MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora - - PowerPoint PPT Presentation

Multiple Integrals

MA102: Multivariable Calculus

Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Maxima/Minima

Let f : E ⊆ Rn → R. If f is continuous on E and E is a closed & bounded, then f attains its maximum and minimum value on E. How to find the (extremum) points at which f attains the maximum value or the minimum value on E? Before finding answer to this question, we formally define maximum and minimum of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f. A point X ∗ ∈ E is said to be a point of relative/local maximum of f if there exists a r > 0 such that f(X) ≤ f(X ∗) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local maximum of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f. A point X ∗ ∈ E is said to be a point of relative/local maximum of f if there exists a r > 0 such that f(X) ≤ f(X ∗) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local maximum of f. A point X ∗ ∈ E is said to be a point of absolute/ global minimum of f if f(X ∗) ≤ f(X) for all X ∈ E.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f. A point X ∗ ∈ E is said to be a point of relative/local maximum of f if there exists a r > 0 such that f(X) ≤ f(X ∗) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local maximum of f. A point X ∗ ∈ E is said to be a point of absolute/ global minimum of f if f(X ∗) ≤ f(X) for all X ∈ E. A point X ∗ ∈ E is said to be a point of absolute/ global maximum of f if f(X) ≤ f(X ∗) for all X ∈ E.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Extremum Points and Extremum Values

A point X ∗ ∈ E is said to be a point of extremum of f if it is either (local/global) minimum point or maximum point of f. The function value f(X ∗) at the extremum point X ∗ is called an extremum value of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Absolute/Global Extremum

f(x, y) = −x2 − y2 has an absolute maximum at (0, 0) in R2. f(x, y) = x2 + y2 has an absolute minimum at (0, 0) in R2.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Relative/Local Extremum and Saddle Point

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Critical Points

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a critical point of f if

• either

∂f ∂x1 (X ∗) = ∂f ∂x2 (X ∗) = · · · = ∂f ∂xn (X ∗) = 0 ,

• or at least one of the first order partial derivatives of f

does not exist. The function value f(X ∗) at the critical point X ∗ is called a critical value of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Critical Points: Examples

The point (0, 0) is the critical point of the function f(x, y) = x2 + y2 and the critical value is 0 corresponding to this critical point. The point (0, 0) is a critical point of the function h(x, y) =

• x sin(1/x) + y

if x = 0 , y if x = 0 . Here, hx(0, 0) does not exist and hy(0, 0) = 1.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Saddle Points

Let X ∗ be a critical point of f. If every neighborhood N(X ∗) of the point X ∗ contains points at which f is strictly greater than f(X ∗) and also contains points at which f is strictly less than f(X ∗). That is, f attains neither relative maximum nor relative minimum at the critical point X ∗. Example: Let f(x, y) = y2 − x2 for (x, y) ∈ R2. Then (0, 0) is a saddle point of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

To find the extremum points of f, where to look for?

If f : E ⊆ Rn → R, then where to look in E for extremum values of f? The maxima and minima of f can occur only at

• boundary points of E,
• critical points of E
• interior point of E where all the first order partial derivatives of f are

zero,

• interior point of E where at least one of the first order partial

derivatives of f does not exist.

Example: In the closed rectangle R = {(x, y) ∈ R2 : −1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1}, the function f(x, y) = x2 + y2 attains

• its minimum value at (0, 0),
• its maximum value at (±1, ±1).
• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Necessary Condition for Extremum

Let f : E ⊆ Rn → R. If an interior point X ∗ of E is a point

• f relative/absolute extremum of f, and if the first order

partial derivatives of f at X ∗ exists then ∂1f(X ∗) = · · · = ∂nf(X ∗) = 0. That is, the gradient vector at X ∗ is the zero vector. Further, the directional derivative of f at X ∗ in all directions is zero, if f is differentiable at X ∗.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Quadratic Forms

Definition

Let H = [aij] be an n × n symmetric matrix. A function of the form Q(X) = X T H X for X T = (x1, x2, · · · , xn) ∈ Rn from Rn into R is called a quadratic form (or bilinear form). Examples of Quadratic Forms: Q(x1, x2) = (x1, x2)

• a

b/2 b/2 c x1 x2

• = ax2

1 + bx1x2 + cx2 2 .

