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CHAPTER 9 VECTOR CALCULUS-PART 3

WEN-BIN JIAN (簡紋濱)

DEPARTMENT OF ELECTROPHYSICS

NATIONAL CHIAO TUNG UNIVERSITY

OUTLINE

10.DOUBLE INTEGRALS

11.DOUBLE INTEGRALS IN POLAR COORDINATES

12.GREEN’S THEOREM

13.SURFACE INTEGRALS

DEFINITION Double IntegralLet be a function of two variables defined on a closed region of 2D space. The double integral of over is given by

.

Integratability: When is continuous on , it is integrable.

Calculation of Area: When , the area is .Calculation of Volume: When , the volume is

.

10. DOUBLE INTEGRALS

Type of : Regions of Type I (bounded by ) and Type II (bounded by )

Calculation of Area & Volume

THEOREM Fubini’s Theorem

Type I Region: ,

Center of Mass

Type II Region: ,


If the function is the area density of a 2D object, then the mass and the COM are calculated by

Center of Mass


Example: Evaluate the double integral over the region

bounded by the graphs of , , , .

This is a Type II region.

Calculation of Double Integral


Example: Evaluate over the region in the first quadrant

bound by the graphs of , , .

Type I:

Type II:



Example: A lamina has the shape of the region in the first quadrant that is bounded by the graphs of , , between and . Find its center of mass if the density is .

/ /

/

/ /

/



Example: Find the moment of inertia about the y-axis of the thin homogeneous disk of m shown in the right figure.

Let

=2𝑚𝑟

3𝜋

1 + 2 cos 2𝜃

4+1 + cos 4𝜃

8

/

/

𝑑𝜃 =2𝑚𝑟

3𝜋

3𝜋

8=𝑚𝑟

4

=2𝑚𝑟

3𝜋

1 + cos 2𝜃

2

/

/

𝑑𝜃


𝐼 =2𝑚𝑟

3𝜋cos 𝜃𝑑𝜃

/

/

11. DOUBLE INTEGRALS IN POLAR COORDINATES

Change of Variables from to

to : ,

to : ,

Remember that we are studying “vector analysis”, the calculation is based on the “length”. The small area in polar coordinate is:

The 2D integration based on the area in the polar coordinate is

Change of Variables


Example: Find the center of mass of the lamina that corresponds to the region bounded by one leaf of the rose in the first quadrant if the density at a point P in the lamina is directly proportional to the distance from the origin. The area is in the range of

.

let , calculate the total m as a first step/

/ /

/

Change of Variables


Example: Find the center of mass of the lamina …

/

/

= 3663 − 90 + 35

315= 36

8

315=32

35

Change of Variables


Example: Use polar coordinate to evaluate

.

/

/

=𝜋 ln 13 − ln 5

8

Change of Variables


Example: Find the volume of the solid that is under the hemisphere and above the region bounded by the graph of the circle .

choose the polar coordinate for

rewrite the boundary of as

/

Change of Variables


Example: Find the volume of the solid that is under the hemisphere and above the region bounded by the graph of the circle .

//

/

=𝜋

3−2

31 − sin 𝜃 𝑑 sin𝜃 =

𝜋

3−2

3sin𝜃 −

sin 𝜃

3=𝜋

3−4

9

Change of Variables

OUTLINE

10.DOUBLE INTEGRALS




12. GREEN’S THEOREM

Suppose that is a piecewise-smooth simple closed curve bounding

a simply connected region . If , , ,

, and ,

are continuous on , then

1.5 − 1 − 𝑥 − 1,5

𝑑𝑥.

.

1.5 + 1 − 𝑥 − 1,5

𝑑𝑥.

.1.5 − 1 − 𝑥 − 1,5

𝑑𝑥

.

.

+ 1.5 + 1 − 𝑥 − 1,5

𝑑𝑥.

.

Green’s Theorem

𝑔 𝑥 = 1.5 + 1 − 𝑥 − 1.5

𝑔 𝑥 = 1.5 − 1 − 𝑥 − 1.5


Type I Region: Type II Region:

Proof of Green’s Theorem

ℎ 𝑦 = 1.5 − 1 − 𝑦 − 1.5

ℎ 𝑦 = 1.5 + 1 − 𝑦 − 1.5


Type I Region: Type II Region:

Proof of Green’s Theorem


Example: Evaluate , where consists of

the boundary of the region in the first quadrant that is bounded by the graphs of and .

Application of Green’s Theorem


Example: Evaluate , where is the

circle .

−𝜕𝑃

𝜕𝑦+𝜕𝑄

𝜕𝑥= −3 + 2 = −1

𝑥 + 3𝑦 𝑑𝑥 + 2𝑥 − 𝑒 𝑑𝑦



Example: acting along the simple closed curve shown in the right figure.

