2. The area element.

Changing Variables in Multiple Integrals

1. Changing variables.

Double integrals in x, y coordinates which are taken over circular regions, or have inte- grands involving the combination x²+ y², are often better done in polar coordinates:

(1)

Z Z

R

f (x, y) dA = Z Z

R

g(r, θ) r drdθ .

This involves introducing the new variables r and θ, together with the equations relating them to x, y in both the forward and backward directions:

(2) r =px²+ y², θ = tan⁻¹(y/x); x = r cos θ, y = r sin θ . Changing the integral to polar coordinates then requires three steps:

A. Changing the integrand f (x, y) to g(r, θ), by using (2);

B. Supplying the area element in the r, θ system: dA = r dr dθ ;

C. Using the region R to determine the limits of integration in the r, θ system.

In the same way, double integrals involving other types of regions or integrands can sometimes be simplified by changing the coordinate system from x, y to one better adapted to the region or integrand. Let’s call the new coordinates u and v; then there will be equations introducing the new coordinates, going in both directions:

(3) u = u(x, y), v = v(x, y); x = x(u, v), y = y(u, v)

(often one will only get or use the equations in one of these directions). To change the integral to u, v-coordinates, we then have to carry out the three steps A, B, C above. A first step is to picture the new coordinate system; for this we use the same idea as for polar coordinates, namely, we consider the grid formed by the level curves of the new coordinate functions:

(4) u(x, y) = u⁰, v(x, y) = v⁰ .

Once we have this, algebraic and geometric intuition will usually handle steps A and C, but for B we will need a formula: it uses a determinant called the Jacobian, whose notation and definition are

(5) ∂(x, y)

∂(u, v) =    

xu xv

yu yv

    .

u=u u=u

u=u

v=v v=v v=v 0

1 2

0 1 2

Using it, the formula for the area element in the u, v-system is

(6) dA =

∂(x, y)

∂(u, v)    du dv ,

1

2

so the change of variable formula is (7)

Z Z

R

f (x, y) dx dy = Z Z

R

g(u, v)    

∂(x, y)

∂(u, v)    

du dv ,

where g(u, v) is obtained from f (x, y) by substitution, using the equations (3).

We will derive the formula (5) for the new area element in the next section; for now let’s check that it works for polar coordinates.

Example 1. Verify (1) using the general formulas (5) and (6).

Solution. Using (2), we calculate:

∂(x, y)

∂(r, θ) =    

xr xθ

yr yθ

=    

cos θ −r sin θ sin θ r cos θ

= r(cos²θ + sin²θ) = r ,

so that dA = r dr dθ, according to (5) and (6); note that we can omit the absolute value, since by convention, in integration problems we always assume r ≥ 0, as is implied already by the equations (2).

We now work an example illustrating why the general formula is needed and how it is used; it illustrates step C also — putting in the new limits of integration.

Example 2. Evaluate Z Z

R

 x − y x + y + 2

²

dx dy over the region R pictured.

Solution. This would be a painful integral to work out in rectangular coordinates.

But the region is bounded by the lines

1

-1 -1

1

(8) x + y = ±1, x − y = ±1

and the integrand also contains the combinations x − y and x + y. These powerfully suggest that the integral will be simplified by the change of variable (we give it also in the inverse direction, by solving the first pair of equations for x and y):

(9) u = x + y, v = x − y; x = u + v

2 , y = u − v 2 . We will also need the new area element; using (5) and (9) above. we get

(10) ∂(x, y)

∂(u, v) =    

1/2 1/2 1/2 −1/2

    = −1

2 ;

note that it is the second pair of equations in (9) that were used, not the ones introducing u and v. Thus the new area element is (this time we do need the absolute value sign in (6))

(11) dA = 1

2 du dv .

We now combine steps A and B to get the new double integral; substituting into the integrand by using the first pair of equations in (9), we get

(12)

Z Z

R

 x − y x + y + 2

²

dx dy = Z Z

R

 v u + 2

² 1

2 du dv .

3

In uv-coordinates, the boundaries (8) of the region are simply u = ±1, v = ±1, so the integral (12) becomes

Z Z

R

 v u + 2

² 1

2 du dv = Z ¹

−1

Z ¹

−1

 v u + 2

² 1 2 du dv We have

inner integral = − v² 2(u + 2)

u=1 u=−1

= v²

3 ; outer integral = v³ 9

¹

−1

= 2 9 .

