Multivariable Calculus 515 16.9: Change of Variables in Multiple Integrals Monday, November 24 and Tuesday, November 25, 2008
Introduction Consider integrating some function, such as f( x, y) = 4x+ 8 y, over the parallelogram in the xyplane with vertices ( 1,3, ) ( 1, 3, ) ( 3, 1, ) and ( 1,5. ) This would require several separate integrals because of the shape of the domain. In this lesson, we explore a less complicated way to integrate over strangely-shaped regions.
Exercise 1 Find equations in terms of x and y for the lines that comprise the boundary of the region of integration.
Exercise 1 Find equations in terms of x and y for the lines that comprise the boundary of the region of integration. The line from ( 1,3) to (1, 3) is y= 3. x The line from (1, 3) to (3, 1) is y= x 4. The line from (3, 1) to (1,5) is y= 3x+ 8. The line from (1,5) to ( 1,3) is y= x+ 4. (A) (B) (C) (D)
Exercise 2 x= 1( u+ v) Use the substitutions 4 to find y = 1( v 3u) 4 equations in terms of u and v for the lines that comprise the boundary of the region of integration.
Exercise 2 x= 1( u+ v) Use the substitutions 4 to find y = 1( v 3u) 4 equations in terms of u and v for the lines that comprise the boundary of the region of integration. (A): v=0 (B): u=4 (C): v=8 (D): u= 4
The Jacobian Determinant (2D) For a transformation T 2 from ² to ²that takes (x,y) to (u,v) whose inverse is given by x= x( uv, ) and y= y( uv, ), we define the Jacobian determinant as ( xy, ) x u x v = ( uv, ) y u y v
The Jacobian Determinant (3D) For a transformation T 3 from ³ to ³that takes (x,y,z) to (u,v,w) whose inverse is given by x= x( uvw,, ), y= y( uvw,, ), and z= z( uvw,, ), the Jacobian is x u x v x w ( xyz,,) = y u y v y w ( uvw,, ) z u z v z w
With T 2 given by Exercise 3 1( ) x= 4 u+ v y= ( v 3u) 1 4, determinant for the transformation. find the Jacobian
With T 2 given by Exercise 3 1( ) x= 4 u+ v y= ( v 3u) 1 4, determinant for the transformation. find the Jacobian x u x v 14 14 = y u y v 34 14 = 14 14 34 14 ( )( ) ( )( ) = 116+ 316 = 14
Application To integrate a new function f( uv, ) given by making the substitutions from a transformation T 2 with a nonzero Jacobian into the original f( x, y), we evaluate ( ) ( xy, f uv, ) da ( uv, ) R uv over the new region found by transforming the boundaries of the xy-region of integration to that in the uv-plane.
Informal Justification This is a higher-dimensional analogue of the technique of integration by u-substitution. Consider substituting u= u( x) to evaluate an b integral f ( x ) dx : we find a fu ( ), rewrite the limits du using the transformation, and since du= dx dx, we du substitute for dx. We have effectively divided du dx by the size of the transformation : if we had u= 2 x, we would find du= 2 dx; since du is twice dx, we divide by two in the integral that involves fu ( ).
Informal Justification The size of a transformation of several variables is given by the Jacobian determinant, a concept which will be explored further in the semester of linear algebra.
Informal Justification For a simple example, consider the transformation u= 2x which doubles each coordinate, thus multiplying the area of the unit da by four. We thus v= 2y must divide by four to undo this. The transformation is equivalent to x= u 2, and the Jacobian is y= v 2 14. So when we multiply by that, we divide by four to undo the area increase, as we would expect.
Exercise 4 1( ) Substitute x= 4 u+ v into the integrand 1( ) y= 4 v 3u f( x, y) = 4x+ 8y to find f( uv, ). Then evaluate the new integral, including the term of the Jacobian determinant, to find described originally. R xy f x, y da ( ) with R xy as
Exercise 4 Make the given substitution, then evaluate the new integral, including the term of the Jacobian determinant, to find described originally. f x, y da ( ) with R xy as The integrand becomes 5u+ 3 v, and we have already found the limits and the Jacobian. Therefore the integral is which evaluates to 192. R xy 4 4 0 8 ( ) 5u+ 3v 1 4dvdu
Exercise 5 Use a substitution to evaluate xyda; R xy is the square with vertices (0,0), (1,1), (2,0), and (1, 1). R xy
Exercise 5 Use a substitution to evaluate xyda; R xy is the square with vertices (0,0), (1,1), (2,0), and (1, 1). Calling the vertices of the boundary A, B, C, and D, respectively, the lines that enclose the boundary are AB: y= x BC: y= 2 x CD: y= x 2 AD: y= x R xy
Exercise 5 Use a substitution to evaluate xyda; R xy is the square with vertices (0,0), (1,1), (2,0), and (1, 1). This suggests that we might wish to use the substitution u= y x 1 1, which gives x= 2u+ 2v. 1 1 v=+ y x y= 2u+ 2v The domain then becomes u [ 2,0 ] and v [ 0,2 ]. Substituting into R f( x, y) = xy gives f( uv, ) = u² + v². 4 xy
Exercise 5 Use a substitution to evaluate xyda; R xy is the square with vertices (0,0), (1,1), (2,0), and (1, 1). The Jacobian is ( xy, ) 12 12 = = 1, so the ( uv, ) 12 12 2 integral is evaluates to 0. 0 2 0 2 u² + v² 4 12 R xy dvdu which
Exercise 5 Use a substitution to evaluate xyda; R xy is the square with vertices (0,0), (1,1), (2,0), and (1, 1). We might have anticipated this because f( x, y) = xy is an odd function of y, and the region described is symmetric about the x-axis. R xy
Exercise 6 Use the transformation x= rcosθ to show that the y= rsinθ Jacobian determinant for the transformation from Cartesian to polar coordinates is r.
