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