Surface area and surface integrals. (Sect. 16.5) Review ...

[Pages:15]Surface area and surface integrals. (Sect. 16.5)

Review: Arc length and line integrals. Review: Double integral of a scalar function. The area of a surface in space. Next class: Surface integrals of a scalar field. The flux of a vector field on a surface. Mass and center of mass thin shells.

Review: Arc length and line integrals.

The integral of a function f : [a, b] R is

b

n

a

f (x) dx = lim

n

f (xi) x.

i =0

The arc length of a curve r : [t0, t1] R3 in space is

t1

st1,t0 = |r (t)| dt.

t0

The integral of a function f : R3 R along a curve

t1

r : [t0, t1] R3 is f ds = f r(t) |r (t)| dt.

C

t0

The circulation of a function F : R3 R3 along a curve

t1

r : [t0, t1] R3 is F ? u ds = F r(t) ? r (t) dt.

C

t0

The flux of a function F : {z = 0} R3 {z = 0} R3 along

a loop r : [t0, t1] {z = 0} R3 is F = F ? n ds.

C

Surface area and surface integrals. (Sect. 16.5)

Review: Arc length and line integrals. Review: Double integral of a scalar function. The area of a surface in space.

Review: Double integral of a scalar function.

The double integral of a function f : R R2 R on a region R R2, which is the volume under the graph of f and above

the z = 0 plane, and is given by

nn

R

f

dA

=

lim

n

i =0

j =0

f

(xi, yj) x

y .

The area of a plane surface R R2 is the particular case f = 1, that is, A(R) = dA.

R

We now show how to compute: The area of a surface in space. The integral of a scalar function on a surface is space. The flux of a vector-valued function on a surface in space.

Surface area and surface integrals. (Sect. 16.5)

Review: Arc length and line integrals. Review: Double integral of a scalar function. The area of a surface in space.

The area of a surface in space.

Theorem

Given a smooth function f : R3 R, the area of a level surface S = {f (x, y , z) = 0}, over a closed, bounded region R in the plane

{z = 0}, is given by

|f |

A(S) =

dA.

R |f ? k|

z S = { f (x,y,z) = 0 } k f

Remark: Eq. (3), page 1183, in

the textbook is more general than the equation above, since the region R can be located on any plane, not only the plane {z = 0} considered here.

The vector p in the textbook is

y

the vector normal to R. In our

x

R

case p = k.

The area of a surface in space.

Proof: Introduce a partition in R R2, and consider an arbitrary

rectangle R in that partition. We compute the area P.

(x i , yi , z i ) u

P

v (x i ,yi +

y, z i )

It is simple to se that P = |u ? v|,

(x i + x, yi, z i )

k0

S u x v

x

y

R

and

u = x, 0, (zi - z^i ) , v = 0, y , (zi - zi ) . Therefore,

ij

k

u ? v = x 0 (zi - z^i ) = -y (zi - z^i ), -x(zi - zi ), xy .

0 y (zi - zi )

The area of a surface in space.

Proof: Recall: u ? v = -y (zi - z^i ), -x(zi - zi ), xy .

The linearization of f (x, y , z) at (xi , yi , zi ) implies f (x, y , z) f (xi , yi , zi ) + (x f )i x + (y f )i y + (z f )i (z - zi ).

Since f (xi , yi , zi ) = 0, f (xi + x, yi , z^i ) = 0, f (xi , yi + y , zi ) = 0,

0 = (x f )i x + (z f )i (zi - z^i )

(zi

-

z^i )

=

- (x f )i (z f )i

x ,

0 = (y f )i y + (z f )i (zi - zi )

(zi

-

zi)

=

- (y f )i (z f )i

y .

u?v =

(x f )i , (y f )i , (z f )i

x y u?v =

(f )i

xy .

(z f )i

(f ? k)i

P = |(f )i | xy A(S) =

|f | dA.

|(f ? k)i |

R |f ? k|

The area of a surface in space.

