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

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MA102: Multivariable Calculus

Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati

R. Barman & S. Bora

MA-102 (2017)

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Surfaces

1 Locus of a point moving in space with 2 degrees of

freedom.

2 Level curve of a scalar field F : D ⊆ R3 → R. For

example, x2 + y 2 + z2 = c, z = x2 + y 2, etc.

3 Sometimes surfaces can be described by

{(x, y, z) : z = f (x, y), (x, y) ∈ D}. This is called explicit representation.

4 The unit sphere is a union of two such explicit

representations: {(x, y, z =

1 − x2 − y 2) : x2 + y 2 ≤ 1}

∪ {(x, y, z = −

1 − x2 − y 2) : x2 + y 2 ≤ 1}
R. Barman & S. Bora

MA-102 (2017)

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Parametric representation of a surface

A surface may also be described by x = X(u, v), y = Y (u, v), z = Z(u, v), where u, v ∈ D and D is a connected subset of the uv-plane, for example, plane region like circle, rectangle, etc. Definition: A continuous function R : D ⊂ R2 → R3 is called a parametric surface in R3. The image S := R(D) is called a geometric surface in R3. If the surface has an explicit representation given by a continuous function z = f (x, y), (x, y) ∈ D, then R(x, y) = x ˆ i + y ˆ j + f (x, y) ˆ k is a parametric representation.

R. Barman & S. Bora

MA-102 (2017)

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Parametric representation of a sphere of radius a

If we take spherical coordinates, then x = X(θ, φ) = a sin φ cos θ, y = Y (θ, φ) = a sin φ sin θ, z = Z(θ, φ) = a cos φ, where 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π. This gives a parametric representation of the sphere: R(φ, θ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k, where 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.

R. Barman & S. Bora

MA-102 (2017)

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Parametric representation of a cone

We find a parametrization of the cone z =

x2 + y 2, 0 ≤ z ≤ 1

Here cylindrical coordinates provide everything we need. x(r, θ) = r cos θ, y(r, θ) = r sin θ, z =

x2 + y 2 = r.

Also 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. So the required parametrization is R(r, θ) = r cos θ ˆ i + r sin θ ˆ j + r ˆ k, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.

R. Barman & S. Bora

MA-102 (2017)

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Parametric representation of a cylinder

We find a parametrization of the cylinder x2 + (y − 3)2 = 9, 0 ≤ z ≤ 5. We take cylindrical coordinates: x(r, θ) = r cos θ, y(r, θ) = r sin θ, z = z. Substituting in x2 + (y − 3)2 = 9 we get r 2 − 6r sin θ = 0. Therefore, r = 6 sin θ. Thus, we have the following parametrization: R(θ, z) = 3 sin 2θ ˆ i + 6 sin2 θ ˆ j + z ˆ k, where 0 ≤ θ ≤ π, 0 ≤ z ≤ 5.

R. Barman & S. Bora

MA-102 (2017)

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Smooth parametric surface

Let R : D ⊂ R2 → R3 be a parametric surface and let R(u, v) = (x(u, v), y(u, v), z(u, v)). Then the partial derivatives of R, when exist, are given by Ru = (xu, yu, zu) and Rv = (xv, yv, zv). The parametric surface S = R(D) is said to be smooth if R is C 1 and Ru × Rv = 0 for (u, v) ∈ D. Assumptions:

❼ D is connected ❼ R is injective except possibly on the boundary of D ❼ R is C 1 and Ru × Rv = 0 for (u, v) ∈ D.

R. Barman & S. Bora

MA-102 (2017)

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Singular Points

A point on S is called a singular point if S fails to be smooth at that point. That is, either R is not C 1 and/or Ru × Rv = 0 at that point. The parametrization R(x, y) = x ˆ i + y ˆ j +

1 − x2 − y 2 ˆ

k of the interior of the upper hemisphere of the unit sphere has no singular point. We have Rx × Ry =

1 + f 2

x + f 2 y = 0,

∀(x, y) s.t x2 + y 2 < 1. (0, 0, a) is the only singular point of the hemisphere w.r.t. the parametrization: R(φ, θ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k, where 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.

