Double Integrals, Iterated Integrals, Cross-sections

Chapter 14 Multiple Integrals 1 Double Integrals, Iterated Integrals, Cross-sections 2 Double Integrals over more general regions, Definition, Evaluation of Double Integrals, Properties of Double Integrals 3 Area and Volume by Double Integration, Volume by Iterated Integrals, Volume between Two surfaces 4 Double Integrals in Polar Coordinates, More general Regions 5 Applications of Double Integrals, Volume and First Theorem of Pappus, Surface Area and Second Theorem of Pappus, Moments of Inertia 6 Triple Integrals, Iterated Triple Integrals 7 Integration in Cylindrical and Spherical Coordinates 8 Surface Area, Surface Area of Parametric Surfaces, Surfaces Area in Cylindrical Coordinates 9 Change of Variables in Multiple Integrals, Jacobian Matb 21 in 214-215

Chapter 14 Multiple Integrals 148 Parametric Surfaces, Surface Area Integral 149 Differentiable Coordinate Transformations, Jacobian Change of Variables in Double Integrals, Change of Variables in Triple Integrals, Parametric Surfaces, Surface Area Integral Matb 21 in 214-215

148 Surface Area Definition A parametric surface S is the image of a function or transformation r that is defined in uv-plane and has values in xyz-space, ie r : R R 2 S R 3 given by r(u, v) = ( x(u, v), y(u, v), z(u, v) ), where x, y, z are continuous functions defined in a region R in uv-plane The variables u and v are called the parameters of S In case x, y, z are smooth functions, then we call S is a smooth parametric surface Given a function z = f (x, y) defined in a region R R 2, one has a parameterized surface S from the graph of f : r(x, y) = (x, y, f (x, y) ), ie S = {(x, y, f (x, y) ) (x, y) in the domain of f } Similarly, one can also parameterize the same set S as S = {(r cos θ, r sin θ, f (r cos θ, r sin θ, ) ) (r cos θ, r sin θ) lies in region R in the polar coordinate plane } Matb 21 in 214-215

Given a smooth parametric surface S, with parametric function r(u, v) defined in a region R in uvplane, one wants to define the surface area of S First subdivide the region R into smaller rectangles R i (1 i n), each of size u v with lower lefthand corner (u i, v i ) Suppose both u and v are small, then the area S i of curvilinear figure S i can be approximated by the area P i of parallelogram with adjacent sides (in vector form) r u (u i, v i ) u and r v (u i, v i ) v P i = r u (u i, v i ) u r v (u i, v i ) v = N(u i, v i ) u v So the area of S can be approximated by the Riemann sum n i=1 S i n i=1 N(u i, v i ) u v, which converges to N(u, v) du dv Matb 21 in 214-215 R

Definition The surface area of a parametric surface S defined by r(u, v) in a uv-region R is given by N(u, v) du dv = r R R u r du dv v Moreover, the surface area element ds = N(u, v) du dv, in which the right hand side depend on the choice of parameters u and v Theorem If S is parameterized by the graph of a function of z = z(x, y) in R of xy-plane, then the surface area of S is 1 + z 2 x + z 2 y dx dy R Proof As one can parameterize the graph by r(x, y) = (x, y, z(x, y)), so r x = (1,, z x ), and r y = (, 1, z y ) So i j k N(x, y) = r x r y = (1,, z x ) (, 1, z y ) = 1 z x = ( z x, z y, 1) 1 z y Matb 21 in 214-215

Theorem If S is parameterized by a function of z = z(r, θ) in R of polar coordinate plane, then the surface area of S is r 2 + (rz r ) 2 + (z θ ) 2 dr dθ R Proof As one can parameterize the graph by r(r, θ) = (r cos θ, r sin θ, z(r, θ)), so r r = (cos θ, sin θ, z r ), and r θ = ( r sin θ, r cos θ, z θ ) i j k Then N(x, y) = r r r θ = cos θ sin θ z r r sin θ r cos θ z θ = ( z θ sin θ rz r cos θ, (z θ cos θ + rz r sin θ), r ), and hence N(r, θ) = r 2 + r 2 (z r ) 2 + (z θ ) 2 Matb 21 in 214-215

