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### Transcript of Review for Exam 3. Triple integral in spherical ... · PDF fileTriple integral in cylindrical...

Review for Exam 3.

I Sections 15.1-15.4, 15.6.

I 50 minutes.

I 5 problems, similar to homework problems.

I No calculators, no notes, no books, no phones.

I No green book needed.

Triple integral in spherical coordinates (Sect. 15.6).

Example

Use spherical coordinates to find the volume of the region outsidethe sphere = 2 cos() and inside the half sphere = 2 with [0, /2].

Solution: First sketch the integration region.

I = 2 cos() is a sphere, since

2 = 2 cos() x2+y2+z2 = 2z

x2 + y2 + (z 1)2 = 1.

I = 2 is a sphere radius 2 and [0, /2] says we only considerthe upper half of the sphere.

rho = 2 cos ( 0 )

y

z

x

1

2

2

2

rho = 2

Triple integral in spherical coordinates (Sect. 15.6).

Example

Use spherical coordinates to find the volume of the region outsidethe sphere = 2 cos() and inside the sphere = 2 with [0, /2].

Solution:

rho = 2 cos ( 0 )

y

z

x

1

2

2

2

rho = 2

V =

20

/20

22 cos()

2 sin() d d d.

V = 2

/20

(33

22 cos()

)sin() d

=2

3

/20

[8 sin() 8 cos3() sin()

]d.

V =16

3

[( cos()

/20

)

/20

cos3() sin() d].

Triple integral in spherical coordinates (Sect. 15.6).

Example

Use spherical coordinates to find the volume of the region outsidethe sphere = 2 cos() and inside the sphere = 2 with [0, /2].

Solution: V =16

3

[( cos()

/20

)

/20

cos3() sin() d].

Introduce the substitution: u = cos(), du = sin() d.

V =16

3

[1 +

01

u3 du]

=16

3

[1 +

(u44

01

)]=

16

3

(1 1

4

).

V =16

3

3

4 V = 4.

C

Triple integral in cylindrical coordinates (Sect. 15.6).

Example

Use cylindrical coordinates to find the volume of a curved wedgecut out from a cylinder (x 2)2 + y2 = 4 by the planes z = 0 andz = y .

Solution: First sketch the integration region.

I (x 2)2 + y2 = 4 is a circle, since

x2 + y2 = 4x r2 = 4r cos()

r = 4 cos().

I Since 0 6 z 6 y , the integrationregion is on the y 6 0 part of thez = 0 plane.

4

x

y

z

z = y

2

2

(x 2) + y = 42

Triple integral in cylindrical coordinates (Sect. 15.6).

Example

Use cylindrical coordinates to find the volume of a curved wedgecut out from a cylinder (x 2)2 + y2 = 4 by the planes z = 0 andz = y .

Solution:

4

x

y

z

z = y

2

2

(x 2) + y = 42

V =

23/2

4 cos()0

r sin()0

r dz dr d.

V =

23/2

4 cos()0

[r sin() 0

]r dr d

V = 2

3/2

( r33

4 cos()0

)sin() d.

V = 2

3/2

43

3cos3() sin() d.

Triple integral in cylindrical coordinates (Sect. 15.6).

Example

Use cylindrical coordinates to find the volume of a curved wedgecut out from a cylinder (x 2)2 + y2 = 4 by the planes z = 0 andz = y .

Solution: V = 2

3/2

43

3cos3() sin() d.

Introduce the substitution: u = cos(), du = sin() d;

V =43

3

10

u3 du =43

3

(u44

10

)=

43

3

1

4.

We conclude: V =16

3. C

Triple integral in Cartesian coordinates (Sect. 15.4).

Example

Find the volume of a parallelepiped whose base is a rectangle inthe z = 0 plane given by 0 6 y 6 2 and 0 6 x 6 1, while the topside lies in the plane x + y + z = 3.

Solution:z

x3

y3

3

V =

10

20

3xy0

dz dy dx ,

V =

10

20

(3 x y) dy dx ,

=

10

[(3 x)

(y20

) 1

2

(y2

20

)]dx ,

V =

10

[2(3 x) 4

2

]dx .

