Example Determine the new region that we get by applying the given transformation to the region R. (a) R is the ellipse x 2 + y2

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1 .9 Change of Variables in Multiple Integrals ecall: For sinlge variable, we change variables x to u in an integral by the formula (substitution rule) b d f(x)dx = f(x(u)) dx du du a where x = x(u), dx = dx du du, and the interval changes from [a, b to [c, d = [x (a), x (b). c Why do we do change of variables?. We get a simpler integrand.. In addition to converting the integrand into something simpler it will often also transform the region into one that is much easier to deal with. notation: We call the equations that define the change of variables a transformation. Example Determine the new region that we get by applying the given transformation to the region. (a) is the ellipse x + y = and the transformation is x = u, y = 3v. 3 (b) is the region bounded by y = x +, y = x +, and y = x/3 /3 and the transformation is x = (u + v), y = (u v) (a) Plug the transformation into the equation for the ellipse. ( u ) + (3v) 3 = u + 9v 3 = u + v = After the transformation we had a disk of radius in the uv-plane. (b) Plugging in the transformation gives: y = x + (u v) = (u + v) u = y = x + (u v) = (u + v) + v = y = x/3 /3 (u v) = 3 (u + v) /3 v = u + See Fig. and Fig. for the original and the transformed region. Note: We can not always expect to transform a specific type of region (a triangle for example) into the same kind of region. Definition

2 Figure : Figure : The Jacobian of the transformation x = g(u, v), y = h(u, v) is: [ x x (u, v) = det u v = g u h v g v h u u Change of Variables for a Double Integral Assume we want to integrate f(x, y) over the region in the xy-plane. Under the transformation x = g(u, v), y = h(u, v), S is the region transformed into the uv-plane, and the integral becomes f(x, y)da = f(g(u, v), h(u, v)) (u, v) dudv S Note:. The dudv on the right side of the above formula is just an indication that the right side integral is an integral in terms of u and v variables. The real oder of integration depends on the set-up of the problem.. If we look just at the differentials in the above formula we can also say that da = dudv v

3 . 3. Here we take the absolute value of the Jacobian. The one dimensional formula is just the derivative dx du Example Show that when changing to polar coordinates we have da = rdrdθ The transformation here is x = r cos(θ), y = r sin(θ). So we have da = [ cos(θ) r sin(θ) = det sin(θ) r cos(θ) (x,y) (r,θ) (r, θ) = det drdθ = r drdθ = rdrdθ. [ x r r x θ θ = r(cos (θ) + sin (θ)) = r Example Evaluate x+yda where is the trapezoidal region with vertices given by (, ), (5, ), (5/, 5/) and (5/, 5/) using the transformation x = u+3v and y = u 3v Plugging in the transformation gives: Figure 3: y = x v = y = x u = y = x + 5 u = 5/ y = x 5 v = 5/ Therefore the region S in uv-plane is then a rectangle whose sides are given u =, v =, u = 5/ and v = 5/. 3

4 The Jacobian [ 3 (u, v) = det 3 = = x + yda = = ((u + 3v) + (u 3v)) dudv 8ududv = u 5 dv = 75/dv = 5/ Example Compute y da where is the region bounded by xy =, xy =, xy = and xy = xy= xy= xy = xy = The curves intersect in points: Figure : = xy = xy (, ) = xy = xy / (/, ) = xy = xy (, ) = xy/ = xy (, /) We choose a transiformation u = xy and v = xy to transform into a new region S by u and v. Now we solve for x and y to compute the Jacobian: u /v = x, v/u = y

5 y da = [ u/v u (u, v) = det /v v/u /u = /v v u v dudv = [ /u [/v = 3/ Note: In v dudv, we dropped the absolute value sign for Jacobian, since is u v v v positive in the region we were integrating over. Example y da where is the region in the first quadrant bounded by x y =, x y =, y = and y = (3/5)x. 5 x y = x y = y=5/3x y= Figure 5: We choose new variable to transform into a simpler region. Let u = x y = (x y)(x + y). Then two of the boundary curves for the new region S are u = and u =. The integrand e x y is also simplified to e u. We choose v so that we could easily solve for x and y. Let v = x + y, then u/v = x y. The boundaries y = and y = 3/5x becomes: The Jacobian is (u, v) = det v + u/v = x and v u/v = y y = v u/v = u = v y = 3/5x v u/v = (3/5)(v + u/v) u = (/)v [ (/)v ( u/v )/ (/)v ( + u/v )/ = ( + u/v )/(v) + ( u/v )/(v) = v 5

6 v v=sqrt(u) v=sqrt(u) u= u= u Figure : e x y da = e u [ln( u) ln( u)du = S e u v da = e u u u e u v dvdu = ln()du = ln() (e e) Note: In S eu v da, we dropped the absolute value sign for Jacobian, since v positive in the region we were integrating over. Triple Integrals We start with a region and use the transformation x = g(u, v, w), y = h(u, v, w), z = k(u, v, w), and to transform the region into the new region S. The Jabobian is: (u, v, w) = det x u x v x w y u y v y w z u z v z w v is The integral under this transformation is: f(x, y, z)dv = f(g(u, v, w), h(u, v, w), k(u,v, w)) (u, v, w) dudvdw S Note:. dudvdw on the right hand side of the above formula is just an indication that the right hand side integral is an integral in terms of u, v and w variables. The real oder of integration depends on the set-up of the problem.. As with double integrals, dv = (u, v, w) dudvdw

7 Example If x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ), and z = ρ cos(φ), then (x,y,z) (ρ,φ,θ) = ρ sin(φ). Example Find the volume V of the solid ellipsoid x + y + z a b c We choose new variables u = x/a, v = y/b, w = z/c and transform the ellipsoid into a sphere F: u + v + w. The Jacobian is: V = E dv = (u, v, w) = det F a b c = abc (u, v, w) dv = abcdv = abc F 3 π()3 = 3 πabc 7