Q(x1, x2, x3) = (x1, x2, x3)   a d/2 f/2 d/2 b e/2 f/2 e/2 c     x1 x2 x3   = ax2

1 + bx2 2 + cx2 3 + dx1x2 + ex2x3 + fx3x1 .

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Classification of quadratic forms Q

• If Q(X) > 0 for all X = 0, then Q is said to be positive definite.
• If Q(X) < 0 for all X = 0, then Q is said to be negative definite.
• If Q(X) > 0 for some X and Q(X) < 0 for some other X, then Q

is said to be indefinite.

• If Q(X) ≥ 0 for all X and Q(X) = 0 for some X = 0, then Q is

said to be positive semidefinite.

• If Q(X) ≤ 0 for all X and Q(X) = 0 for some X = 0, then Q is

said to be negative semidefinite. All the above terms used to describe quadratic forms Q can also be applied to the corresponding symmetric matrices H.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Examples

• Positive Definite: Q(x1, x2) = x2

1 + x2 2

• Negative Definite: Q(x1, x2) = −x2

1 − x2 2

• Indefinite: Q(x1, x2) = x2

1 − x2 2,

Reasons: Q(1, 0) = 1 > 0 and Q(0, 1) = −1 < 0

• Positive Semidefinite: Q(x1, x2) = x2

2,

Reasons: Q(X) ≥ 0 for all X and Q(1, 0) = 0.

• Negative Semidefinite: Q(x1, x2) = −x2

2,

Reasons: Q(X) ≤ 0 for all X and Q(1, 0) = 0.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Classifying Quadratic Forms from the nature of Eigenvalues

Theorem

Let Q(X) = X T H X for X T = (x1, x2, · · · , xn) ∈ Rn where H is a n × n symmetric matrix.

• If all the eigenvalues of H are positive, then Q (and H) is positive

definite.

• If all the eigenvalues of H are negative, then Q (and H) is

negative definite.

• If H has both positive and negative eigenvalues, then Q (and H)

is indefinite.

• If all eigenvalues of H are non-negative (≥ 0), then H is positive

semidefinite.

• If all eigenvalues of H are non-positive (≤ 0), then H is negative

semidefinite.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Examples

• Positive Definite:

  1 2 3  , Eigenvalues are 1, 2, 3.

• Negative Definite:

  −1 −2 −3  , Eigenvalues are −1, −2, −3.

• Indefinite:

  1 −2 −3  , Eigenvalues are 1, −2, −3.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Examples (Continuation)

• Positive Semidefinite:

  1 2  , Eigenvalues are 1, 2, 0.

• Negative Semidefinite:

  −1 −2  , Eigenvalues are −1, −2, 0.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Results for 2 × 2 Real Symmetric Matrices

Let H =

• a

b b c

• be a 2 × 2 symmetric matrix. Then H is
• positive definite if det H = ac − b2 > 0 and a > 0;
• negative definite if det H = ac − b2 > 0 and a < 0;
• indefinite if det H = ac − b2 < 0;

Example:

• Positive Definite:
• 2

3 3 8

• , det H = 7 > 0 and a = 2 > 0.
• Negative Definite:
• −2

3 3 8

• , det H = 7 > 0 and a = −2 < 0.
• Indefinite:
• 2

−3 −3 1

• , det H = −7 < 0.
• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Second Derivative Test

Motivation/Idea: When f : R2 → R, the Taylors Theorem (upto 2-nd derivative term) takes the form f(X0 + X) − f(X0) = (xfx(X0) + yfy(X0)) +

• x2 fxx(P) + 2xy fxy(P) + y2 fyy(P)
• for X ∈ N(X0).

Here, blue color terms are: Q(X) = X THX Nature of Hessian matrix = ⇒ Nature of Critical Point

Theorem

Let f : E ⊆ Rn → R and let X0 be an interior point of E. Suppose that all the second order partial derivatives of f exist and continuous at X0 and X0 is a critical point of f. Let H denote the Hessian matrix of f.

