Lesson 9 Change of Variables to Compute Double Integrals
Example : A Familiar Double Integral Use a double integral to calculate the area of a circle of radius centered at the origin: da The region is given by x y 6 : 6 x 6 x da Now da means "a small change in area" in -coordinates. We can measure a small change in area with a little rectangle. 6 x 6 x dy dx dy da dx So da = dy dx:
Example : A Familiar Double Integral This integral is actually pretty complicated to evaluate: 6 x dy dx 6 x dx 6 x x 6 x 6 sin 6 What a mess! There must be a better way to use integration to do this... What coordinate system would you prefer to use to when dealing with a region like?
Example : An Attempt at a Change of ight! Let's try polar coordinates: x(r, t) r cos(t) y(r, t) r sin(t) Variables In this coordinate system, the region is described quite elegantly with 0 r and 0 t. We will try to (naively) apply this to our double integral: da r(x, y) x y t(x,y) tan y x (quadrant adjusted) 0 0 0 8 drdt r 0 dt Oh no! 8is not the right Created answer!! by Christopher Grattoni. All We rights reserved. wanted 6...
Detour: Working Out the Change of Variables the ight Way Our misconception was that da dr dt. Our da refers to a small change in -area, not rt-area. We need to delve a bit deeper to sort this out: 6x For dy dx our region was a circle, while dr dt looks more 6 x 0 0 like the integral you'd take for a rectangle: x(r,t) r cos(t) y(r,t) r sin(t) r(x,y) x y t(x,y) tan y x
Detour: Working Out the Change of Variables the ight Way Let's analyze the transformation T(r,t) and verify that the picture shown below makes sense: For the orange red green blue line, line, t Let's r t 0,, map 0, 0 a r few t r points : :: just for practice: x(r,0) x(0,t) x(,t) x(r, r 0 ) rcos t y(r,0) y(0,t) y(r, y(,t) ) 0 0sint rcost,r sint piece by piece This whole process can be called a mapping, a transformation, a change of variables, or a change of coordinates.
Detour: Working Out the Change of Variables the ight Way Try the Mathematica applet I made to see how a small region in rt-space maps into -space: This is called a non-area-preserving map. rt-rectangles map into -sectors. Notice that this size of the sectors varies while the size of the rectangles doesn't... Now we are ready to start thinking about how a small change in area on the rectangular region relates Created to by Christopher a change Grattoni. All rights in reserved. area on the circular region.
Detour: Working Out the Change of Variables the ight Way Now imagine a very small rectangle in the rt-plane: (r, t t) (r r, t t) T(r r, t t) T(r, t t) (r,t) (r r, t) T(r, t) T(r r, t) That is, consider the limit as A, r, and t tend to 0: T(r r, t) T(r, t) T(r r, t) T(r, t) r r T r r T(r, t t) T(r, t) T(r, t t) T(r, t) t t These approximations using partial derivatives let us approximate our sector using a parallelogram: T t t
Detour: Working Out the Change of Variables the ight Way Now imagine a very small rectangle in the rt-plane: (r, t t) (r r, t t) T(r r, t t) (r,t) (r r, t) Theorem: The area of the parallelogram formed by two vectors V and W equals V W. T T A r t r t T T r t r t T(r, t t) T t t T(r, t) (We can take a scalar out of T r either vector in a cross product) r T(r r, t) i j k T r T t T r T 0 rt t 0 T r T t T r T t rt
The Area Conversion Factor: If we let r and t tend to zero, we can use this to get da T r T t T r d rdt T t T T T T r r is called the Jacobian matrix, and r r T T T T t t t t T T You can let A (r,t) r r. Think of the Jacobian determinant as an T T Hence, f(x,y)da f(x(r,t),y(r,t)) A (r,t) dr dt Created by rt Christopher Grattoni. All rights reserved. is called the Jacobian determin ant. t t Area Conversion Factor that lets us compute an -space integral in rt-space:
The Area Conversion Factor: f(x,y)da f(x(r,t),y(r,t)) A (r,t) dr dt rt Let T(r, t) be a transformation from rt-space to -space. That is, T(r,t) T (r,t),t (r,t) (x(r,t), y(r, t)). Then A T T (r,t) r r. T T t t Note: Since we derived A (r, t) as the magnitude of a cross product, we need it to be positive. This is why we Created put by Christopher absolute Grattoni. All rights value reserved. bars into the formula above.