2. The area element.

In polar coordinates, we found the formula dA = r dr dθ for the area element by drawing the grid curves r = r⁰ and θ = θ⁰ for the r, θ-system, and determining (see the picture) the infinitesimal area of one of the little elements of the grid.

rd dr dA θ

dθ

x y

For general u, v-coordinates, we do the same thing. The grid curves (4) divide up the plane into small regions ∆A bounded by these contour curves. If the contour curves are close together, they will be approximately parallel, so that the grid element will be approximately a small parallelogram, and

(13) ∆A ≈ area of parallelogram PQRS = |P Q × P R|

In the uv-system, the points P, Q, R have the coordinates

P : (u⁰, v⁰), Q : (u⁰+ ∆u, v⁰), R : (u⁰, v⁰+ ∆v) ;

to use the cross-product however in (13), we need PQ and PR in i j - coordinates.

Consider PQ first; we have

P R

u=u u=u v=v

+ +

0 0 0

∆u

∆v

∆A S

Q v=v₀

x y

(14) P Q = ∆x i + ∆y j ,

where ∆x and ∆y are the changes in x and y as you hold v = v⁰and change u⁰to u⁰+ ∆u.

According to the definition of partial derivative,

∆x ≈  ∂x

∂u



0

∆u, ∆y ≈  ∂y

∂u



0

∆u;

so that by (14),

(15) P Q ≈  ∂x

∂u



0

∆u i +  ∂y

∂u



0

∆u j .

In the same way, since in moving from P to R we hold u fixed and increase v⁰ by ∆v,

(16) P R ≈  ∂x

∂v



0

∆v i +  ∂y

∂v



0

∆v j .

4

We now use (13); since the vectors are in the xy-plane, P Q × P R has only a k -component, and we calculate from (15) and (16) that

k-component of P Q × P R ≈    

xu∆u yu∆u xv∆v yv∆v    ₀

=    

xu xv

yu yv

₀∆u∆v , (17)

where we have first taken the transpose of the determinant (which doesn’t change its value), and then factored the ∆u and ∆v out of the two columns. Finally, taking the absolute value, we get from (13) and (17), and the definition (5) of Jacobian,

∆A ≈    

∂(x, y)

∂(u, v)    ₀

∆u∆v ;

passing to the limit as ∆u, ∆v → 0 and dropping the subscript 0 (so that P becomes any point in the plane), we get the desired formula for the area element,

dA =    

∂(x, y)

∂(u, v)    

du dv .

3. Examples and comments; putting in limits.

If we write the change of variable formula as (18)

Z Z

R

f (x, y) dx dy = Z Z

R

g(u, v)    

∂(x, y)

∂(u, v)    

du dv , where

(19) ∂(x, y)

∂(u, v) =    

xu xv

yu yv

, g(u, v) = f (x(u, v), y(u, v)),

it looks as if the essential equations we need are the inverse equations:

(20) x = x(u, v), y = y(u, v)

rather than the direct equations we are usually given:

(21) u = u(x, y), v = v(x, y) .

If it is awkward to get (20) by solving (21) simultaneously for x and y in terms of u and v, sometimes one can avoid having to do this by using the following relation (whose proof is an application of the chain rule, and left for the Exercises):

(22) ∂(x, y)

∂(u, v)

∂(u, v)

∂(x, y) = 1

The right-hand Jacobian is easy to calculate if you know u(x, y) and v(x, y); then the left- hand one — the one needed in (19) — will be its reciprocal. Unfortunately, it will be in terms of x and y instead of u and v, so (20) still ought to be needed, but sometimes one gets lucky. The next example illustrates.

5

Example 3. Evaluate Z Z

R

y

x dx dy, where R is the region pictured, having as boundaries the curves x²− y²= 1, x²− y²= 4, y = 0, y = x/2 .

x -y = y=x/

x -y =^{2 2}

2 2 1 2

R 4

Solution. Since the boundaries of the region are contour curves of x²− y² and y/x , and the integrand is y/x, this suggests making the change of variable

(23) u = x²− y², v =y

x .

We will try to get through without solving these backwards for x, y in terms of u, v. Since changing the integrand to the u, v variables will give no trouble, the question is whether we can get the Jacobian in terms of u and v easily. It all works out, using (22):

∂(u, v)

∂(x, y) =    

2x −2y

−y/x² 1/x   

 = 2 − 2y²/x² = 2 − 2v²; so ∂(x, y)

∂(u, v) = 1 2(1 − v²) , according to (22). We use now (18), put in the limits, and evaluate; note that the answer is positive, as it should be, since the integrand is positive.