Example Find the area of the triangular surface G with vertices P(1,, ), Q(, 2, ), and Q(, 1, 1) Solution It is easy to write down the equation of the surface G x 1 y z as = 1 2 = 2(x 1) + y + z, 1 1 1 ie G is given by part of the graph of z = 2(x 1) y Project the surface G onto the xy-plane, and its image is still a triangular region with vertices A(1, ), B(, 1), and C(, 2) Recall that projection is the same as dropping the last coordinate of the point It follows that the surface area of G is given by G da = ABC 1 + z 2 x + z 2 y dxdy = 1 x= 2(1 x) 1 x 1 + 2 2 + 1 2 dydx = Matb 21 in 214-215

Example Find the area of the surface S that is the graph of z = x + y 2 for x 1, y 2 Solution It follows from z x = 1 and z y = 2y, that the surface area of S 1 2 = 1 + 1 2 + (2y) 2 dydx = 2 2 = 2 1 + ( 2y) 2 d( 2y) = 2 2 1 + 2y 2 dy 1 + u 2 du = Matb 21 in 214-215

Example If S is the part of the paraboloid z = x 2 + y 2 with z 2, find the surface area of S Solution Suppose the plane z = 2 meets the paraboloid z = x 2 + y 2 at a curve C, and let (x, y, 2) be any point on C, then x 2 + y 2 = 2, so we know that by projecting the surface S onto xy-plane, its shadow is given by the point in the circular disc D = { (x, y) x 2 + y 2 2 } Then S is the graph of the function z(x, y) = x 2 + y 2, with parametrization r(x, y) = (x, y, z(x, y)), with (x, y) D, and so z x = 2x, z y = 2y The area of surface S is given by 1 + z 2 x + z 2 y da = 1 + 4(x 2 + y 2 ) da = D 2π 2 = 2π 8 1 + 4r 2 rdrdθ = 2π 8 [ ] 2(1 + 4r 2 ) 3/2 2 3 D 2 = π 6 (33 1) = 13π 3 1 + 4r 2 d(1 + 4r 2 ) Matb 21 in 214-215

Example Find the area of the upper hemisphere S : z = a 2 x 2 y 2 of radius a Solution For any point P(x, y, z) in the surface S, we have z = a 2 x 2 y 2, ie x 2 + y 2 a 2, ie the point (x, y, ) satisfies x 2 + y 2 a 2, which means that the image of point P after projecting onto xy-plane lies on the circular disc D = { (x, y) x 2 + y 2 a 2 } It follows that the hemisphere lies above the circular disk D which can be described in terms of polar coordinates as { (r, θ) θ 2π, r a } Let z = f (x, y) = a 2 x 2 y 2 x, then z x (x, y) = = x z and y a 2 x 2 y 2 z y (x, y) = = y z So 1 + z2 x + z 2 y = 1 + x2 It follows that the area of S = 1 + z 2 x + z 2 yda D a 2π a = D a 2 x 2 y da = a 2 a 2 r rdrdθ [ 2 a a (a a 2 r 2 d(a2 r 2 ) = 2 r 2 ) 1/2 aπ = 2π 2 1/2 a 2 x 2 y 2 + y2 = x2 +y 2 +z 2 z 2 z 2 z 2 ] a = 2πa 2 = a2 z 2 Matb 21 in 214-215