Example 3: Fixing Example emember we wanted da for a circle of radius centered at (0,0). T(r, t) (x(r, t), y(r, t)) (r cos(t),r sin(t)) A (r, t) x y r r x y t t cos(t) sin(t) r sin(t) r cos(t) rcos (t) rsin (t) r cos (t) sin (t) r da 0 0 r r drdt 0 0 0 8 dt 6 dt Phew! We did it!
Summary of Change of Variables for Polar Coordinates T(r,t) (x(r,t),y(r,t)) (rcos(t),rsin(t)) A (r,t) r f(x,y)da f(x(r,t),y(r,t)) r dr dt rt
Example : More With Polar Coordinates By now you probably already asked yourself why this change of variables is useful. This was just a circle of course! We could have used A = πr or the Gauss-Green formula. Hence, we should look at an example where a double integral in -coordinate space would be horribly messy, the boundary region is hard to parameterize for Gauss-Green, and we can t just plug into a familiar area formula
Example : More With Polar Coordinates Compute x y da for the region satisfying the following inequalities: This would have been HOIBLE in rectangular coordinates. But using polar coordinates, x r cos(t) and y r sin(t) this is the rt-rectangle with 0 t and r 5 : x 5 0 5 x y 5, x y, y 0 y da r cos(t) r sin(t) r dr dt 0 0 3 r dr dt 609 dt Substitute x r cos(t) and y r sin(t) 609
Analogy Time: This whole change of variables thing isn't just for polar coordinates. We can change from -space to any uv-coordinate space we want: f(x,y)da f(x(u,v),y(u,v)) A (u,v) du dv uv This should be reminiscent of change of variables for single-variable calculus: x(b) x(a) f(x) dx b a f(x(u)) x'(u) du So you can think of the Jacobian Determinant, A (u,v), as a higher-dimensional analogue of x'(u) in our old friend, u-substitution. Let's try one in a single variable:
Example 5: Change of Variables for Single Variable Calculus Compute e e 6 6 ln(x) x dx. f(x(u)) x'(u) du 6 6 u u f(e ) e du u u u e du e u 6 6 x(b) x(a) f(x) dx f(x) x(u) b a x'(u) f(x(u)) x'(u) du ln(x) It works! This is the same as your more familiar method: 6 u du 6 e u e x u u du ln(x) x dx
Note for the Literacy Sheet ow imagine a very small rectangle in the uv-plane: u,v v) (u u, v v) T(u u, v v) u,v) (u u, v) Theorem: The area of the parallelogram formed by two vectors V and W equals V W. T T A u v u v T T u v u v T(u, v v) T (v b) v (a,b) (We can take a scalar out of T (u a) u either vector in a cross product) T(u u, v) i j k T u T v T u T 0 uv v 0 T u u u v T T v T v
A Note for the Literacy Sheet ow imagine a very small rectangle in the uv-plane: u,v v) (u u, v v) T(u u, v v) T(u, v v) u,v) (u u, v) T (v b) v (a,b) T (u a) u T(u u, v) We are using linearizations to approximate the area of the sector by creating a parallelogram. You will be responsible for explaining how this works on your literacy sheet. Make sure you read B. and B.3 in the Basics.