• If H is positive definite, then f has a local minimum at X0.
• If H is negative definite, then f has a local maximum at X0.
• If H is indefinite, then f has a saddle point at X0.
• If H is semidefinite, then the test is inconclusive.
• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Case n = 2: f : E ⊆ R2 → R

Let f : E ⊆ R2 → R and let X0 be an interior point of E. Suppose that all the second order partial derivatives of f exist and continuous at X0 and X0 is a critical point of f. Then, the Hessian matrix of f is given by H = fxx(X0) fxy(X0) fxy(X0) fyy(X0)

• .
• If H is positive definite (that is, fxx > 0 and fxxfyy − f 2

xy > 0 at X0),

then f has a local minimum at X0.

• If H is negative definite (that is, fxx < 0 and fxxfyy − f 2

xy > 0 at X0),

then f has a local maximum at X0.

• If H is indefinite (that is, fxxfyy − f 2

xy < 0 at X0), then f has a

saddle point at X0.

• If H is semidefinite (that is, fxxfyy − f 2

xy = 0 at X0), then the test is

inconclusive.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example 1

Let f(x, y) = x2 − 2xy + y3

3 − 3y. Determine at which points f have

relative extremum values and at which points f has saddle points. Step 1: Finding Critical Points: Observe that fx(x, y) = 2x − 2y and fy(x, y) = −2x + y2 − 3. The critical points of f are the solutions of fx(x, y) = 0 and fy(x, y) = 0. fx(x, y) = 2x − 2y = 0 and fy(x, y) = −2x + y2 − 3 = 0 = ⇒ (x, y) = (3, 3) or (−1, −1) . Thus, the points (3, 3) and (−1, −1) are critical points of f. Step 2: Determining Nature of Each Critical Point: At (3, 3), fxxfyy − f 2

xy = 8 > 0 and fxx = 2 > 0. Therefore, f has a

relative minimum value at the critical point (3, 3). At (−1, −1), fxxfyy − f 2

xy = −8 < 0. Therefore, the function f has a

saddle point at the critical point (−1, −1).

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example 2

Let f(x, y, z) = 1 − x2 − 2y2 − 3z2 for (x, y, z). Determine all critical points of f in R3 and find its nature. Step 1: Finding Critical Points of f fx = −2x = 0, fy = −4y = 0 and fz = −6z = 0 implies that (0, 0, 0) is the critical point of f. Step 2: Determining the nature of each Critical Point At (0, 0, 0), the Hessian matrix H =   −2 −4 −6  . The eigenvalues of H are −2, −4 and −6. Since all eigenvalues of H are negative, H is negative definite. Hence (0, 0, 0) is a local maximum point of f.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example 3

Find the maxima, minima and saddle points of f(x, y) := (x2 − y2)e−(x2+y2)/2.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example 3

Find the maxima, minima and saddle points of f(x, y) := (x2 − y2)e−(x2+y2)/2. We have fx = [2x − x(x2 − y2)]e−(x2+y2)/2 = 0, fy = [−2y − y(x2 − y2)]e−(x2+y2)/2 = 0, so the critical points are (0, 0), (± √ 2, 0) and (0, ± √ 2). Point fxx fxy fyy det(H) Type — (0, 0) 2 −2 −4 saddle ( √ 2, 0) −4/e −4/e 16/e2 maximum (− √ 2, 0) −4/e −4/e 16/e2 maximum (0, √ 2) 4/e 4/e 16/e2 minimum (0, − √ 2) 4/e 4/e 16/e2 minimum

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Global extrema

Let f : E ⊆ Rn → R be continuous. If E is a closed & bounded, then it is known that f attains its maximum and minimum value on E. Find global maximum and global minimum of the function f : [−2, 2]×[−2, 2] → R given by f(x, y) := 4xy −2x2 −y4.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Global extrema

Let f : E ⊆ Rn → R be continuous. If E is a closed & bounded, then it is known that f attains its maximum and minimum value on E. Find global maximum and global minimum of the function f : [−2, 2]×[−2, 2] → R given by f(x, y) := 4xy −2x2 −y4. To find global extrema, find extrema of f in the interior and then on the boundary.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Global extrema

Let f : E ⊆ Rn → R be continuous. If E is a closed & bounded, then it is known that f attains its maximum and minimum value on E. Find global maximum and global minimum of the function f : [−2, 2]×[−2, 2] → R given by f(x, y) := 4xy −2x2 −y4. To find global extrema, find extrema of f in the interior and then on the boundary. Solving fx = 4y − 4x = 0 and fy = 4x − 4y3 = 0 we obtain the critical points (0, 0), (1, 1) and (−1, −1). We have f(1, 1) = f(−1, −1) = 1. (0, 0) is a saddle point.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Global extrema