R. Barman & S. Bora

MA-102 (2017)

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Surface area

Let R : D ⊂ R2 → R3 be a smooth parametric surface. Let (u0, v0) ∈ D and consider the rectangle T formed by the vertices (u0, v0), (u0 + △u, v0), (u0, v0 + △v), (u0 + △u, v0 + △v). Each side of the rectangle T maps to a curve on the surface

S. Let C1 and C2 be the curves corresponding to the sides

v = v0 and u = u0, respectively. These two curves meet at P0 = R(u0, v0). Drawing the other two curves determined by v = v0 + △v and u = u0 + △u, we find that the rectangle T is mapped to the portion of the surface bounded by these four curves. We denote the surface area of this curved patch by △σuv.

R. Barman & S. Bora

MA-102 (2017)

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Surface area

Ru(u0, v0) is tangent to C1 at P0 and Rv(u0, v0) is tangent to C2 at P0, where P0 = R(u0, v0). We approximate the surface area △σuv by the area of the parallelogram on the tangent plane whose sides are determined by the vectors △u · Ru(u0, v0) and △v · Rv(u0, v0). Hence the surface area of S denoted by

S dσ is
S

dσ =

D

Ru × Rvdudv. Example: Find the surface area of the hemisphere S := {(x, y, z) : z =

a2 − x2 − y 2}, a > 0.
R. Barman & S. Bora

MA-102 (2017)

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Surface area

Solution: The parametric representation of the surface is R(φ, θ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k, where φ ∈ [0, π/2], θ ∈ [0, 2π]. Now, Rφ × Rθ =

i

j k a cos θ cos φ a sin θ cos φ −a sin φ −a sin θ sin φ a cos θ sin φ

=

a2 cos θ sin2 φ ˆ i + a2 sin θ sin2 φ ˆ j + a2 sin φ cos φ ˆ k Thus, Rφ × Rθ = a2 sin φ. Hence the surface area is π/2

φ=0

2π

θ=0

Rθ × Rφdφdθ = π/2

φ=0

2π

θ=0

a2 sin φdφdθ = 2πa2.

R. Barman & S. Bora

MA-102 (2017)

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Surface area of a surface z = f (x, y)

Let f : D ⊆ R2 → R. Then, we consider the parametrization R(x, y) = x ˆ i + y ˆ j + f (x, y) ˆ k, (x, y) ∈ D. We have Rx × Ry =

i

j k 1 fx 1 fy

= −fx ˆ

i − fy ˆ j + ˆ k. Hence,

S

dσ =

D
1 + f 2

x + f 2 y dxdy

R. Barman & S. Bora

MA-102 (2017)

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Figure : Surface element

R. Barman & S. Bora

MA-102 (2017)

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Examples

Find the surface area of the surface of the cone z =

x2 + y 2, 0 ≤ z ≤ 1.

Solution (method-1): We consider the following smooth parametrization of the cone: R(r, θ) = r cos θ ˆ i + r sin θ ˆ j + r ˆ k, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. We find that Rr × Rθ = −r cos θ ˆ i − r sin θ ˆ j + r ˆ k and Rr × Rθ =

r 2 cos2 θ + r 2 sin2 θ + r 2 =

√ 2r. Therefore, Surface Area = 2π 1 √ 2rdrdθ = π √ 2

R. Barman & S. Bora

MA-102 (2017)

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Examples

Solution (method-2): Here z = f (x, y) =

x2 + y 2. We

consider the following smooth parametrization of the cone: R(x, y) = x ˆ i + y ˆ j + f (x, y) ˆ k, (x, y) ∈ D, where D = {(x, y) ∈ R2 : x2 + y 2 ≤ 1}. Now, fx = x

x2 + y 2 and fy =

y

x2 + y 2. Therefore,

Surface Area =

D
1 + f 2

x + f 2 y dxdy

= √ 2

x2+y2≤1

dxdy = √ 2π.

