Review for Exam 3 Double integrals in Cartesian ...

[Pages:10]Review for Exam 3

Tuesday Recitations: 14.7, 15.1-15.5, half 15.7. Thursday Recitations: 15.1-15.5, 15.7. 50 minutes. From five 10-minute problems to ten 5-minutes problems. Problems similar to homework problems. No calculators, no notes, no books, no phones.

Double integrals in Cartesian coordinates (Section 15.2)

Example

3

2(1-

x 3

)

Switch the integration order in I =

f (x, y ) dy dx.

0

-2

q 1-

x2 32

Solution:

We first draw the integration region. Start with the outer limits.

x [0, 3]. x2

y 2 - 2x/3 and y 2 1 - 32 . The lower limit is part of the ellipse

x2 y2 32 + 22 = 1.

y

2

3x

-2

Double integrals in Cartesian coordinates (Section 15.2)

Example

3

2(1-

x 3

)

Switch the integration order in I =

f (x, y ) dy dx.

0

-2

q 1-

x2 32

Solution:

Split the integral at y = 0.

y

2

3x

-2

In y [-2, 0], holds 0 x.

The upper limit comes from x2 y2 32 + 22 = 1,

y2 so, x = +3 1 - 22 .

In y [0, 2], holds 0 x. The upper limit comes from

x

y

y = 2 1 - , that is, x = 3 1 - . We then conclude:

3

2

0

q 3 1-

y2 22

2

3(1-

y 2

)

I=

f (x, y ) dx dy +

f (x, y ) dx dy .

-2 0

00

Areas as double integrals (Section 15.3)

Example

Compute the are of the region on the xy -plane below the curve y = 4 - x2 and above y = x2. Also switch the integration order.

Solution: First, sketch the integration region.

It is simpler integrating dy dx.

y

4

y = x2

2 4-x2

A=

dy dx.

-2 x2

-2

A=

y = 4 - x2

2

x

2

A = (4 - x2) - x2 dx

-2

2 (4 - 2x2) dx = 4x 2

- 2x3 2

2 = (8 + 8) - (8 + 8)

-2

-2 3 -2

3

16 A= .

3

Areas as double integrals (Section 15.3)

Example

Compute the are of the region on the xy -plane below the curve y = 4 - x2 and above y = x2. Also switch the integration order.

Solution: We now interchage the integration region to dx dy .

y

4

y = x2

y = 4 - x2

We need to divide the y -interval at y such that

4 - x2 = x2 x = ? 2.

-2

2

x

That is, y = 2. Then,

2 y

4

4-y

A=

0

dx dy +

-y

2

dx dy .

- 4-y

Double integrals in polar coordinates. (Sect. 15.4)

Example

Transform to polar coordinates and then evaluate the integral

-2

4-x 2

2

4-x 2

I=

x2 + y 2 dy dx +

x2 + y 2 dy dx.

-2

- 4-x2

-2 x

Solution: First sketch the integration region.

x [-2, 2].

y

y = x

For x [-2, - 2], we have |y | 4 - x2, so the curve is part

of the circle x2 + y 2 = 4.

For x [- 2, 2], we have that y

-2

- 2

2 2x

is between the line y = x and the upper side of the circle

x

2

+

y

2

=

4

x2 + y 2 = 4.

Double integrals in polar coordinates. (Sect. 15.4)

Example

Transform to polar coordinates and then evaluate the integral

-2

4-x 2

2

4-x 2

I=

x2 + y 2 dy dx +

x2 + y 2 dy dx.

-2

- 4-x2

-2 x

Solution:

y

y = x

-2

- 2

2 2x

x

2

+

y

2

=

4

5/4 2

I=

r 2 rdr d

/4 0

5 I= -

2

r 3 dr

4 40

r4 2 I =

40

I = 4.

Double integrals in polar coordinates. (Sect. 15.4)

Example

Transform to polar coordinates and then evaluate the integral

0

4-x 2

2

4-x 2

I=

x2 + y 2 dy dx +

x2 + y 2 dy dx

-2 0

0x

Solution: First sketch the integration region.

x [-2, 2].

For x [-2, 0], we have y 0 and y 4 - x2. The latter curve is

part of the circle x2 + y 2 = 4.