For the boundary, consider f(x, 2), f(x, −2), f(2, y), f(−2, y) and find their extrema on [−2, 2]. The global minimum is attained at (2, −2) and (−2, 2) with f(2, −2) = −40. The global maximum is attained at (1, −1) and (−1, 1). Example: Find the absolute maxima and minima of f(x, y) = 2 + 2x + 2y − x2 − y2 on the triangular region in the first quadrant bounded by the lines x = 0, y = 0, y = 9 − x. We have fx = 2 − 2x = 0, fy = 2 − 2y = 0 which implies that x = 1, y = 1 is the only critical point (local maxima).

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

We next study the behaviour of f on the boundary which is a triangle. Case 1. On the segment y = 0, ϕ(x) := f(x, 0) = 2 + 2x − x2 is defined on I = [0, 9]. We have ϕ(0) = f(0, 0) = 2, ϕ(9) = f(9, 0) = −61. On (0, 9), ϕ′(x) = 2 − 2x = 0 gives x = 1. Thus, x = 1 is the

• nly critical point of ϕ(x) in (0, 9) and

ϕ(1) = f(1, 0) = 3. Case 2. On the segment x = 0, ψ(y) = f(0, y) = 2 + 2y − y2 and ψ′(y) = 2 − 2y = 0 implies y = 1 and ψ(1) = f(0, 1) = 3. Case 3. On the segment y = 9 − x, we have f(x, 9 − x) = −61 + 18x − 2x2 and the critical point is x = 9/2. At this point f(9/2, 9/2) = −41/2. Finally, f(0, 0) = 2, f(9, 0) = f(0, 9) = −61.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Constrained extrema of f : Rn → R

In the earlier two examples, the boundary of the domains are lines in R2. We now discuss a method to deal with more general boundary. Let U ⊂ Rn be open and f, g : U ⊂ Rn → R be continuous. Then Maximize or Minimize f(X) Subject to the constraint g(X) = α.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Constrained extrema of f : Rn → R

In the earlier two examples, the boundary of the domains are lines in R2. We now discuss a method to deal with more general boundary. Let U ⊂ Rn be open and f, g : U ⊂ Rn → R be continuous. Then Maximize or Minimize f(X) Subject to the constraint g(X) = α. Example: Find the extreme values of f(x, y) = x2 − y2 on the circle x2 + y2 = 1. It turns out that f attains minimum at (0, ±1) and maximum at (±1, 0) although ∇f(0, ±1) = 0 and ∇f(±1, 0) = 0.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Test for constrained extrema of f : R2 → R

Theorem: Let f, g : U ⊂ R2 → R be C1. Suppose that f has an extremum at (a, b) ∈ U such that g(a, b) = α and that ∇g(a, b) = (0, 0). Then there is a λ ∈ R, called Lagrange multiplier, such that ∇f(a, b) = λ∇g(a, b).

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Test for constrained extrema of f : R2 → R

Theorem: Let f, g : U ⊂ R2 → R be C1. Suppose that f has an extremum at (a, b) ∈ U such that g(a, b) = α and that ∇g(a, b) = (0, 0). Then there is a λ ∈ R, called Lagrange multiplier, such that ∇f(a, b) = λ∇g(a, b). Proof: Let r(t) be a local parametrization of the curve g(x, y) = α such that r(0) = (a, b). Then f(r(t)) has an extremum at t = 0. Therefore df(r(t)) dt |t=0 = ∇f(a, b) • r ′(0) = 0.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Test for constrained extrema of f : R2 → R

Theorem: Let f, g : U ⊂ R2 → R be C1. Suppose that f has an extremum at (a, b) ∈ U such that g(a, b) = α and that ∇g(a, b) = (0, 0). Then there is a λ ∈ R, called Lagrange multiplier, such that ∇f(a, b) = λ∇g(a, b). Proof: Let r(t) be a local parametrization of the curve g(x, y) = α such that r(0) = (a, b). Then f(r(t)) has an extremum at t = 0. Therefore df(r(t)) dt |t=0 = ∇f(a, b) • r ′(0) = 0. Now g(r(t)) = α ⇒ ∇g(a, b) • r ′(0) = 0. This shows that r ′(0) ⊥ ∇g(a, b) and r ′(0) ⊥ ∇f(a, b). Hence ∇f(a, b) = λ∇g(a, b) for some λ ∈ R.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Method of Lagrange multipliers for f : R2 → R