Example Evaluate the surface area element of a sphere x 2 + y 2 + z 2 = a in terms of spherical coordinates Solution Let r(ϕ, θ) = ( a cos θ sin ϕ, a sin θ sin ϕ, a cos ϕ ), then r ϕ (ϕ, θ) = (a cos θ cos ϕ, a sin θ cos ϕ, a sin ϕ), and r θ (ϕ, θ) = ( a sin θ sin ϕ, a cos θ sin ϕ, ) Then N(ϕ, θ) = (a 2 cos θ sin 2 ϕ, a 2 sin θ sin 2 ϕ, a 2 sin ϕ cos ϕ), thus N(ϕ, θ) 2 = a 4 (cos 2 θ + sin 2 θ) sin 4 ϕ + a 4 sin 2 ϕ cos 2 ϕ = a 4 sin 2 ϕ Thus the area element ds = N(ϕ, θ) dϕ dθ = a 2 sin ϕ dϕ dθ Ẹxample Determine the surface area element of a sphere x 2 + y 2 + z 2 = a Solution Let R = { (ϕ, θ) ϕ π, θ 2π }, which is transformed to the sphere D given by x 2 + y 2 + z 2 = a 2 The surface area of a sphere of 2π π radius a is given by 1 ds = a 2 sin ϕ dϕ dθ = a 2 sin ϕ dϕ dθ π = 2πa 2 a 2 sin ϕ dϕ = 4πa 2 D R Matb 21 in 214-215

Example Find the area of the portion T of the unit sphere inside the cylinder x 2 + y 2 = 2 1 and z > Solution The intersection of the unit sphere and the cylinder is a circle, and the angle between the z-axis and a line from the origin to the circle is π/4 We parameterize the surface T by means of spherical coordinates as follows r(ϕ, θ) = (sin θ sin ϕ, cos θ sin ϕ, cos ϕ) where R : ϕ π/4, and θ 2π is a region in ϕθ-plane In this case, the intrinsic surface area element ds = r ϕ r θ dϕ dθ = sin ϕ dϕ dθ Note that i j k r ϕ r θ = sin θ cos ϕ cos θ cos ϕ sin ϕ cos θ sin ϕ sin θ sin ϕ = sin 2 ϕ sin θi + sin 2 ϕ cos θj sin ϕ cos ϕk And hence r ϕ r θ = sin ϕ sin 2 θ sin 2 ϕ + cos 2 θ sin 2 ϕ + cos 2 ϕ = sin ϕ Area of T = = 2π 1 ds = r ϕ r θ dϕ dθ T R π/4 sin ϕ dϕd θ = 2π [ cos ϕ] π/4 = (2 2)π Matb 21 in 214-215

Example Find the surface area of the torus generated by revolving the circle C = { (x,, z) (x b) 2 + z 2 = a 2 } ( < a < b) in the xz-plane y = around the z-axis Solution For any point P(x,, z) lying on the circle C, parameterize P by z = a sin ψ and x b = a cos ψ, where ψ 2π Let O be the projection of P onto z-axis As one rotates the vector O P about the z-axis with angle from to 2π, then the locus of the moving point P will trace out a circle of radius l = O P = b + a cos ψ In this case, we first note that z-coordinate of the moving point from P(x,, z) to Q(x, y, z), given by the same height z = a sin ψ Then under the rotation of the angle θ = PO Q from to 2π about the z-axis, it only changes the x and y coordinates of the moving point Q just like in polar coordinates, then the position of the moving point is given by r(θ, ψ) = l cos θi + l sin θj + a sin ψk = ( (b + a cos ψ) cos θ, (b + a cos ψ) sin θ, a sin ψ ), where R : θ 2π and ψ 2π is a rectangular region in θψ-plane Matb 21 in 214-215

Example Find the surface area of the torus generated by revolving the circle C : (x b) 2 + z 2 = a 2 ( < a < b) in the xz-plane around the z-axis Solution As the torus is parameterized by the following r(θ, ψ) = l cos θi + l sin θj + a sin ψk = ( (b + a cos ψ) cos θ, (b + a cos ψ) sin θ, a sin ψ ), defined in the region R = { (θ, ψ) θ 2π and ψ 2π } Then r θ = ( (b + a cos ψ) sin θ, (b + a cos ψ) cos θ, ), and r ψ = ( a sin ψ cos θ, a sin ψ sin θ, a cos ψ ) Then N(θ, ψ) = r θ r ψ = (a(b + a cos ψ) cos θ cos ψ, a(b + a cos ψ) sin θ cos ψ, a(b + a cos ψ) sin ψ ), and N(θ, ψ) = a(b + a cos ψ) The surface of a torus is given by 2π 2π N(θ, ψ) dθ dψ = a(b + a cos ψ) dθ dψ R = 2πa [bψ + a sin ψ] 2π = 4π 2 ab Matb 21 in 214-215