= −16𝑦 + sin 𝑥 𝑑𝑥 + 4𝑒 + 3𝑥 𝑑𝑦

use polar coordinate

/

/

//



Green’s theorem can only be used if , , ,

, and ,

are continuous on the region .

Green’s theorem can also be used in a multiply connected region where the functions are continuous without any divergences.

The region is bounded by two simple closed curves and . That is . If , , , and are continuous in , then Green’s theorem is applicable, even though these functions are not continuous in the holes.

Apply Green’s Theorem in The Region with Holes


Example: Evaluate , where is

the boundary of the shaded region in the figure.

=𝑥 + 𝑦 − 2𝑦

𝑥 + 𝑦+𝑥 + 𝑦 − 2𝑥

𝑥 + 𝑦= 0

∪



Example: Evaluate .

= −𝑦

𝑥 + 𝑦𝑑𝑥 +

𝑥

𝑥 + 𝑦𝑑𝑦


OUTLINE

10.DOUBLE INTEGRALS




13. SURFACE INTEGRALS

For a curve in the 2D plane, the curve can be described as , then the length of the curve is

For a plane in the 3D space, the plane can be described as , what is the area of the curved surface?

The length of one side of the curved surface is

The length of the other side of the curved surface is

Calculation of Length and Area

For a plane in the 3D space, the plane can be described as , what is the area of the curved surface?


The area of the curved surface is

Standard Derivation:

The vector of one side of the curved surface is and the vector of the other side of the curved surface is .

The area of the curved surface is .

𝑢 × �� =𝚤 𝚥 𝑘𝑑𝑥 0 𝑧 𝑑𝑥0 𝑑𝑦 𝑧 𝑑𝑦

= −𝑧 𝑑𝑥𝑑𝑦𝚤 − 𝑧 𝑑𝑥𝑑𝑦𝚥 + 𝑑𝑥𝑑𝑦𝑘

= 1 + 𝑧 + 𝑧

𝑑𝑥𝑑𝑦


Differential of a Surface Area:


Surface Integral of over is:

Example: Find the surface area of that portion of the sphere that is above the xy plane and within the cylinder

, .


Example: Find the surface area of that portion of the sphere that is above the xy plane and within the cylinder

, .


Let



DEFINITION Surface Integral

Let be a function of three variables defined over a region of 3D space containing the surface S. Then the surface integral over

S is

Evaluate the surface Integral:

𝐺 𝑥, 𝑦, 𝑧 𝑑𝑠 = 𝐺 𝑥, 𝑦, 𝑧 1 + 𝑧 + 𝑧

𝑑𝑥𝑑𝑦

𝐺 𝑥, 𝑦, 𝑧 𝑑𝑠 = 𝐺 𝑥, 𝑦, 𝑧 1 + 𝑥 + 𝑥

𝑑𝑦𝑑𝑧

𝐺 𝑥, 𝑦, 𝑧 𝑑𝑠 = 𝐺 𝑥, 𝑦, 𝑧 1 + 𝑦 + 𝑦

𝑑𝑧𝑑𝑥

Surface Integral

Application of Surface Integral


Mass of a Surface - Let represents the density of a surface at any point, or mass per unit

area; then the mass m of the surface is

Example: Find the mass of the surface of the paraboloid in the first octant for if the density at a point P is directly

proportional to its distance from xy-plane ( ).

region:

density:

Example: Find the mass of the surface of the paraboloid in the first octant for if the density at a point P is directly

proportional to its distance from xy-plane ( ).


/let

/ /

/ /


Example: Evaluate , where is that portion of the cylinder

in the first octant bounded by x=0, x=2, z=4, and z=8.


evaluate on xz planethe surface is

𝑥𝑧 𝑑𝑠 = 𝑥𝑧 1 + 16𝑥

𝑑𝑥𝑑𝑧

/

/



If a smooth surface S is defined by , then recall that a unit normal is

. If S is defined by , then we can use

or depending on the orientation of S. The two orientations of an orientable closed surface are outward and inward.

Example: Consider the sphere of radius : . If we define , what are the two orientations of the surface.

Calculation of Surface Normal


If is the velocity field of a fluid, then, the volume of the fluid passing through an element of surface area per unit

time is . The total volume of a fluid passing through S per

unit time is called the flux of through S and is given by

. In the case of a closed surface S, if is the outer

normal, then the flux gives the volume of fluid flowing out through S per unit time.

Surface Integral – Calculation of Flux in Vector Fields

Example: Let represent the flow of a liquid. Find the flux of through the surface S given by that portion of the plane

in the first octant oriented upward.


Calculation of Flux in Vector Fields

Documents

Week14 - National Chiao Tung Universitynqpl.ep.nctu.edu.tw/wp-content/uploads/2018/05/Week14.pdf'(),1,7,21 'RXEOH ,QWHJUDO /HW EH D IXQFWLRQ RI WZR YDULDEOHV GHILQHG RQ D FORVHG UHJLRQ