[PDF] [PDF] Review for Exam 3 Triple integral in spherical coordinates (Sect

[PDF] Review for Exam 3 Triple integral in spherical coordinates (Sect

Since sin(θ)=1/2, we get θ1 = 5π/6 and θ0 = π/6 Double integrals in polar coordinates (Sect 15 3) Example Find the area of the region 

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Review for Exam 3.

?Sections 15.1-15.4, 15.6. ?50 minutes. ?5 problems, similar to homework problems. ?No calculators, no notes, no books, no phones. ?No green book needed.Triple integral in spherical coordinates (Sect. 15.6).

Example

Use spherical coordinates to find the volume of the region outside the sphereρ= 2cos(φ) and inside the half sphereρ= 2 with φ?[0,π/2].Solution:First sketch the integration region. ?ρ= 2cos(φ) is a sphere,since

2= 2ρcos(φ)?x2+y2+z2= 2zx

2+y2+ (z-1)2= 1.?ρ= 2 is a sphere radius 2 and

φ?[0,π/2] says we only consider

the upper half of the sphere.rho = 2 cos ( 0 ) yz x 1 2 22
rho = 2 Triple integral in spherical coordinates (Sect. 15.6).

Example

Use spherical coordinates to find the volume of the region outside the sphereρ= 2cos(φ) and inside the sphereρ= 2 with

φ?[0,π/2].Solution:rho = 2 cos ( 0 )

yz x 1 2 22
rho = 2V=? 2π 0?

π/2

0? 2

2cos(φ)ρ2sin(φ)dρdφdθ.V= 2π?

π/2

0?

ρ33

??2

2cos(φ)?

sin(φ)dφ

2π3

π/2

0?

8sin(φ)-8cos3(φ) sin(φ)?

dφ.V=16π3?? -cos(φ)???π/2 0?

π/2

0 cos3(φ)sin(φ)dφ? .Triple integral in spherical coordinates (Sect. 15.6).

Example

Use spherical coordinates to find the volume of the region outside the sphereρ= 2cos(φ) and inside the sphereρ= 2 with

φ?[0,π/2].Solution:V=16π3

-cos(φ)???π/2 0?

π/2

0 cos3(φ)sin(φ)dφ? Introduce the substitution:u= cos(φ),du=-sin(φ)dφ.V=16π3 1 +? 0 1 u3du?=

16π3

1 +?u44

??0 1??=

16π3

1-14 .V=16π3 34
?V= 4π.? Triple integral in cylindrical coordinates (Sect. 15.6).

Example

Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x-2)2+y2= 4 by the planesz= 0 and z=-y.Solution:First sketch the integration region. ?(x-2)2+y2= 4 is a circle,since x

2+y2= 4x?r2= 4rcos(θ)r= 4cos(θ).?Since 0?z?-y, the integration

region is on they?0 part of the z= 0 plane.4 xyz z = - y 2 2 (x - 2) + y = 4

2Triple integral in cylindrical coordinates (Sect. 15.6).

Example

Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x-2)2+y2= 4 by the planesz= 0 and z=-y.Solution: 4 xyz z = - y 2 2 (x - 2) + y = 4 2V=? 2π

3π/2?

4cos(θ)

0? -rsin(θ) 0 r dz dr dθ.V=? 2π

3π/2?

4cos(θ)

0?-rsin(θ)-0?r dr dθ

V=-? 2π

3π/2?

r33 ??4cos(θ) 0? sin(θ)dθ.V=-? 2π

3π/24

33
cos3(θ)sin(θ)dθ. Triple integral in cylindrical coordinates (Sect. 15.6).

Example

Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x-2)2+y2= 4 by the planesz= 0 and z=-y.Solution:V=-? 2π

3π/24

33
cos3(θ)sin(θ)dθ.Introduce the substitution:u= cos(θ),du=-sin(θ)dθ;V=433 1 0 u3du= 433
u44 ??1 0?= 433
14 .We conclude:V=163.?Triple integral in Cartesian coordinates (Sect. 15.4).

Example

Find the volume of a parallelepiped whose base is a rectangle in thez= 0 plane given by 0?y?2 and 0?x?1, while the top side lies in the planex+y+z= 3.Solution:z x3y 33V=?
1 0? 2 0? 3-x-y 0 dz dy dx,V=? 1 0? 2 0 (3-x-y)dy dx, 1 0? (3-x)? y???2 0? -12 y2???2 0?? dx,V=? 1 0?

2(3-x)-42

dx.V=? 1

0?4-2x?dx=

4? x???1 0? x2???1

0??= 4-1?V= 3.

Double integrals in polar coordinates. (Sect. 15.3)

Example

Find the area of the region in the plane inside the curve r= 6sin(θ) and outside the circler= 3, wherer,θare polar coordinates in the plane.Solution:First sketch the integration region. ?r= 6sin(θ) is a circle,since r

2= 6rsin(θ)?x2+y2= 6yx

2+ (y-3)2= 32.?The other curve is a circler= 3centered

at the origin.r = 3 xy 3 3-36

r = 6 cos ( 0 )The condition 3 =r= 6sin(θ) determines the range inθ.Since sin(θ) = 1/2,we getθ

1= 5π/6andθ

0=π/6.Double integrals in polar coordinates. (Sect. 15.3)

Example

Find the area of the region in the plane inside the curve r= 6sin(θ) and outside the circler= 3, wherer,θare polar coordinates in the plane.Solution:Recall:θ?[π/6,5π/6].A=?