Example Find the area of the part of the paraboloid S : z = 9 x 2 y 2 that lies above the plane z = 5 Solution For any point P(x, y, 9 x 2 y 2 ) of the graph lying above the plane z = 5, we have 9 x 2 y 2 5, so its projection point Q(x, y) on xy-plane satisfies x 2 + y 2 2 2, which is a circular disk D of radius 2 in xy-plane Since z = 9 x 2 y 2, so z x = 2x, and z y = 2y, hence z 2 x + z 2 y = 4(x 2 + y 2 ) Then the surface area of S is 1 ds = 1 + z 2 x + z 2 y dxdy S D = 1 + 4(x 2 + y 2 ) dxdy ( in term of polar coordinates) D 2π 2 = r 1 + 4r 2 drdθ = 2π 2 1 + 4r 8 2 d(1 + 4r 2 ) [ ] 2 = π (1 + 4r 2 ) 1/2+1 = π 4 1 + 1/2 6 (93/2 1 3/2 ) = π 13 26 = 6 3 π Matb 21 in 214-215

Example Find the area of the part of the surface 2z = x 2 that lies directly above the triangle in the xy-plane with vertices at A(, ), B(1, ) and C(1, 1) Solution The surface is obviously given by the graph z(x, y) = x 2 /2, then z x = x and z y = It follows that the surface area is given by 1 x 1 1 + z 2 x + x 2 y dxdy = 1 + x 2 dydx = x 1 + x 2 dx ABC [ ] = 1 1 1 1 + x 2 2 d(1 + x 2 ) = 1 (1 + x 2 ) 1/2+1 = 1 2 1 + 1/2 3 (2 2 1) Matb 21 in 214-215

149 Change of Variables in Multiple Integrals Definition Let D be an open subset of uv-plane, and f : D R 2 be a vector-valued function defined by f (u, v) = (x(u, v), y(u, v)) Sometimes such a function f is called transformation Example Let D = { (r, θ) < r, < θ 2π} be a infinite rectangle with some boundary removed in rθ-plane, and R = R 2 \ {(, )} be the xy-plane without the origin (, ) 1 Define f : D R as follows: f (r, θ) = (r cos θ, r sin θ), which transforms the region D to R 2 Define g : D R as follows: g(r, θ) = (r 3 cos θ, r 3 sin θ), in which one can check g transforms the region D to R Matb 21 in 214-215

149 Change of Variables in Multiple Integrals Definition Let D be an open subset of uv-plane, and f : D R 2 be a vector-valued function defined by f (u, v) = (x(u, v), y(u, v)) Sometimes such a function f is called transformation (i) f is called continuously differentiable on D if the component functions x : D R and y : D R are continuously differentiable on D, ie the first order partial derivatives of x and y exist and are continuous on D (ii) f is called one-to-one, if the following condition holds: f takes different points in D to different points in R, ie If f (u, v) = f (a, b) then (u, v) = (a, b), ie (iii) Let R be a region in xy-plane, f is called from D onto R, if the following condition holds: the range of f is equal to R, ie for any point (x, y) R, one can always find a point (a, b) D such that f (a, b) = (x, y) Matb 21 in 214-215

Example Let f be a transformation defined by the equations (x, y) = f (u, v) = (u + v, v) Describe the image of the rectangular region S = { (u, v) u 2, v 1 } under transformation f Solution For those who know linear algebra, f is a linear transform, so it maps a line or a segment to a line or a segment respectively As a segment is given by AB = { toa + (1 t) OB t 1 }, it follows from f (toa + (1 t) OB) = tf (O)f (A) + (1 t) f (O)f (B) that the image of segment AB by f is also { to A + (1 t) O B t 1 }, where O, A, B are the image of O, A, B by f respectively It suffices to determine the images of 4 vertices (, ), (2, ), (2, 1), (, 1) of the rectangle S by f Remark This method works for the linear and affine transformations Matb 21 in 214-215