To find extremum of f subject to the constraint g(x, y) = α, define L(x, y, λ) := f(x, y) − λ(g(x, y) − α) and solve the equations

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Method of Lagrange multipliers for f : R2 → R

To find extremum of f subject to the constraint g(x, y) = α, define L(x, y, λ) := f(x, y) − λ(g(x, y) − α) and solve the equations ∂L ∂x = 0 ⇒ ∂f ∂x = λ∂g ∂x , ∂L ∂y = 0 ⇒ ∂f ∂y = λ∂g ∂y , ∂L ∂λ = 0 ⇒ g(x, y) = α.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Method of Lagrange multipliers for f : R2 → R

To find extremum of f subject to the constraint g(x, y) = α, define L(x, y, λ) := f(x, y) − λ(g(x, y) − α) and solve the equations ∂L ∂x = 0 ⇒ ∂f ∂x = λ∂g ∂x , ∂L ∂y = 0 ⇒ ∂f ∂y = λ∂g ∂y , ∂L ∂λ = 0 ⇒ g(x, y) = α.

• The auxiliary function L(x, y, λ) is called Lagrangian

which converts constrained extrema to unconstrained extrema.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Method of Lagrange multipliers for f : R2 → R

To find extremum of f subject to the constraint g(x, y) = α, define L(x, y, λ) := f(x, y) − λ(g(x, y) − α) and solve the equations ∂L ∂x = 0 ⇒ ∂f ∂x = λ∂g ∂x , ∂L ∂y = 0 ⇒ ∂f ∂y = λ∂g ∂y , ∂L ∂λ = 0 ⇒ g(x, y) = α.

• The auxiliary function L(x, y, λ) is called Lagrangian

which converts constrained extrema to unconstrained

• extrema. • Critical points of L are eligible solutions for

constrained extrema.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Find the extreme values of f(x, y) = x2 − y2 on the circle x2 + y2 = 1.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Find the extreme values of f(x, y) = x2 − y2 on the circle x2 + y2 = 1. The equations fx = λgx, fy = λgy and g(x, y) = 1 give 2x = λ2x, −2y = λ2y and x2 + y2 = 1. The first equation shows either x = 0 or λ = 1.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Find the extreme values of f(x, y) = x2 − y2 on the circle x2 + y2 = 1. The equations fx = λgx, fy = λgy and g(x, y) = 1 give 2x = λ2x, −2y = λ2y and x2 + y2 = 1. The first equation shows either x = 0 or λ = 1. If x = 0 then y = ±1 ⇒ λ = −1. Thus (x, y, λ) := (0, ±1, −1) are possible solutions.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Find the extreme values of f(x, y) = x2 − y2 on the circle x2 + y2 = 1. The equations fx = λgx, fy = λgy and g(x, y) = 1 give 2x = λ2x, −2y = λ2y and x2 + y2 = 1. The first equation shows either x = 0 or λ = 1. If x = 0 then y = ±1 ⇒ λ = −1. Thus (x, y, λ) := (0, ±1, −1) are possible solutions. If λ = 1 then y = 0 ⇒ x = ±1. Thus (x, y, λ) := (±1, 0, 1) are also possible solutions.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Find the extreme values of f(x, y) = x2 − y2 on the circle x2 + y2 = 1. The equations fx = λgx, fy = λgy and g(x, y) = 1 give 2x = λ2x, −2y = λ2y and x2 + y2 = 1. The first equation shows either x = 0 or λ = 1. If x = 0 then y = ±1 ⇒ λ = −1. Thus (x, y, λ) := (0, ±1, −1) are possible solutions. If λ = 1 then y = 0 ⇒ x = ±1. Thus (x, y, λ) := (±1, 0, 1) are also possible solutions. Now f(0, 1) = f(0, −1) = −1 and f(1, 0) = f(−1, 0) = 1 so that minimum and maximum values are −1 and 1.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Finding global extrema of f : R2 → R

Let f : R2 → R and U ⊂ R2 be a region with smooth closed boundary curve C. To find global emtremum of f in U :