V =

10

(4 2x

)dx =

[4(x10

)

(x2

10

)]= 4 1 V = 3.

Double integrals in polar coordinates. (Sect. 15.3)

Example

Find the area of the region in the plane inside the curver = 6 sin() and outside the circle r = 3, where r , are polarcoordinates in the plane.

Solution: First sketch the integration region.

I r = 6 sin() is a circle, since

r2 = 6r sin() x2 + y2 = 6y

x2 + (y 3)2 = 32.

I The other curve is a circle r = 3 centeredat the origin.

r = 3

x

y

3

33

6 r = 6 cos ( 0 )

The condition 3 = r = 6 sin() determines the range in .Since sin() = 1/2, we get 1 = 5/6 and 0 = /6.

Double integrals in polar coordinates. (Sect. 15.3)Example

Find the area of the region in the plane inside the curver = 6 sin() and outside the circle r = 3, where r , are polarcoordinates in the plane.

Solution: Recall: [/6, 5/6].

A =

5/6/6

6 sin()3

rdr d =

5/6/6

( r22

6 sin()3

)d

A =

5/6/6

[622

sin2() 32

2

]d =

5/6/6

[6222

(1cos(2)

) 3

2

2

]d

A = 32(5

6

6

) 3

2

2

(sin(2)

5/6/6

) 3

2

2

(56

6

).

A = 6 3 32

2

(

3

2

3

2

), hence A = 3 + 9

3/2. C

Double integrals in Cartesian coordinates. (Sect. 15.2)

Example

Find the y -component of the centroid vector in Cartesiancoordinates in the plane of the region given by the diskx2 + y2 6 9 minus the first quadrant.

Solution: First sketch the integration region.

3

y

x3

y =1

A

R

y dA, where A = R2(3/4), with

R = 3. That is, A = 27/4. We use polarcoordinates to compute y .

y =4

27

2/2

30

r sin() rdr d.

y =4

27

( cos()

2/2

)( r33

30

)=

4

27(1)(9) y = 4

3.

Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral

I =

22

4x2

4x2

(x2+y2

)dy dx +

2

2

4x2x

(x2+y2

)dy dx .

Solution: First sketch the integration region.

I x [2,

2].

I For x [2,

2], we have|y | 6

4 x2, so the curve is part

of the circle x2 + y2 = 4.

I For x [

2,

2], we have that yis between the line y = x and theupper side of the circlex2 + y2 = 4.

2

y

x

x + y = 4

y = x

22 2 2

2

Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral

I =

22

4x2

4x2

(x2+y2

)dy dx +

2

2

4x2x

(x2+y2

)dy dx .

Solution:

2

y

x

x + y = 4

y = x

22 2 2

2

I =

5/4/4

20

r2 rdr d

I =(5

4

4

) 20

r3 dr

I = ( r4

4

20

)We conclude: I = 4. C

Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral

I =

02

4x20

(x2 + y2

)dy dx +

20

4x2x

(x2 + y2

)dy dx

Solution: First sketch the integration region.

I x [2,

2].

I For x [2, 0], we have 0 6 y andy 6

4 x2. The latter curve is

part of the circle x2 + y2 = 4.

I For x [0,

2], we have x 6 y andy 6

4 x2.

2

y

x

x + y = 4

y = x

22 2

2

Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral

I =

02

4x20

(x2 + y2

)dy dx +

20

4x2x

(x2 + y2

)dy dx

Solution:

2

y

x

x + y = 4

y = x

22 2

2

I =

/4

20

r2 rdr d

I =3

4

( r44

20

)We conclude: I = 3. C

Integrals along a curve in space. (Sect. 16.1)

I Line integrals in space.

I The addition of line integrals.

I Mass and center of mass of wires.

Line integrals in space.

DefinitionThe line integral of a function f : D R3 R along a curveassociated with the function r : [t0, t1] R D R3 is given by

C

f ds =

s1s0

f(r(s)

)ds,

where r(s) is the arc length parametrization of the function r, ands(t0) = s0, s(t1) = s1 are the arc lengths at the points t0, t1,respectively.

( f r )

r ( s )rf

f ( r (s ) )

s00

Line integrals in space.

Remarks:

I A line integral is an integral of a function along a curved