-2

For x [0, 2], we have y x and

y 4 - x2.

y

y = x

2 2x x2+ y2 = 4

Double integrals in polar coordinates. (Sect. 15.4)

Example

Transform to polar coordinates and then evaluate the integral

0

4-x 2

2

4-x 2

I=

x2 + y 2 dy dx +

x2 + y 2 dy dx

-2 0

0x

Solution:

y

y = x

-2

2 2x

x2+ y2 = 4

2

I=

r 2 rdr d

/4 0

3 r 4 2 I=

4 40

We conclude: I = 3.

Triple integral in Cartesian coordinates (Sect. 15.5)

Example

Find the volume of the region in the first octant below the plane 2x + y - 2z = 2 and x 1, y 2.

Solution: First sketch the integration region.

The plane contains the points (1, 0, 0), (0, 2, 0), (1, 2, 1).

We choose the order dz dy dx. The integral is

z 2x+y-2z = 2

(1,2,1)

2y

12

-1+x+y /2

V=

dz dy dx.

0 2-2x 0

1

x 2x+y = 2

12

y

V=

(-1 + x) + dy dx,

0 2-2x

2

V=

1

-

(1

-

x )[2

-

2(1

-

x )]

+

1 [4

-

4(1

-

x )2]

dx .

0

4

Triple integral in Cartesian coordinates (Sect. 15.5)

Example

Find the volume of the region in the first octant below the plane 2x + y - 2z = 2 and x 1, y 2.

Solution: V = 1 - (1 - x)[2 - 2(1 - x)] + 1 [4 - 4(1 - x)2] dx.

0

4

1

V = - 2(1 - x) + 2(1 - x)2 + 1 - (1 - x)2 dx,

0

1

1

V = - 1 + 2x + (1 - x)2 dx = - 1 + 2x + 1 + x2 - 2x dx

0

0

V=

1

x2 dx

=

x3

1

1 V= .

0

30

3

Triple integral in Cartesian coordinates (Sect. 15.5)

Example

Find the volume of the region in the first octant below the plane x + y + z = 3 and y 1.

Solution: First sketch the integration region.

The plane contains the points (1, 0, 0), (0, 2, 0), (1, 2, 1).

z x+y+z=3

3

We choose the order dz dy dx. We need x + y = 3 at z = 0.

2 1 3-x-y

V=

dz dy dx

000

3 3-x 3-x-y

1

3y

+

dz dy dx.

20

0

3

x

21

3 3-x

V=

(3 - x - y ) dy dx +

(3 - x - y ) dy dx.

00

20

Triple integral in Cartesian coordinates (Sect. 15.5)

Example

Find the volume of the region in the first octant below the plane x + y + z = 3 and y 1.

Solution:

21

3 3-x

V=

(3 - x - y ) dy dx +

(3 - x - y ) dy dx.

00

20

2

1

y2 1

(3-x )

y 2 (3-x)

V = (3-x) y -

+(3-x) y

-

dx

0

0

20

0

20

V=

2

(3

-

x)

-

1

+

(3

-

x )2

-

1 (3

-

x )2

dx

0

2

2

V=

2

5

-

x

+

1 (3

-

x )2

dx

22 V= .

02

2

3

Triple integral in spherical coordinates (Sect. 15.7)

Example

Use spherical coordinates to find the volume of the region below the paraboloid z = 9 - x2 - y 2 below the xy -plane and outside the cylinder x2 + y 2 = 1.

Solution: First sketch the integration region.

z

z = 9 - x2 - y2

In cylindrical coordinates, z = 9 - x2 - y2 z = 9 - r2.

1

3

y

2

2

x

x + y =1

2 3

V=

01

x2 + y 2 = 1 r = 1.

9-r 2

3

r dz dr d = 2 (9 - r 2) r dr

0

1

9r 2 r 4 3

V = 2 -

V = 32.

2 41

Triple integral in spherical coordinates (Sect. 15.7)

Example

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

Solution: First sketch the integration region. = 2 cos() is a sphere, since

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

x2 + y 2 + (z - 1)2 = 1.

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

z

2

rho = 2

1

2y

2

x rho = 2 cos ( 0 )

Triple integral in spherical coordinates (Sect. 15.7)

Example

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

Solution:

z

2

rho = 2

1

2y

2

x rho = 2 cos ( 0 )

2 /2 2

V=

2 sin() d d d.

00

2 cos()

/2 3 2

V = 2

sin() d

0

3 2 cos()

2 V=

/2

8 sin() - 8 cos3() sin() d.

30

V = 16

/2

- cos() -

/2

cos3() sin() d .

3

0

0

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download