Definition Let D be an open subset of uv-plane, and f : D R 2 be a continuously differentiable transformation, given by f (u, v) = (x(u, v), y(u, v)) The Jacobian of f at (u, v), denoted by (x, y) x x J f (u, v) = (u, v) = u v y y = x u y v x v y u u v Example Define the transformation from the polar coordinates to rectangular coordinates by f (r, θ) = (x, y) = (r cos θ, r sin θ) f is defined on D = { (r, θ) < r, < θ < 2π } to R = R 2 \ { (x, ) x } In this case, the Jacobian of f is given by (x, y) (r, θ) cos θ r sin θ = sin θ r cos θ = r Matb 21 in 214-215

Theorem Change of Variables in Double Integral Let f : D R be a continuously differentiable transformation, and let F(x, y) be a continuous function defined on D Suppose that f is one-to-one and and maps the interior of D onto the interior of R, then F(x, y)dxdy = F(x(u, v), y(u, v)) (x, y) R D (u, v) dudv Matb 21 in 214-215

Example Let R be a parallelogram with corners (, ), (1, 2), (2, 1), (3, 3) in xy-plane Evaluate the double integral ( x + y) da, using the change of variables x = 2u + v, y = u + 2v Solution Let D = { (u, v) u 1, v 1 }, and define a transformation from D to xy-plane by (x, y) = f (u, v) = (2u + v, u + 2v) It follows that (u, v) = g(x, y) = ( 1 3 (2x y), 1 3 ( x + 2y) ) is the inverse transformation from xy-plane back to uv-plane, such that g f (u, v) = g(2u + v, u + 2v) = ( 1 3 (2(2u + v) (u + 2v)), 1 3 ( (2u + v) + 2(u + 2v)) ) = (u, v), and similarly f g(x, y) = (x, y) One can easily check that f is one-to-one from D onto its image, denoted by R The jacobian of f is given by (x,y) (u,v) = 2 1 1 2 = 3 (> ) ( x + y) dxdy = ( (2u + v) + (u + 2v) ) (x, y) (u, v) du dv = R 1 1 = 3 1 D 1 1 ( u + v) (3) du dv = 3 ( u + v) du dv ( 12 ) ( + v dv = 3 1 2 + 1 ) = 2 R Matb 21 in 214-215

Theorem Change of Variables in Triple Integral Let f : D R be a continuously differentiable transformation, defined in terms of coordinate functions by f (u, v, w) = ( x(u, v, w), y(u, v, w), z(u, v, w) ) for all (u, v, w) D Let F(x, y, z) be a continuous function defined on R Suppose that f is one-to-one and and maps the interior of D onto the interior of R, then F(x, y, z) dx dy dz R = F( x(u, v, w), y(u, v, w), z(u, v, w) ) (x, y, z) D (u, v, w) du dv dw x x x (x, y, z) (u, v, w) = u v w y y y 3 3 determinant u v w z z y y z u v y v w y w = x v w u z z x v z z + x v w w z z v w v w v w Remark Note that the sign of the jacobian will change by a negative sign if one interchanges the order of any two variables in the same list y y Matb 21 in 214-215

Example Let D = { (ρ, ϕ, θ) < ρ, < ϕ < π, < θ < 2π } be an open subset of R 3 The coordinate transformation given by spherical coordinates are given by f (ρ, ϕ, θ) = (ρ sin ϕ cos θ, ρ sin ϕ cos θ, ρ cos ϕ) for all (ρ, ϕ, θ) D One can check that f transform the subset D onto the domain R = R 3 \ { (x,, z) z R, x }, ie removing the xz-half-plane with non-negative x-coordinate from the space In this case, the Jacobian J f of f is (x, y, z) given by (ρ, ϕ, θ) = ρ2 sin θ for all (ρ, ϕ, θ) D sin ϕ cos θ ρ cos ϕ cos θ ρ sin ϕ sin θ J f = sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ cos ϕ ρ sin ϕ = cos ϕ ρ cos ϕ cos θ ρ sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ = ρ 2 sin ϕ cos 2 ϕ(cos 2 θ ( sin 2 θ)) + ρ 2 sin ϕ sin 2 ϕ(cos 2 θ ( sin 2 θ)) = ρ 2 sin ϕ ( ρ sin ϕ) sin ϕ cos θ sin ϕ sin θ ρ sin ϕ sin θ ρ sin ϕ cos θ Matb 21 in 214-215