R. Barman & S. Bora

MA-102 (2017)

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Surface area and Jacobian determinants

Let R(u, v) = X(u, v)ˆ ı + Y (u, v) ˆ  + Z(u, v) ˆ k , (u, v) ∈ D be a smooth parametrization of a surface S. We have Ru × Rv =

ˆ

ı ˆ  ˆ k Xu Yu Zu Xv Yv Zv

=
Yu

Zu Yv Zv

ˆ

ı +

Zu

Xu Zv Xv

ˆ

 +

Xu

Yu Xv Yv

ˆ

k = ∂(Y , Z) ∂(u, v) ˆ ı + ∂(Z, X) ∂(u, v) ˆ  + ∂(X, Y ) ∂(u, v) ˆ k Thus, the surface area is

S

dσ =

D

∂(Y , Z) ∂(u, v) 2 + ∂(Z, X) ∂(u, v) 2 + ∂(X, Y ) ∂(u, v) 2 dudv.

R. Barman & S. Bora

MA-102 (2017)

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Implicit surfaces

Let S be a surface implicitly given by F(x, y, z) = c, where F is C 1. We assume that either ∇F • ˆ k = 0 or ∇F • ˆ j = 0 or ∇F •ˆ i = 0 on S. Suppose that ∇F • ˆ k = 0, that is, Fz = 0 on S. Then we can write the surface S explicitly as z = h(x, y) (due to Implicit function theorem). Now, we take the parametrization of S given by R(x, y) = x ˆ ı + y ˆ  + h(x, y) ˆ k . Then, Rx = ˆ ı + hx ˆ k = ˆ ı − Fx Fz ˆ k Ry = ˆ  + hy ˆ k = ˆ  − Fy Fz ˆ k .

R. Barman & S. Bora

MA-102 (2017)

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Implicit surfaces

Hence, Rx × Ry = Fx Fz ˆ ı + Fy Fz ˆ  + ˆ k = Fx ˆ ı + Fy ˆ  + Fz ˆ k Fz = ∇F Fz = ∇F ∇F • ˆ k Thus, the surface area is given by

S

dσ =

D
∇F

∇F • ˆ k

dxdy,

where D is the projection of the surface on the xy-plane. Remark: If Fx = ∇F • ˆ ı = 0, then we have to take projection

f the surface on the yz-plane. If Fy = 0, then we take

projection of the surface on the xz-plane.

R. Barman & S. Bora

MA-102 (2017)

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Example: Implicit surfaces

Find the surface area of the upper hemisphere of the sphere x2 + y 2 + z2 = a2. Here, F(x, y, z) = x2 + y 2 + z2 = a2, ∇F = 2x ˆ ı + 2y ˆ  + 2z ˆ k , and ∇F • ˆ k = 2z. Thus, ∇F =

4x2 + 4y 2 + 4z2 = 2a.
S

dσ =

D

2a 2z dxdy =

D

a

a2 − x2 − y 2dxdy

= a

r=0

2π

θ=0

a √ a2 − r 2rdrdθ (using polar co-ordinates) = aπ a2

t=0

dt √t = 2πa2.

R. Barman & S. Bora

MA-102 (2017)

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Example: Implicit surfaces

Example: Find the surface area of the cap obtained by cutting the hemisphere x2 + y 2 + z2 = 2 by the cone z =

x2 + y 2.