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Finding global extrema of f : R2 → R

Let f : R2 → R and U ⊂ R2 be a region with smooth closed boundary curve C. To find global emtremum of f in U :

• Locate all critical points of f in U.
• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Finding global extrema of f : R2 → R

Let f : R2 → R and U ⊂ R2 be a region with smooth closed boundary curve C. To find global emtremum of f in U :

• Locate all critical points of f in U.
• Find eligible global extremum of f on the curve C by

using Lagrange multipliers or parametrization.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Finding global extrema of f : R2 → R

Let f : R2 → R and U ⊂ R2 be a region with smooth closed boundary curve C. To find global emtremum of f in U :

• Locate all critical points of f in U.
• Find eligible global extremum of f on the curve C by

using Lagrange multipliers or parametrization.

• Choose points among eligible solutions in C and the

critical points at which f attains extreme values. These extreme values are global extremum.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example: global extrema

Find global maximum and global minimum of the function f(x, y) := (x2 + y2)/2 such that x2/2 + y2 ≤ 1.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example: global extrema

Find global maximum and global minimum of the function f(x, y) := (x2 + y2)/2 such that x2/2 + y2 ≤ 1. Since fx = x and fy = y, (0, 0) is the only critical point.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example: global extrema

Find global maximum and global minimum of the function f(x, y) := (x2 + y2)/2 such that x2/2 + y2 ≤ 1. Since fx = x and fy = y, (0, 0) is the only critical point. Next, consider L(x, y, λ) := (x2 + y2)/2 − λ(x2/2 + y2 − 1). Then Lagrange multiplier equations are x = λx, y = 2λy, x2/2 + y2 = 1. If x = 0 then y = ±1 and λ = 1/2. If y = 0 then x = ± √ 2 and λ = 1. If xy = 0 then λ = 1 and λ = 1/2 -which is not possible.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example: global extrema

Find global maximum and global minimum of the function f(x, y) := (x2 + y2)/2 such that x2/2 + y2 ≤ 1. Since fx = x and fy = y, (0, 0) is the only critical point. Next, consider L(x, y, λ) := (x2 + y2)/2 − λ(x2/2 + y2 − 1). Then Lagrange multiplier equations are x = λx, y = 2λy, x2/2 + y2 = 1. If x = 0 then y = ±1 and λ = 1/2. If y = 0 then x = ± √ 2 and λ = 1. If xy = 0 then λ = 1 and λ = 1/2 -which is not possible. Thus (0, ±1) and (± √ 2, 0) are eligible solutions for the boundary curve. We have f(0, ±1) = 1/2, f(± √ 2, 0) = 1 and f(0, 0) = 0.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Test for constrained extrema of f : Rn → R

Theorem: Let f, g : U ⊂ Rn → R be C1. Suppose that f has an extremum at X0 ∈ U such that g(X0) = α and ∇g(X0) = (0, . . . , 0). Then there is a λ ∈ R such that ∇f(X0) = λ∇g(X0).

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Test for constrained extrema of f : Rn → R

Theorem: Let f, g : U ⊂ Rn → R be C1. Suppose that f has an extremum at X0 ∈ U such that g(X0) = α and ∇g(X0) = (0, . . . , 0). Then there is a λ ∈ R such that ∇f(X0) = λ∇g(X0). If g(X) = α is a closed surface then global extremum is

• btained by finding all points where ∇f(X) = λ∇g(X) and

choosing those where f is largest or smallest.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Test for constrained extrema of f : Rn → R

Theorem: Let f, g : U ⊂ Rn → R be C1. Suppose that f has an extremum at X0 ∈ U such that g(X0) = α and ∇g(X0) = (0, . . . , 0). Then there is a λ ∈ R such that ∇f(X0) = λ∇g(X0). If g(X) = α is a closed surface then global extremum is

• btained by finding all points where ∇f(X) = λ∇g(X) and

choosing those where f is largest or smallest. The Lagrangian is given by L(X, λ) := f(X) − λ(g(X) − α). So, the multiplier equations are ∇L(X, λ) = (0, 0, . . . , 0).