Example

Find the area of the surface in space given by the paraboloid z = x2 + y 2 between the planes z = 0 and z = 4.

Solution: The surface is the level surface of the function f (x, y , z) = x2 + y 2 - z. The region R is the disk z = x2 + y 2 4.

|f |

A(S) =

dx dy , f = 2x, 2y , -1 , f ? k = -1,

R |f ? k|

A(S) =

1 + 4x2 + 4y 2 dx dy .

R

Since R is a disk radius 2, it is convenient to use polar coordinates in R2. We obtain

2 2

A(S) =

1 + 4r 2 r dr d.

00

The area of a surface in space.

Example

Find the area of the surface in space given by the paraboloid z = x2 + y 2 between the planes z = 0 and z = 4.

2 2

Solution: Recall: A(S) =

00

2

A(S) = 2

1 + 4r 2 r dr ,

0

1 + 4r 2 r dr d. u = 1 + 4r 2, du = 8r dr .

2 A(S) =

17

u1/2 du

=

2

2

u3/2 17 .

81

83

1

We conclude: A(S) = (17)3/2 - 1 . 6

The area of a surface in space.

Remark: The formula for the area of a surface in space can be

generalized as follows.

Theorem

The area of a surface S given by f (x, y , z) = 0 over a closed and bounded plane region R in space is given by

z

f

f (x,y,z) = 0

|f |

A(S) =

dA,

R |f ? p|

R

y k

p

x

where p is a unit vector normal to the region R and f ? p = 0.

The area of a surface in space.

Example

Find the area of the region cut from the plane x + 2y + 2z = 5 by the cylinder with walls x = y 2 and x = 2 - y 2.

Solution:

z x + 2y + 2z = 5

The surface is given by f = 0 with f (x, y , z) = x + 2y + 2z - 5.

The region R is in the plane z = 0,

2

x

x = y2 y

1

x = 2 - y2

(x, y , z) : z = 0, y [-1, 1]

R=

x [y 2, (2 - y 2)]

.

|f |

Recall: A(S) =

dA. Here p = k, f = 1, 2, 2 .

R |f ? p|

The area of a surface in space.

Example

Find the area of the region cut from the plane x + 2y + 2z = 5 by the cylinder with walls x = y 2 and x = 2 - y 2.

|f |

Solution: A(S) =

dA. Here p = k, f = 1, 2, 2 .

R |f ? p|

Therefore: |f | = 1 + 4 + 4 = 3, and |f ? k| = 2.

And the region R = {(x, y ) : y [-1, 1], x [y 2, (2 - y 2)]}.

So we can write down the expression for A(S) as follows,

A(S) =

3

3 1 2-y 2

dx dy =

dx dy .

R2

2 -1 y 2

The area of a surface in space.

Example

Find the area of the region cut from the plane x + 2y + 2z = 5 by the cylinder with walls x = y 2 and x = 2 - y 2.

3 1 2-y 2

Solution: A(S) =

dx dy .

2 -1 y 2

3 A(S) =

1

2 - y2 - y2

3 dy =

1

2 - 2y 2 dy

2 -1

2 -1

A(S) = 3

1

1 - y2

dy = 3

y3 y-

1

1

1

=3 1- +1-

-1

3 -1

3

3

2

4

A(S) = 3 2 - = 3 A(S) = 4.

3

3

Surface area and surface integrals. (Sect. 16.5)

Review: The area of a surface in space. Surface integrals of a scalar field. The flux of a vector field on a surface. Mass and center of mass thin shells.

Review: The area of a surface in space.

Theorem

Given a smooth function f : R3 R, the area of a level surface S = {f (x, y , z) = 0}, over a closed, bounded region R in the plane

{z = 0}, is given by

|f |

A(S) =

dA.

R |f ? k|

z S = { f (x,y,z) = 0 } k f

Remark: Eq. (3), page 1183, in

the textbook is more general than the equation above, since the region R can be located on any plane, not only the plane {z = 0} considered here.

The vector p in the textbook is

y

the vector normal to R. In our

x

R

case p = k.

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