5π/6

π/6?

6sin(θ)

3 rdr dθ=

5π/6

π/6?

r22 ??6sin(θ) 3? dθA=?

5π/6

π/6?

622
sin2(θ)-322 dθ=

5π/6

π/6?

622

2?1-cos(2θ)?-322

dθA= 32?5π6 -π6 -322 sin(2θ)???5π/6

π/6?

-322

5π6

-π6

A= 6π-3π-322?

-⎷3

2-⎷3

2? , henceA= 3π+ 9⎷3/2.? Double integrals in Cartesian coordinates. (Sect. 15.2)

Example

Find they-component of the centroid vector in Cartesian coordinates in the plane of the region given by the disk x

2+y2?9 minus the first quadrant.Solution:First sketch the integration region.3

y x 3y=1A R y dA, whereA=πR2(3/4), with

R= 3.That is,A= 27π/4.We use polar

coordinates to computey.y=427π? 2π

π/2?

3 0 rsin(θ)rdr dθ.y=427π? -cos(θ)???2π

π/2??

r33 ??3 0?=

427π(-1)(9)?y=-43π.Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral I=? -⎷2 -2? ⎷4-x2 ⎷4-x2?x2+y2?dy dx+? ⎷2 ⎷2 ⎷4-x2 x?x2+y2?dy dx.Solution:First sketch the integration region. ?x?[-2,⎷2]. ?Forx?[-2,-⎷2], we have |y|?⎷4-x2,so the curve is part of the circlex2+y2= 4.?Forx?[-⎷2,⎷2], we have thaty is between the liney=xand the upper side of the circle x

2+y2= 4.

2 y x x + y = 4y = x

2-22- 2

2 Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral I=? -⎷2 -2? ⎷4-x2 ⎷4-x2?x2+y2?dy dx+? ⎷2 ⎷2 ⎷4-x2 x?x2+y2?dy dx.Solution:2 y x x + y = 4y = x

2-22- 2

2I=?

5π/4

π/4?

2 0 r2rdr dθ

I=?5π4

-π4 ?2 0 r3dr

I=π?r44

??2 0? We conclude:I= 4π.?Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral I=? 0 -2? ⎷4-x2

0?x2+y2?dy dx+?

⎷2 0? ⎷4-x2 x?x2+y2?dy dxSolution:First sketch the integration region. ?x?[-2,⎷2]. ?Forx?[-2,0], we have 0?yand y?⎷4-x2. The latter curve is part of the circlex2+y2= 4.?Forx?[0,⎷2], we havex?yand y?⎷4-x2. 2 y x x + y = 4y = x 2-22 2 Double integrals in polar coordinates. (Sect. 15.2)

Example

Transform to polar coordinates and then evaluate the integral I=? 0 -2? ⎷4-x2

0?x2+y2?dy dx+?

⎷2 0? ⎷4-x2 x?x2+y2?dy dxSolution:2 y x x + y = 4y = x 2-22 2I=?

π/4?

2 0 r2rdr dθI=3π4 r44 ??2

0?We conclude:I= 3π.?Integrals along a curve in space. (Sect. 16.1)

?Line integrals in space. ?The addition of line integrals. ?Mass and center of mass of wires.

Line integrals in space.

Definition

Theline integralof a functionf:D?R3→Ralong a curve associated with the functionr: [t0,t1]?R→D?R3is given by? C f ds=? s1 s

0f?ˆr(s)?ds,whereˆ

r(s)is the arc length parametrization of the functionr, and s(t0) =s0,s(t1) =s1are the arc lengths at the pointst0,t1, respectively.( f r ) r ( s )rf f ( r (s ) ) s

00Line integrals in space.

Remarks:

?A line integral is an integral of a function along a curved path.

?Why is the functionrparametrized with its arc length?(1)Because in this way the line integral isindependent of the

original parametrization of the curve.Given two different parametrizations of the curve, we have switch them to the unique arc length parametrization and compute the integral above.(2)Because this is the appropriate generalization of the integral of a functionF:R→R.Recall:? b a

F(x)dx= limn→∞n

i=0F(x?i)Δxi, where

Δxi=xi+1-xiis thedistancefromxi+1tox1.This Δxigeneralizes to Δsion a curved path. This is why the

arc length parametrization is needed in the line integral.

Line integrals in space.

Theorem (Arbitrary parametrization.)

The line integral of a continuous function f:D?R3→Ralong a differentiable curver: [t0,t1]?R→D?R3is given by? C f ds=? t1 t

0f(r(t))|r?(t)|dt,where t is any parametrization of the vector-valued functionr.Proof:The integration by substitution formula says

s1 s

0f?ˆr(s)?ds=?

t1 t

0f?ˆr(s(t))?s?(t)dt,s

0=s(t0),

s

1=s(t1).The arc length function iss(t) =?t

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