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Maximize the function f(x, y, z) := x + z subject to the constraint x2 + y2 + z2 = 1.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Maximize the function f(x, y, z) := x + z subject to the constraint x2 + y2 + z2 = 1. The multiplier equations are 1 = 2xλ, 0 = 2yλ, 1 = 2zλ, x2 + y2 + z2 = 1. From first and 3rd equation, λ = 0. Thus, by second equation, y = 0. By first and 3rd equations x = z ⇒ x = z = ±1/ √ 2.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Maximize the function f(x, y, z) := x + z subject to the constraint x2 + y2 + z2 = 1. The multiplier equations are 1 = 2xλ, 0 = 2yλ, 1 = 2zλ, x2 + y2 + z2 = 1. From first and 3rd equation, λ = 0. Thus, by second equation, y = 0. By first and 3rd equations x = z ⇒ x = z = ±1/ √ 2. Hence X0 := (1/ √ 2, 0, 1/ √ 2) and X1 := (−1/ √ 2, 0, −1/ √ 2) are eligible solutions. This shows that f(X0) = √ 2 and f(X1) = − √ 2.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Riemann sum for double integral

Consider the rectangle R := [a, b] × [c, d] and a bounded function f : R → R. Let P be a partition of R into mn sub-rectangles Rij and cij ∈ Rij for i = 1, 2, . . . , m, j = 1, 2, . . . , n. Also let ∆Aij = area(Rij) = ∆xi∆yj and µ(P) := max

ij

∆Aij. Consider the Riemann sum S(P, f) :=

m

• i=1

n

• j=1

f(cij)∆Aij =

m

• i=1

n

• j=1

f(cij)∆xi∆yj.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Double integral

Definition: If limµ(P)→0 S(P, f) exists then f is said to be Riemann integrable and the (double) integral of f over R is given by

• R

f(x, y)dA =

• R

f(x, y)dxdy = lim

µ(P)→0 S(P, f).

• If f(x, y) ≥ 0 then
• R f(x, y)dA gives the volume of

the region bounded by R and the graph of f. Theorem: If f : R → R is continuous then f is Riemann integrable over R.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Theorem: Let f, g : R → R be Riemann integrable. Then

• f + αg is Riemann integrable for α ∈ R and
• R

(f + αg)dA =

• R

fdA + α

• R

gdA

• |f| is Riemann integrable and

|

• R

f(x, y)dA| ≤

• R

|f(x, y)|dA.

R dA = Area(R).

• If R = R1 + R2 then
• R

f(x, y)dA =

• R1

f(x, y)dA +

• R2

f(x, y)dA.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Iterated integrals

Let f : R → R. Suppose that for each fixed x ∈ [a, b] φ(x) := d

c

f(x, y)dy

• exists. If φ is Riemann integrable on [a, b] then

b

a

φ(x)dx = b

a

d

c

f(x, y)dy

• dx

is called an iterated integral of f over R. Similarly d

c

b

a f(x, y)dx

• dy, when exists, is another

iterated integral of f over R.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Fubini’s Theorem

Theorem: Let f : R → R be continuous. Then both the iterated limits exist and

• R

f(x, y)dA = b

a

d

c

f(x, y)dy

• dx

= d

c

b

a

f(x, y)dx

• dy.

Example: Evaluate

• R xexydA, where R = [0, 1] × [0, 1].

Since the function is continuous,

• R

xexydA = 1 ( 1 xexydy)dx = 1 (ex − 1)dx = e − 2.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Double integrals over general domains

Definition: Let D ⊂ R2 be bounded and f : D → R be a bounded function. Then f is said to be integrable over D if for some rectangle R containing D the function F(x, y) :=

• f(x, y)

if (x, y) ∈ D

• therwise

is Riemann integrable over R. The double integral of f

• ver D is then defined by
• D

f(x, y)dA :=

• R

F(x, y)dA. Remark: Since F is zero outside D the choice of R is unimportant in defining double integral of f over D.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Special Regions: Type-I Regions

• Type-I Region:

R = {(x, y) ∈ R2 : x ∈ [a, b] and g1(x) ≤ y ≤ g2(x)} where g1(x) and g2(x) are continuous functions on [a, b] and g1(x) ≤ g2(x) for all x ∈ [a, b].

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Type-II Regions

• Type-II Region:

R = {(x, y) ∈ R2 : y ∈ [c, d] and h1(y) ≤ x ≤ h2(y)} where h1(y) and h2(y) are continuous functions on [c, d] and h1(y) ≤ h2(y) for all y ∈ [c, d].