Exercise 6 Use the transformation x= rcosθ to show that the y= rsinθ Jacobian determinant for the transformation from Cartesian to polar coordinates is r. The Jacobian is ( xy, ) x r x θ = ( r, θ) y r y θ cosθ r sinθ = = r( cos² θ+ sin² θ) = r. sinθ r cosθ
Exercise 7 x= ρsinϕcosθ Use the transformation y = ρsinϕsinθ to show z = ρcosϕ that the Jacobian for the transformation from Cartesian to spherical coordinates is ρ²sin ϕ.
Exercise 7 Use the given transformation to show that the Jacobian for the transformation from Cartesian to spherical coordinates is ρ²sin ϕ. x ρ x θ x ϕ The Jacobian is ( xyz,,) = y ρ y θ y ϕ ( ρθϕ,, ) z ρ z θ z ϕ sinϕcosθ ρsinϕsinθ ρcosϕcosθ = sinϕsinθ ρsinϕcosθ ρcosϕsinθ cosϕ 0 ρsinϕ
Exercise 7 Use the given transformation to show that the Jacobian for the transformation from Cartesian to spherical coordinates is ρ²sin ϕ. which is easiest to evaluate by taking minors across the bottom row: cos ϕ( ρ²sinϕcosϕsin² θ ρ²sinϕcosϕcos² θ) ρsinϕ( ρsin² ϕcos² θ+ ρsin² ϕsin² θ) = ρ²sinϕcos² ϕ() 1 ρ²sin³ ϕ= ρ²sin ϕ, of which we take the absolute value to get ρ²sin ϕ.
Exercise 8 Use the transformation of the general ellipsoid x= au y = bv to find the volume z = cw x² + y² + z² = 1. a² b² c²
Exercise 8 Use the given transformation to find the volume of the general ellipsoid x² a² + y² b² + z² c² = 1. a 0 0 The Jacobian is ( xyz,,) = 0 b 0 = abc. The ( uvw,, ) 0 0 c ellipsoid s equation transforms to u² + v² + z² = 1, which is a sphere in uvw-space with radius 1 and volume 4π 3. We multiply that by the Jacobian to give the ellipsoid s volume V= 4 π abc. 3
Exercise 9 A transformation from the uv-plane to the xy-plane 3 2 is given by u= x+ y. v= x+ 4y (a) Find the Jacobian for this transformation.
Exercise 9 A transformation from the uv-plane to the xy-plane 3 2 is given by u= x+ y. v= x+ 4y (a) Find the Jacobian for this transformation. Since 2 1 x= 5u 5v, 1 3 y= 10u+ 10v the Jacobian is 2 1 5 5 1 3 10 10 =1 10.
Exercise 9 With the previously given transformation, (b) Find the image of the triangular region in the first quadrant of the xy-plane bounded by the coordinate axes and the line x+= y 1.
Exercise 9 With the previously given transformation, (b) Find the image of the triangular region in the first quadrant of the xy-plane bounded by the coordinate axes and the line x+= y 1. The region is bounded by the lines v= u 3, v= 2 u, and and v= 10 3 u.
Exercise 9 With the previously given transformation, (c) evaluate R xy 3x + 14xy+ 8y da ( 2 2) where R xy is the region in the xy-plane bounded by the lines y= x+ 1, y= x+ 3, y= x, and y= x+ 1. 3 3 1 1 2 2 4 4
Exercise 9 With the previously given transformation, (c) evaluate the given integral over the given region. The boundaries transform to, respectively, u=2, u= 6, v= 0, and v= 4, and making the substitutions into f x, y = 3 x² + 14xy+ 8 y² gives fuv (, ) = uv, so ( ) the integral is 2 6 0 4 uv 1 dvdu= 64 5. 10