Z Z

R

y

x dx dy = Z Z

R

v

2(1 − v²) du dv

= Z ¹/2

0

Z ⁴

1

v

2(1 − v²) du dv

= −3

4 ln(1 − v²)

^1/2

0

= −3 4 ln 3

4 . Putting in the limits

In the examples worked out so far, we had no trouble finding the limits of integration, since the region R was bounded by contour curves of u and v, which meant that the limits were constants.

If the region is not bounded by contour curves, maybe you should use a different change of variables, but if this isn’t possible, you’ll have to figure out the uv-equations of the boundary curves. The two examples below illustrate.

Example 4. Let u = x + y, v = x − y; change Z ¹

0

Z x 0

dy dx to an iterated integral du dv.

Solution. Using (19) and (22), we calculate ∂(x, y)

∂(u.v) = −1/2, so the Jacobian factor in the area element will be 1/2.

To put in the new limits, we sketch the region of integration, as shown at the right. The diagonal boundary is the contour curve v = 0; the horizontal and vertical boundaries are not contour curves — what are their uv-equations? There are two ways to answer this; the first is more widely applicable, but requires a separate

calculation for each boundary curve. ¹

1

Method 1 Eliminatex and y from the three simultaneous equations u = u(x, y), v = v(x, y), and thexy-equation of the boundary curve. For the x-axis and x = 1, this gives







u = x + y v = x − y y = 0

⇒ u = v;







u = x + y v = x − y x = 1

⇒

u = 1 + y

v = 1 − y ⇒ u + v = 2.

6

Method 2 Solve forx and y in terms of u, v; then substitute x = x(u, v), y = y(u, v) into thexy-equation of the curve.

Using this method, we get x = ¹2(u+v), y =¹2(u−v); substituting into the xy-equations:

y = 0 ⇒ 1

2(u − v) = 0 ⇒ u = v; x = 1 ⇒ 1

2(u + v) = 1 ⇒ u + v = 2.

To supply the limits for the integration orderRR du dv, we

1. first hold v fixed, let u increase; this gives us the dashed lines shown;

2. integrate with respect to u from the u-value where a dashed line enters R (namely, u = v), to the u-value where it leaves (namely, u = 2 − v).

3. integrate with respect to v from the lowest v-values for which the dashed lines intersect the region R (namely, v = 0), to the highest such v- value (namely, v = 1).

Therefore the integral is Z ¹

0

Z ²−v v

1

2 du dv . ¹

1

v=

v=

v=v u=2-v

0

1 0

u=v

(As a check, evaluate it, and confirm that its value is the area of R. Then try setting up the iterated integral in the order dv du; you’ll have to break it into two parts.)

Example 5. Using the change of coordinates u = x²− y², v = y/x of Example 3, supply limits and integrand for

Z Z

R

dxdy

x² , where R is the infinite region in the first quadrant under y = 1/x and to the right of x²− y²= 1.

Solution. We have to change the integrand, supply the Jacobian factor, and put in the right limits.

To change the integrand, we want to express x² in terms of u and v; this suggests eliminating y from the u, v equations; we get

u = x²− y², y = vx ⇒ u = x²− v²x² ⇒ x²= u 1 − v² . From Example 3, we know that the Jacobian factor is 1

2(1 − v²); since in the region R we have by inspection 0 ≤ v < 1, the Jacobian factor is always positive and we don’t need the absolute value sign. So by (18) our integral becomes

Z Z

R

dx dxy x² =

Z Z

R

1 − v²

2u(1 − v²)du dv = Z Z

R

du dv 2u

Finally, we have to put in the limits. The x-axis and the left-hand boundary curve x²− y²= 1 are respectively the contour curves v = 0 and u = 1; our problem is the upper boundary curve xy = 1. To change this to u − v coordinates, we follow Method 1:







u = x²− y² y = vx xy = 1

⇒

u = x²− 1/x²

v = 1/x² ⇒ u = 1 v − v .

7

The form of this upper limit suggests that we should integrate first with respect to u. Therefore we hold v fixed, and let u increase; this gives the dashed ray shown in the picture; we integrate from where it enters R at u = 1 to where it leaves, at u = 1

v − v.

xy=

x -y =

1 2 2 1

u= /v-v1 u=1 P

O v=a

The rays we use are those intersecting R: they start from the lowest ray, corresponding to v = 0, and go to the ray v = a, where a is the slope of OP. Thus our integral is

Z a 0

Z ¹_/v−v

1

du dv 2u .

To complete the work, we should determine a explicitly. This can be done by solving xy = 1 and x²− y² = 1 simultaneously to find the coordinates of P . A more elegant approach is to add y = ax (representing the line OP) to the list of equations, and solve all three simultaneously for the slope a. We substitute y = ax into the other two equations, and get

 ax²= 1

x²(1 − a²) = 1 ⇒ a = 1 − a² ⇒ a =−1 +√ 5

2 ,

by the quadratic formula.