Example 6: Beyond Polar Coordinates ompute y da for the region given by the ellipse y. Let's let x ucos(v ) and y usin(v) for 0 v and 0 u. These are NOT polar coordinates, so we need to compute A (u, v) : x A (u, v) x u x v y u y v cos(v) sin(v) usin(v) ucos(v) ucos (v) usin (v) u cos (v) sin (v) u
Example 6: Beyond Polar Coordinates ompute y da for the region given by the ellipse y. Let's let x ucos(v ) and y usin(v) for 0 v and 0 u. These are NOT polar coordinates, so we need to compute A (u, v) : x y A (u, v) u u x y v v cos(v) sin(v) usin(v) ucos(v) ucos (v) usin (v) u In general, you will start to notice that a u cos (v) sin (v) If x(u, v) u cos(v), then x x, x gradx[u, v]. u v If y( u, v) usin(v), then y y, u v y grady[u, v]. So you can think of the area conversion factor as a determinant of a pair of gradient vectors from x(u,v) and y(u,v). change of variables can often be found by parameterizing your curve/surface/etc.
Example 6: Beyond Polar Coordinates Compute y da for the region given by the ellipse y. y da u sin (v) u du dv 0 0 3 u sin (v) du dv 0 0 u u sin (v) dv sin (v)dv 0 u0 0 x Substitute x ucos( v) and y usin(v).
Example 7: Mathematica-Aided Change of Variables (Parallelogram egion) Use Mathematica to compute These lines are given by y x, y x, 3 y x,and y x 6. We can rewrite them as y x, y y e da for given by the parallelogram: 3 x y,and x y 6. 3 Then we can let u yx and v x y for u and v 6! But to compute our integral, we need the map from uv-space to -space... That is, we have u(x, y) and v(x,y), but we need x(u, v) an dy(u,v). Let Mathematica do the work: Solve[ { u u[ x, y], v v[ x, y]},{ x, y}] x (u v) and y (u v) 5 5 x,
Example 7: Mathematica-Aided Change of Variables (Parallelogram egion) Use Mathematica to compute These lines are given by y x, y x, 3 y x,and y x 6. We can rewrite them as y x, yx, y e da for given by the parallelogram: What is this Mathematica command doing? What will YOU do on a quiz/test without Mathematica? 3 x y,and x y 6. 3 Then we can let u yx and v x y for u and v 6! But to compute our integral, we need the map from uv-space to -space... That is, we have u(x, y) and v(x,y), but we need x(u, v) an dy(u,v). Let Mathematica do the work: Solve[ { u u[ x, y], v v[ x, y]},{ x, y}] x (u v) and y (u v) 5 5
Example 7: Mathematica-Aided Change of Variables (Parallelogram egion) Use Mathematica to compute x(u, v) (u v ) and y( u, v) (u v) 5 5 y e da for given by the parallelogram: A (u, v) x u x v y u y v 5 5 5 5 5 So we use 5
Example 7: Mathematica-Aided Change of Variables (Parallelogram egion) Use Mathematica to compute x(u, v) (u v ) and y( u, v) (u v) 5 5 A (u, v) x u x v y u y v 5 5 5 5 5 If x(u, v) (u v), then 5 x x, x gradx[u, v]. u v y e da for given by the parallelogram: If y(u, v) (u v), then 5 y y, y grady[u, v]. u v So you can think of the area conversion factor as a determinant of a pair of gradient vectors from x(u,v) and y(u,v). So we use 5 Again, you can see that our change of variables is found by parameterizing our region (in this case, a parallelogram).
Example 7: Mathematica-Aided Change of Variables (Parallelogram egion) Use Mathematica to compute y e da for given by the parallelogram: A (u,v), x (u v), y ( u v), 5 5 5 3 for u and v 6 : e y da 6 3/ (u v)/5 e du dv 5 7. (from Mathematica)
An Important Connection A parameterization (x(u,v),y(u,v)) for a region with a u b and c v d is actually a map between a uv-rectangle,, and your messy -region, : uv uv (uv-rectangle) (x(u,v),y(u,v)) Long story short: A parameterization IS a change of coordinates. If you can parameterize a region, you can (probably) integrate over it.