Example Find the volume of the solid torus R obtained by revolving around the z-axis the circular disk (x b) 2 + z 2 a 2 with < a < b in the yz-plane Solution Let D = { (w, ψ, θ) < θ < 2π, < ψ < 2π, < w < a } One can parameterize the solid torus as follows: f (w, ψ, θ) = (x, y, z) = ( (b + w cos ψ) cos θ, (b + w cos ψ) sin θ, w sin ψ ), and f transforms the open subset D one-to-one and onto the solid torus R with the central circle removed Then the jacobian of f is given by (x, y, z) J f (w, ψ, θ) = = w(b + w cos ψ) w(b a) > (w, ψ, θ) Then the volume of the solid R is given by 1 dx dy dz = (x, y, z) (w, ψ, θ) dw dθ dψ R 2π 2π a = w(b + w cos ψ)dw dθ dψ 2π ( 1 = 2π 2 a2 b + 1 ) 3 a3 cos ψ dψ = 2π 2 a 2 b D Matb 21 in 214-215

Example Find the area of the region R bounded by the curves xy = 1, xy = 3, xy 14 = 1, xy 14 = 2 Solution Define a transformation from xy-plane to uv-plane as follows (u, v) = g(x, y) = ( u(x, y), v(x, y) ) = (xy, xy 14 ) Then v u = xy14 xy and x = x(u, v) = u y = u u 5 Define f (u, v) = = y4, so y = y(u, v) = ( v u ) 1/4 = v 5 v 5 = ( u 7 v 5 ) ( ) u 7 v, 5, which transforms from the rectangle D ( in v 5 v 5 uv-plane) D = [1, 3] [1, 2] = { (u, v) 1 u 3, 1 v 2} one-to-one and (x, y) onto the region R on xy-plane J f (u, v) = = x u x v = 25 (u, v) y u y v v Area of R = 1 da = (x, y) 2 3 R D (u, v) du dv = 25 du dv = 5 ln 2 1 1 v Remark Instead of trying to determine f explicitly as above, one can use ( ) (x, y) (u, v) 1 = = u 1 x u y = y x 1 1 25 (u, v) (x, y) v x v y y 14 14xy 4 = = 4xy14 v Matb 21 in 214-215 u 5,

Example Let R be the region in the first quadrant of xy-plane bounded by the hyperbolas xy = 1, xy = 9 and the lines y = x, y = 4x Use the transformation ( y x = u/v, y = uv with u >, v > to evaluate R x + ) xy dxdy Solution y x = uv u/v = v2, it follows that 1 v 2 4, and hence 1 v 2 And xy = u v uv = u2, it follows that 1 u 2 9, and hence 1 u 3 Define a transformation (u, v) = g(x, y) = ( xy, y/x), which transforms region R in xy-plane 1-1 and onto rectangle S = { (u, v) 1 u 3, 1 v 2 } in uv-plane Reversing this viewpoint by inverting the transform g to f, the region R in xy-plane can be described by the region S in uv-plane by this coordinate change (or transformation) (x, y) = f (u, v) = ( u v, uv) Moreover, the Jacobian of this transformation f is (x,y) (u,v) = x u y u = 1/v v x v y v u/v 2 u = 2u v ( y So the integral x + ) (x, y) xy dxdy = (v + u) (u, v) dudv = 2 3 1 1 R (v + u) 2u v dudv = S Matb 21 in 214-215