4. Changing coordinates in triple integrals

u = u(x, y, z), v = v(x, y, z) w = w(x, y, z)

but the general principles remain the same. The new coordinates u, v, and w give a three- dimensional grid, made up of the three families of contour surfaces of u, v, and w. Limits are put in by the kind of reasoning we used for double integrals. What we still need is the formula for the new volume element dV .

To get the volume of the little six-sided region ∆V of space bounded by three pairs of these contour surfaces, we note that nearby contour surfaces are approximately parallel, so that ∆V is approximately a parallelepiped, whose volume is (up to sign) the 3 × 3 determinant whose rows are the vectors forming the three edges of ∆V meeting at a corner.

These vectors are calculated as in section 2; after passing to the limit we get

(24) dV =

∂(x, y, z)

∂(u, v, w)    

du dv dw , where the key factor is the Jacobian

(25) ∂(x, y, z)

∂(u, v, w) =      

xu xv xw

yu yv yw

zu zv zw

      .

As an example, you can verify that this gives the correct volume element for the change from rectangular to spherical coordinates:

(26) x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ;

while this is a good exercise, it will make you realize why most people prefer to derive the volume element in spherical coordinates by geometric reasoning.

More on Change of Variables in Multiple Integrals

Recall: For sinlge variable, we change variables x to u in an integral by the formula (substitution rule)

Z

f (x)dx = Z

c

f (x(u)) dx du du

where x = x(u), dx =

du, and the interval changes from [a, b] to [c, d] = [x

(a), x

(b)].

Why do we do change of variables?

1. We get a simpler integrand.

2. 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 R.

(a) R is the ellipse x

+

= 1 and the transformation is x =

, y = 3v.

(b) R is the region bounded by y = −x + 4, y = x + 1, and y = x/3 − 4/3 and the transformation is x =

(u + v), y =

(u − v)

Soln:

(a) Plug the transformation into the equation for the ellipse.

( u

2 )

+ (3v)

36 = 1 u

4 + 9v

36 = 1 u

+ v

= 4

After the transformation we had a disk of radius 2 in the uv-plane.

(b)

Plugging in the transformation gives:

y = −x + 4 ⇒ 1

2 (u − v) = − 1

2 (u + v) ⇒ u = 4 y = x + 1 ⇒ 1

2 (u − v) = 1

2 (u + v) + 1 ⇒ v = −1 y = x/3 − 4/3 ⇒ 1

2 (u − v) = 1 3

1

2 (u + v) − 4/3 ⇒ v = u 2 + 2 See Fig. 1 and Fig. 2 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

1

Figure 1:

Figure 2:

The Jacobian of the transformation x = g(u, v), y = h(u, v) is:

∂(x, y)

∂(u, v) = det



∂u

∂x

∂y ∂v

∂u

∂y

∂v



= g

h

− g

h

Change of Variables for a Double Integral

Assume we want to integrate f (x, y) over the region R in the xy-plane. Under the transformation x = g(u, v), y = h(u, v), S is the region R transformed into the uv-plane, and the integral becomes

Z Z

R

f (x, y)dA = Z Z

S

f (g(u, v), h(u, v))    

∂(x, y)

∂(u, v)    

dudv

Note: 1. 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.

2. If we look just at the differentials in the above formula we can also say that dA =

∂(x, y)

∂(x, y)    

dudv

2

.

3. Here we take the absolute value of the Jacobian. The one dimensional formula is just the derivative

Example Show that when changing to polar coordinates we have dA = rdrdθ Soln:

The transformation here is x = r cos(θ), y = r sin(θ).

∂(x, y)

∂(r, θ) = det



∂r

∂x

∂y ∂θ

∂r

∂y

∂θ



= det  cos(θ) −r sin(θ) sin(θ) r cos(θ)



= r(cos

(θ) + sin

(θ)) = r

So we have dA =   

∂(x,y)

∂(r,θ)

 drdθ = |r|drdθ = rdrdθ.

Example Evaluate RR

R

x + ydA where R is the trapezoidal region with vertices given by (0, 0), (5, 0), (5/2, 5/2) and (5/2, −5/2) using the transformation x = 2u+3v and y = 2u−3v

Soln:

Figure 3:

Plugging in the transformation gives:

y = x ⇒ v = 0 y = −x ⇒ u = 0 y = −x + 5 ⇒ u = 5/4

y = x − 5 ⇒ v = 5/6

Therefore the region S in uv-plane is then a rectangle whose sides are given u = 0, v = 0, u = 5/4 and v = 5/6.