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Type-III Regions (Both Type-I and Type-II)

R is called Type-III region if R is simultaneously of Type-I and Type-II.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Double integral over special domains

Theorem: Let f : D ⊂ R2 → R be continuous. If D is Type-I and D = {(x, y) : x ∈ [a, b] and φ1(x) ≤ y ≤ φ2(x)} then f is integrable over D and

• D

f(x, y)dA = b

a

φ2(x)

φ1(x)

f(x, y)dy

• dx.

If D is Type-II and D = {(x, y) : ψ1(y) ≤ x ≤ ψ2(y) and y ∈ [c, d]} then f is integrable over D and

• D

f(x, y)dA = d

c

ψ2(y)

ψ1(y)

f(x, y)dx

• dy.
• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Area and Volume

Let D ⊂ R2 be a special (Type-I or Type-II or Type-III) domain and f : D → R be continuous. Then Area(D) =

• D

dA. If f(x, y) ≥ 0 then the volume of the solid S bounded by D and the graph of z = f(x, y) is given by Volume(S) =

• D

f(x, y)dA.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Find the volume of the solid S bounded by elliptic paraboloid x2 + 2y2 + z = 16, the planes x = 2, y = 2, and the coordinate planes. Volume(S) =

• R

(16 − x2 − 2y2)dA = 2 2 (16 − x2 − 2y2)dxdy = 48.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Evaluate

• D(x + 2y)dA, where D is the region bounded

by the parabolas y = 2x2 and y = 1 + x2. The region D is Type-I and

• D

(x + 2y)dA = 1

−1

1+x2

2x2

(x + 2y)dy

• dx

= 1

−1

(−3x4 − x3 + 2x2 + x + 1)dx = 32 15.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Riemann sum for Triple integral

Consider the rectangular cube V := [a1, b1] × [a2, b2] × [a3, b3] and a bounded function f : V → R. Let P be a partition of V into sub-cubes Vijk and cijk ∈ Vijk for i = 1, . . . , m; j = 1, . . . , n; k = 1, . . . , p. Also let ∆Vijk := Volume(Vijk) = ∆xi∆yj∆zk and µ(P) := max

ijk

∆Vijk. Consider the Riemann sum S(P, f) :=

m

• i=1

n

• j=1

p

• k=1

f(cijk)∆Vijk.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Triple integral

If limµ(P)→0 S(P, f) exists then f is said to be Riemann integrable and the (triple) integral of f over V is given by

• V

f(x, y, z)dV =

• V

f(x, y, z)dxdydz = lim

µ(P)→0 S(P, f).

Theorem: Let f : V → R is continuous. Then

• f is Riemann integrable over V.
• Fubini’s theorem holds, i.e, the iterated integrals exist

and are equal to

• V fdV.
• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Example

Evaluate

• V xyz2dV where V = [0, 1] × [−1, 2] × [0, 3].

By Fubini’s theorem,

• V

fdV = 3 2

−1

1 xdx

• ydy
• z2dz = 27

4 .

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Triple integrals over general domains

Let D ⊂ R3 be bounded and f : D → R be a bounded

• function. Then f is said to be integrable over D if for

some rectangular cube V containing D the function F(x, y, z) :=

• f(x, y, z)

if (x, y, z) ∈ D

• therwise

is Riemann integrable over V. Then

• D

f(x, y, z)dV :=

• V

F(x, y, z)dV and Volume(D) :=

• D

dV.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Type-I domain:

A domain V ⊂ R3 is Type-I if V = {(x, y, z) : (x, y) ∈ D and u1(x, y) ≤ z ≤ u2(x, y)} for some D ⊂ R2 and continuous functions ui : D → R. If f : V → R be continuous and D is a special domain (e.g.,Type-I, Type-II, Type-III) then

• V

f(x, y, z)dV =

• D

u2(x,y)

u1(x,y)

f(x, y, z)dz

• dxdy.

Similar results hold for Type-II and Type-III domains.

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Figure : Type-I domain

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Figure : Type-II domain

• R. Barman & S. Bora

MA-102 (2017)

Multiple Integrals

Figure : Type-III domain

• R. Barman & S. Bora

MA-102 (2017)