Solution: The equation of surface is F(x, y, z) = x2 + y 2 + z2 − 2 = 0 and we can take the projection onto xy-plane. The projection is obtained by solving x2 + y 2 + z2 = 2, z =

x2 + y 2 which gives

R = {(x, y) ∈ R2 : x2 + y 2 ≤ 1}. We have, ∇F = 2x ˆ ı + 2y ˆ  + 2z ˆ k and ∇F = 2

x2 + y 2 + z2 = 2

√

2. The required surface area is
S

dσ =

R

√ 2

2 − x2 − y 2 dA

= 2π 1 √ 2 √ 2 − r 2 r dr dθ = 2π(2 − √ 2).

R. Barman & S. Bora

MA-102 (2017)

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Surface Integration: Scalar surface integrals

Let S be a surface parametrized by R : D → R3 such that R is C 1. Let f : S → R be bounded on S. Then the surface integral of f over S is given by

S

f (x, y, z)dσ :=

D

f (R(u, v))Ru × Rv dudv whenever the double integral on the right exists. Example: Evaluate the surface integral

S(x + y + z)dσ over

the surface of the cylinder x2 + y 2 = 9, 0 ≤ z ≤ 4. Solution: Using the cylindrical coordinates, the surface can be represented as R(θ, z) = 3 cos θ ˆ i + 3 sin θ ˆ j + z ˆ k

ver the parameter domain {(θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 4}.
R. Barman & S. Bora

MA-102 (2017)

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Surface Integration: Scalar surface integrals

Then Rθ × Rz =

9 cos2 θ + 9 sin2 θ = 3. The given

integral is equal to

S

(x + y + z)dσ =

S

(3 cos θ + 3 sin θ + z)Rθ × Rzdθdz = 3 4

z=0

2π

θ=0

(3 cos θ + 3 sin θ + z)dθdz = 6π 4 zdz = 48π.

R. Barman & S. Bora

MA-102 (2017)

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Example

Evaluate

S x2dσ over the sphere S : x2 + y 2 + z2 = a2.

We take the following parametric representation of the sphere R(φ, θ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k, where 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π. Then, we have Rφ × Rθ = a2 sin φ and

S

x2dσ = a4 2π π sin3 φ cos2 θ dφdθ = a4 π sin3 φdφ 2π cos2 θdθ = 4a4π/3.

R. Barman & S. Bora

MA-102 (2017)

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Application

Consider a mass spread over a thin surface S. Let δ(x, y, z) be the density at (x, y, z) ∈ S.

1 Mass M =

S δ(x, y, z)dσ.

2 Moments about co-ordinate planes:

Myz =

S

xδdσ, Mxz =

S

yδdσ, Mxy =

S

zδdσ.

3 Co-ordinates of the center of mass:

x = Myz/M, y = Mxz/M, z = Mxy/M.

4 Moments of inertia:

Ix =

S

(y 2 + z2)δdσ, Iy =

S

(x2 + z2)δdσ, Iz =

S

(x2 + y 2)δdσ, IL =

S

r 2δdσ, where r(x, y, z) is the perpendicular distance from (x, y, z) ∈ S to L.

R. Barman & S. Bora

MA-102 (2017)

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Oriented surface

Let S be a surface parametrized by a C 1 function R : D ⊆ R2 → R3. Let ˆ n denote the unit normal to S. Then, ˆ n = Ru × Rv Ru × Rv. If ˆ n is a continuous function on D, then S together with ˆ n is called an oriented surface, that is, the pair (S, ˆ n) is called an

riented surface.

Example: R1(φ, θ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k, where 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π. Then ˆ n1 =

∂R1 ∂φ × ∂R1 ∂θ

∂R1

∂φ × ∂R1 ∂θ = 1

aR1(φ, θ) is the normal.

R. Barman & S. Bora

MA-102 (2017)

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Figure : Orientation induced by R1(φ, θ)

R. Barman & S. Bora

MA-102 (2017)

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Oriented surface

But, if R2(φ, θ) = −a sin φ cos θ ˆ i − a sin φ sin θ ˆ j + a cos φ ˆ k, where 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π, then ˆ n2 =

∂R2 ∂φ × ∂R2 ∂θ

∂R2

∂φ × ∂R2 ∂θ = 1

aR2(φ, θ) is the associated normal.