3

The Jacobian

∂(x, y)

∂(u, v) = det  2 3 2 −3



= −6 − 6 = −12

Z Z

R

x + ydA = Z

0

Z

0

((2u + 3v) + (2u − 3v))| − 12|dudv

= Z

0

Z

0

48ududv = Z

0

24u

|

5 4

0

dv = Z

0

75/2dv = 125/4 Example Compute RR

R

y

dA where R is the region bounded by xy = 1, xy = 2, xy

= 1 and xy

= 2

Soln:

0.5 1 1.5 2 2.5 3 3.5 4

0.5 1 1.5 2

xy=1 xy=2 xy²=1 xy²=2

Figure 4:

The curves intersect in 4 points:

1 = xy = xy

⇒ (1, 1) 1 = xy = xy

/2 ⇒ (1/2, 2)

2 = xy = xy

⇒ (2, 1) 1 = xy/2 = xy

⇒ (4, 1/2)

We choose a transiformation u = xy and v = xy

to transform R into a new region S by 1 ≤ u ≤ 2 and 1 ≤ v ≤ 2.

Now we solve for x and y to compute the Jacobian:

u

/v = x, v/u = y

4

∂(x, y)

∂(u, v) = det

 2u/v −u

/v

−v/u

1/u



= 1/v

Z Z

R

y

dA = Z

1

Z

v

u

· 1

v dudv = [−1/u]

· [1/2v

]

= 3/4 Note: In R

1

R

v²

u²

·

dudv, we dropped the absolute value sign for Jacobian

, since

is positive in the region we were integrating over.

Example RR

R

y

dA where R is the region in the first quadrant bounded by x

− y

= 1, x

− y

= 4, y = 0 and y = (3/5)x.

Soln:

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1 0 1 2 3 4 5

x²−y²=1 x²−y²=4 y=5/3x y=0

Figure 5:

We choose new variable to transform R 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 = 1 and u = 4.

The integrand e

is also simplified to e

.

We choose v so that we could easily solve for x and y. Let v = x + y, then u/v = x − y.

v + u/v = 2x and v − u/v = 2y The boundaries y = 0 and y = 3/5x becomes:

y = 0 ⇒ v − u/v = 0 ⇒ u = v

y = 3/5x ⇒ v − u/v = (3/5)(v + u/v) ⇒ u = (1/4)v

The Jacobian is

∂(x, y)

∂(u, v) = det

 (1/2)v −(1 − u/v

)/2

−(1/2)v (1 + u/v

)/2



= (1 + u/v

)/(4v) + (1 − u/v

)/(4v) = 1 2v

5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

u

v

v=sqrt(u) v=2sqrt(u) u=1 u=4

Figure 6:

Z Z

R

e

dA = Z Z

S

e

1

2v dA

= Z

1

Z

√u

e

1

2v dvdu = Z

1

e

2 [ln(2 √

u) − ln( √

u)]du = Z

1

e

2 ln(2)du = ln(2)

2 (e

− e) Note: In RR

S

e

dA

, we dropped the absolute value sign for Jacobian

, since

is positive in the region we were integrating over.

Triple Integrals

We start with a region R and use the transformation x = g(u, v, w), y = h(u, v, w), z = k(u, v, w), and to transform the region R into the new region S.

The Jabobian is:

∂(x, y, z)

∂(u, v, w) = det





x

x

x

y

y

y

z

z

z





The integral under this transformation is:

Z Z Z

R

f (x, y, z)dV = Z Z Z

S

f (g(u, v, w), h(u, v, w), k(u, v, w))    

∂(x, y, z)

∂(u, v, w)    

dudvdw Note: 1. 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.

2. As with double integrals,

dV =    

∂(x, y, z)

∂(u, v, w)    

dudvdw

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Example If x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ), and z = ρ cos(φ), then

= ρ

sin(φ).

Example Find the volume V of the solid ellipsoid

+

+

≤ 1 Soln:

We choose new variables u = x/a, v = y/b, w = z/c and transform the ellipsoid into a sphere F : u

+ v

+ w

≤ 1.

The Jacobian is:

∂(x, y, z)

∂(u, v, w) = det





a 0 0 0 b 0 0 0 c



 = abc

V = Z Z Z

E

1dV = Z Z Z

F

| ∂(x, y, z)

∂(u, v, w) |dV

= Z Z Z

F

abcdV

= abc 4

3 π(1)

= 4 3 πabc

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