α)]. z R θ τ α r HELM (VERSION 1: March 18, 2): Workbook Level 1 12

13 1. a) τ :to1, θ :to2π, w :to5, b) 2(1 w 5 )2 τ, c) 2 1 τ= d) 5π π π 2 Rr 2 2π θ= 5 (1 w w= 5 ) τ 3 ( cos 2 θ + sin 2 θ) dw dθ dτ, 13 HELM (VERSION 1: March 18, 2): Workbook Level 1