(R1, ˆ n1) describes the upper hemisphere of the sphere of radius a with different orientation to that of (R2, ˆ n2). Notice that on the edge of the hemisphere (φ = π/2), ˆ n1 and ˆ n2 point in opposite directions as R1(θ, π/2) and R2(θ, π/2) point in opposite directions. ˆ n1 is the outer normal.

R. Barman & S. Bora

MA-102 (2017)

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Non-oriented surface

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MA-102 (2017)

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Orientation with normal pointing downward

If we walk on the boundary along the orientation prescribed by the parametrization and find the surface under consideration on our right hand side, then the parametrization gives an orientation to the surface with normal pointing inward.

R. Barman & S. Bora

MA-102 (2017)

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Surface integrals of vector fields

Let (S, ˆ n) be an oriented surface in R3 and let F : S → R3 be a continuous vector field. Then F • ˆ n is the normal component

f F.

The surface integral of F (also called the flux integral) over the oriented surface (S, ˆ n) is defined as

S

F • ˆ ndσ. If S = R(D), where R is a smooth parametrization of S over the parameter domain D, then

S

F • ˆ ndσ =

D

F(R(u, v)) • Ru × Rv Ru × RvRu × Rvdudv =

D

F(R(u, v)) • (Ru × Rv)dudv.

R. Barman & S. Bora

MA-102 (2017)

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Example

Let F(x, y, z) := (z, y, x). Evaluate the flux integral

S F • ˆ

ndσ over the unit sphere S : x2 + y 2 + z2 = 1. We have R(u, v) = (sin u cos v, sin u sin v, cos u), (u, v) ∈ [0, π]×[0, 2π], Ru × Rv = (sin2 u cos v, sin2 u sin v, sin u cos u). Thus

S

F • ˆ ndσ = 2π π (2 sin2 u cos u cos v + sin3 u sin2 v)dudv = 4π 3 .

R. Barman & S. Bora

MA-102 (2017)

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One more example: Surface over the xz-plane

Find the outward flux of F = yz ˆ ı + x ˆ  − z2 ˆ k through the parabolic cylinder y = x2, 0 ≤ x ≤ 1, 0 ≤ z ≤ 4. Step 1: Writing the Parametric Equation of S We parameterize S by the equation φ(x, y) =

x, x2, z
for (x, z) ∈ D

where D = {(x, z) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ z ≤ 4} in the xz-plane. Step 2: Computing φx × φz φx = (1, 2x, 0) and φz = (0, 0, 1) . φx × φz =

ˆ

ı ˆ  ˆ k 1 2x 1

= 2x ˆ

ı − ˆ 

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MA-102 (2017)

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Example (cont.)

Step 3: Evaluation of the Vector Surface Integral

S

F • ˆ n dσ =

D

F(φ(x, z)) • (φx × φz) dx dz = 1

x=0

4

z=0

(2xyz − x) dz dx = 1

x=0

4

z=0

(2x3z − x) dz dx (By putting y = x2 ) = 1

x=0

(16x3 − 4x) dx = 2.

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MA-102 (2017)

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An application to fluid flow

Let V (x, y, z) be the velocity of a fluid at any point (x, y, z). Let δ(x, y, z) be the density of the fluid at point (x, y, z) and let F(x, y, z) = V (x, y, z) · δ(x, y, z). Since V is a vector field and δ is a scalar field, Therefore F is a vector

field. Its dimension is

mass unit volume × length unit time = mass unit area × unit time Let S be an orientable surface parametrized by R(u, v), (u, v) ∈ D and ˆ n be the associated unit normal. Then the mass of fluid flowing through S in unit time in the direction of ˆ n is given by the flux integral

S

F • ˆ ndσ.

