Triple Integrals in Cylindrical and Spherical Coordinates
Note: Remember that in polar coordinates dA = r dr d . EX 1 Find the volume of the solid bounded above by the sphere x2 + y2 + z2 = 9 below by the plane z
Review for Exam 3. Triple integral in spherical coordinates (Sect
2. ) hence A = 3π + 9. √. 3/2. <. Page 6. Double integrals in Cartesian coordinates. (Sect. 15.2). Example. Find the y-component of the centroid vector in
MATH 20550 Triple Integrals in cylindrical and spherical coordinates
into a spherical coordinate iterated integral. (from here example 2.) Let us start by describing the solid. Note ∫. 3. 0. ∫. √.
Triple Integrals in Cylindrical Coordinates Many applications involve
In particular there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example
15.8: Triple Integrals in Spherical Coordinates
We have: Page 6. Example. Let's get a better handle on things by graphing some basic functions given in spherical coordinates. Let c be a constant. Sketch the
April 8: Triple Integrals via Spherical and Cylindrical Coordinates
Apr 8 2020 Example 1. Let's begin as we did with polar coordinates. We want a. 3-dimensional analogue of integrating over a circle. So we integrate ...
Triple Integrals in Spherical Coordinates
For example the sphere with center the origin and radius c has the simple equation ρ = c (see Figure 2); this is the reason for the name “spherical”
Triple Integrals in Cylindrical and Spherical Coordinates
Triple Integrals in Cylindrical and Spherical Coordinates. 29/67. Page 30. How to Integrate in Spherical Coordinates - An Example. Example 5. Find the volume of
3.6 Integration with Cylindrical and Spherical Coordinates
In this section we describe
TRIPLE INTEGRALS IN SPHERICAL COORDINATES EXAMPLE A
TRIPLE INTEGRALS IN SPHERICAL COORDINATES. EXAMPLE A Find an equation in spherical coordinates for the hyperboloid of two sheets with equation . SOLUTION
Review for Exam 3. Triple integral in spherical coordinates (Sect
Line integrals in space. Example. Evaluate the line integral of the function f (xy
Page 1 Section 15.8: Triple Integrals in Spherical Coordinates
Spherical Coordinates: A Cartesian point (x y
15.8: Triple Integrals in Spherical Coordinates
We have: Page 6. Example. Let's get a better handle on things by graphing some basic functions given in spherical coordinates. Let c be a constant. Sketch the
MATH 20550 Triple Integrals in cylindrical and spherical coordinates
into a spherical coordinate iterated integral. (from here example 2.) Let us start by describing the solid. Note ?. 3. 0. ?. ?.
Integrals in cylindrical spherical coordinates (Sect. 15.7) Cylindrical
Notice the extra factor ?2 sin(?) on the right-hand side. Triple integral in spherical coordinates. Example. Find the volume of a sphere of radius R.
Triple Integrals in Cylindrical Coordinates Many applications involve
In particular there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example
Triple Integrals in Spherical Coordinates
For example the sphere with center the origin and radius c has the simple equation ? = c (see Figure 2); this is the reason for the name “spherical”
Triple Integrals in Cylindrical or Spherical Coordinates
xyz dV as an iterated integral in cylindrical coordinates. x y z. Solution. This is the same problem as #3 on the worksheet “Triple Integrals”
15.7 Triple Integrals in Cylindrical and Spherical Coordinates
Figure 15.44 Page 894. Example. Page 901
Section 9.7/12.8: Triple Integrals in Cylindrical and Spherical
from rectangular to spherical coordinates. Solution: ·. Example 7: Convert the equation ? ? sec2. =.
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,since2= 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 22rho = 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 withφ?[0,π/2].Solution:rho = 2 cos ( 0 )
yz x 1 2 22rho = 2V=? 2π 0?
π/2
0? 22cos(φ)ρ2sin(φ)dρdφdθ.V= 2π?
π/2
0?ρ33
??22cos(φ)?
sin(φ)dφ2π3
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 withφ?[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 x2+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 = 42Triple 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=? 2π3π/2?
4cos(θ)
0? -rsin(θ) 0 r dz dr dθ.V=? 2π3π/2?
4cos(θ)
0?-rsin(θ)-0?r dr dθ
V=-? 2π3π/2?
r33 ??4cos(θ) 0? sin(θ)dθ.V=-? 2π3π/24
33cos3(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=-? 2π3π/24
33cos3(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=? 10?4-2x?dx=
4? x???1 0? 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 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 r2= 6rsin(θ)?x2+y2= 6yx
2+ (y-3)2= 32.?The other curve is a circler= 3centered
at the origin.r = 3 xy 3 3-36r = 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?
622sin2(θ)-322 dθ=
5π/6
π/6?
6222?1-cos(2θ)?-322
dθA= 32?5π6 -π6 -322 sin(2θ)???5π/6π/6?
-3225π6
-π6A= 6π-3π-322?
-⎷32-⎷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 x2+y2?9 minus the first quadrant.Solution:First sketch the integration region.3
y x 3y=1A R y dA, whereA=πR2(3/4), withR= 3.That is,A= 27π/4.We use polar
coordinates to computey.y=427π? 2ππ/2?
3 0 rsin(θ)rdr dθ.y=427π? -cos(θ)???2ππ/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 x2+y2= 4.
2 y x x + y = 4y = x2-22- 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 = x2-22- 2
2I=?5π/4
π/4?
2 0 r2rdr dθI=?5π4
-π4 ?2 0 r3drI=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-x20?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-x20?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 ??20?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 s0f?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 ) ) s00Line 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 aF(x)dx= limn→∞n
i=0F(x?i)Δxi, whereΔxi=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 t0f(r(t))|r?(t)|dt,where t is any parametrization of the vector-valued functionr.Proof:The integration by substitution formula says
s1 s0f?ˆr(s)?ds=?
t1 t0f?ˆr(s(t))?s?(t)dt,s
0=s(t0),
s1=s(t1).The arc length function iss(t) =?t
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