Example Let R be the region in xy-plane bounded by the hyperbolas xy = 1, xy = 3, x 2 y 2 = 1, and x 2 y 2 = 4 Find the polar moment of the inertia of the region R Solution Define a transformation g from xy-plane to uvplane (u, v) = g(x, y) = ( u(x, y), v(x, y) ) = (xy, x 2 y 2 ) ( ) (x, y) (u, v) 1 1 Then (u, v) = = y x (x, y) 2x 2y 1 1 = 2(x 2 + y 2 = ) 2 4u 2 + v 2 Let D = { (u, v) 1 u 3, 1 v 4 }, one can check that g : R D transform R one-to-one and onto D, so apply the inverse transform f : D R of g in integration, such that g f (x, y) = (x, y) and f g(u, v) = (u, v) The polar moment of inertia of the region R is given by (x 2 + y 2 ) dx dy = 4u 2 + v 2 du dv 4 3 R D 2 4u 2 + v = du dv = 3 2 1 1 2 Remark In, we have used the following equality: 4u 2 + v 2 = 4(u(x, y)) 2 + (v(x, y)) 2 = 4(xy) 2 + (x 2 y 2 ) 2 = (x 2 + y 2 ) 2 Matb 21 in 214-215

Example Let D be the region in the first quadrant of xy-plane bounded by the ( curve C : x2 + y ) 4 3 = xy 6 Evaluate xydxdy D Solution Define coordinate transformation (x, y) = f (r, θ) = (x(r, θ), y(r, θ)) by x = x(r, θ) = 2r cos 2 θ, y = y(r, θ) = 3r sin 2 θ Then the defining equation of the the curve C becomes r 4 = r 2 sin 2 θ cos 2 θ, ie r = sin θ cos θ, as C lies in 1st quadrant So region D = { (r, θ) < θ < π/2, < r < sin θ cos θ } in rθ-plane is transformed by f one-to-one and onto D in xy-plane Then the Jacobian of the transformation is J f (r, θ) = (x,y) = 12r sin θ cos θ > (r,θ) xydxdy = 6r sin θ cos θ J drdθ D D π/2 sin θ cos θ = 12 6r 2 sin 2 θ cos 2 θdrdθ θ= r= π/2 = 12 [ r 2 ] sin θ cos θ 6 sin 2 θ cos 2 θ dθ θ= 3 r= = 4 π/2 6 sin 4 θ cos 4 θdθ = 4 π/2 ( ) sin 2θ 4 6 dθ 2 6 π/2 = (sin 2θ) 4 dθ = 3 6 4 64 π Matb 21 in 214-215

Example The transformation x = 2u v, y = 2u + v maps the unit disk G = { (u, v) u 2 + v 2 1 } in the uv-plane to the region R in the xy-plane bounded by the ellipse 5x 2 6xy + 5y 2 = 16 Evaluate the double integral exp(1x 2 12xy + 1y 2 ) dxdy R Solution As the rule x = 2u v, y = 2u + v of change of variables are given, the jacobian factor (x,y) (u,v) = xu xv = 2 1 yu yv 2 1 = 4 Moreover, 1x 2 12xy + 1y 2 = 2(5x 2 6xy + 5y 2 ) = 2[5(2u v) 2 6(2u v)(2u + v) + 5(2u + v) 2 ] = 2[5(4u 2 4uv + v 2 ) 6(4u 2 v 2 ) + 5(4u 2 + 4uv + v 2 )] = 2[16u 2 + 16v 2 ] = 32(u 2 + v 2 ) It follows from the formula of change of variables for integral that exp(1x 2 12xy + 1y 2 ) dxdy = exp(32(u 2 + v 2 )) (x, y) R G (u, v) dudv 2π 1 = 4 exp(32 (u 2 + v 2 ) )dudv = 4 exp(32r 2 )rdrdθ u 2 +v 2 1 1 = 4 2π 2 32 r= exp(32r 2 )d(32r 2 ) = 4π [ ] 1 exp(32r 2 ) 32 = π r= 8 (e32 1) Matb 21 in 214-215