R. Barman & S. Bora

MA-102 (2017)

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Curl of a vector field

F(x, y, z) = M(x, y, z) ˆ ı + N(x, y, z) ˆ  + P(x, y, z) ˆ k for (x, y, z) ∈ D ⊂ R3. The curl of F is denoted by curl(F) and is defined by curlF = ∇ × F =

ˆ

ı ˆ  ˆ k

∂ ∂x ∂ ∂y ∂ ∂z

M(x, y, z) N(x, y, z) P(x, y, z)

=

∂P ∂y − ∂N ∂z

ˆ

ı − ∂P ∂x − ∂M ∂z

ˆ

 + ∂N ∂x − ∂M ∂y

ˆ

k

R. Barman & S. Bora

MA-102 (2017)

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Stoke’s Theorem (3-D version of Green’s Theorem)

Stoke’s Theorem: Assume that S is a smooth parametric surface, say S = R(D), where D is a region in the uv-plane bounded by a closed, simple, piecewise smooth curve Γ. Assume that R is C 2 and one-to-one

n some open set containing D ∪ Γ. Let C denote the image of Γ, that

is, C = R(Γ). Let F(x, y, z) = M(x, y, z)ˆ ı + N(x, y, z) ˆ  + P(x, y, z) ˆ k be a continuously differentiable vector field on S. Then

C

F • dr =

C

M dx + N dy + P dz =

S

(curl(F) • ˆ n) dσ, where ˆ n = (Ru × Rv)/Ru × Rv. The curve Γ is traversed in the positive (counterclockwise) direction and the curve C is traversed in the direction inherited from Γ through R.

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MA-102 (2017)

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Example

Verify Stoke’s Theorem for the vector field F(x, y, z) = 2z ˆ ı + 3x ˆ  + 5y ˆ k taking S to be the portion of the paraboloid z = 4 − x2 − y 2 for which z ≥ 0. The boundary curve of S is the circle C : x2 + y 2 = 4 in the xy-plane with counterclockwise direction. C : r(t) = 2 cos t ˆ ı + 2 sin t ˆ  + 0 ˆ k for t ∈ [0, 2π].

C

F • dr =

C

2z dx + 3x dy + 5y dz = 2π

t=0

(0 + (6 cos t)(2 cos t) + 0) dt = 2π

t=0

12 cos2 t dt = 12 t 2 + sin(2t) 4 2π

t=0

= 12π .

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MA-102 (2017)

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Example (cont.)

curl(F) = 5ˆ ı + 2 ˆ  + 3 ˆ k . ˆ n = 2x ˆ ı + 2y ˆ  + ˆ k

1 + 4x2 + 4y 2 .
S

(curl(F) • ˆ n) dσ =

S

(5ˆ ı + 2 ˆ  + 3 ˆ k ) • (2x ˆ ı + 2y ˆ  + ˆ k )

1 + 4x2 + 4y 2

dσ =

S

10x + 4y + 3

1 + 4x2 + 4y 2 dσ

= 2π

θ=0

2

r=0

(10r cos θ + 4r sin θ + 3) r dr dθ = 2π

θ=0

80 3 cos θ + 32 3 sin θ + 6

dθ = 12π .

Thus,

C

F • dr = 12π =

S

(curl(F) • ˆ n) dσ .

R. Barman & S. Bora

MA-102 (2017)

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Example

Find

S curl(F) • ˆ

ndσ, where F(x, y, x) = (y 2, xy, xz) and S is the upper hemisphere of the unit sphere x2 + y 2 + z2 = 1. Solution: Using Stoke’s theorem, we have

S

(curl(F) • ˆ n) dσ =

C

[y 2dx + xydy + xzdz] (C := {(x, y, 0) : x2 + y 2 = 1}) = 2π

θ=0

[sin2 θ(− sin θ) + cos2 θ sin θ] dθ = 0.

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MA-102 (2017)

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Gauss’s Divergence Theorem

Divergence Theorem: Let V ⊂ R3 be a solid region bounded by an

riented closed surface S, and let ˆ

n be the unit outward normal to S. Let F = (P, Q, R) be a C 1 vector field on any open subset of R3 containing V ∪ S. Then

S

F • ˆ ndσ =

V

div(F) dV , where div(F) = ∂P

∂x + ∂Q ∂y + ∂R ∂z .

Example: Let D be the region bounded by the hemisphere x2 + y 2 + (z − 1)2 = 9, 1 ≤ z ≤ 4 and the plane z = 1. Let F(x, y, z) = x ˆ ı + y ˆ  + (z − 1) ˆ k . Illustrate the Divergence theorem. Solution: Here, divF = 1 + 1 + 1 = 3 and hence

D

divFdV = 3

D

dV = 3 × volume of the hemisphere = 54π.

R. Barman & S. Bora

MA-102 (2017)

SLIDE 41

Gauss’s Divergence Theorem

Here, S = S1 ∪ S2, where S1 = {(x, y, z) : x2 + y 2 + (z − 1)2 = 9, 1 ≤ z ≤ 4} and S2 = {(x, y, 1) : x2 + y 2 ≤ 9}. Let ˆ n1 and ˆ n2 be the outer normals to S1 and S2, respectively. Then,

S2

F • ˆ n2dσ =

S2

(x ˆ ı + y ˆ  + (1 − 1) ˆ k ) • (− ˆ k )dσ = 0. Now, we find the unit outward normal to the hemisphere S1. We can find it by considering the parametrization R(x, y) = x ˆ ı + y ˆ  + (1 +

9 − x2 − y 2) ˆ

k and using the formula ˆ n1 =

Rx×Ry Rx×Ry.

Note that if we put G(x, y, z) = x2 + y 2 + (z − 1)2 − 9, then ˆ n1 =

∇G ∇G = x 3 ˆ

ı + y

3 ˆ

 + z−1

3

ˆ k . Hence, F • ˆ n1 = x2/3 + y 2/3 + (z − 1)2/3 = 9/3 = 3 on S1.

R. Barman & S. Bora

MA-102 (2017)

SLIDE 42

Gauss’s Divergence Theorem

Now,

S1

F • ˆ n1dσ = 3

S1

dσ = 3

x2+y 2≤9
∇G

∇G • ˆ k

dxdy

=

x2+y 2≤9

3

9 − x2 − y 2 dxdy

= 9 2π

θ=0

3

r=0

1 √ 9 − r 2 rdrdθ = 9π 9 dt √t = 54π. Hence,

S

F • ˆ ndσ =

S1

F • ˆ n1dσ +

S2

F • ˆ n2dσ = 54π.

R. Barman & S. Bora

MA-102 (2017)

SLIDE 43

Gauss’s Divergence Theorem

Example: Evaluate

S F • ˆ

ndσ using Divergence Thm, where F(x, y, z) = (x + y)ˆ ı + z2 ˆ  + x2 ˆ k and S is the surface of x2 + y 2 + z2 = 1, z ≥ 0, ˆ n being the outer normal. Solution: By Divergence Theorem,

D

divF dV =

S

F • ˆ ndσ +

S1

F • ˆ n1dσ, where ˆ n is the outer normal to S and ˆ n1 is the outer normal to S1 = {(x, y, 0) : x2 + y 2 ≤ 1}. We have divF = 1 and hence

D divF dV = 2

3π. Now,

S1

F• ˆ n1dσ =

S1

[(x+y)ˆ ı +z2 ˆ  +x2 ˆ k ]•(− ˆ k )dσ = −

S1

x2dσ = −π/4. Hence,

S

F • ˆ ndσ =

D

divF dV −

S1

F • ˆ n1dσ = 2π 3 + π 4 = 11 12π.

R. Barman & S. Bora

MA-102 (2017)