find polar coordinates ri and θi,
with ri>0 and 0≤θi<2π, for the point (xi,yi).
Answer+
The left hand sketch below contains the points, (x1,y1), (x3,y3),
(x5,y5), that are on the axes. The right hand sketch below contains the
points, (x2,y2), (x4,y4), that are not on the axes.
The left hand sketch below contains the points, (x1,y1), (x3,y3),
(x5,y5), that are on the axes. The right hand sketch below contains the
points, (x2,y2), (x4,y4), that are not on the axes.
Recall that the polar coordinates r, θ
are related to the cartesian coordinates x, y, by x=rcosθ,
y=rsinθ. So r=x2+y2 and tanθ=xy
(assuming that x=0 and r>0) and
(x1,y1)(x2,y2)(x3,y3)(x4,y4)(x5,y5)=(3,0)=(1,1)=(0,1)=(−1,1)=(−2,0)⟹⟹⟹⟹⟹r1=3,tanθ1=0r2=2,tanθ2=1r3=1,cosθ3=0r4=2,tanθ4=−1r5=2,tanθ5=0⟹θ1⟹θ2⟹θ3⟹θ4⟹θ5=0 as (x1,y1) is on the positive x-axis=4π as (x2,y2) is in the first octant=2π as (x3,y3) is on the positive y-axis=43π as (x4,y4) is in the third octant=π as (x5,y5) is on the negative x-axis
find all pairs (ri,θi) such that
(xi,yi)=(ricosθi,risinθi), and
in particular, find a pair (ri,θi) with ri<0 and
0≤θi<2π such that
(xi,yi)=(ricosθi,risinθi)
(x1,y1)=(−2,0)
(x2,y2)=(1,1)
(x3,y3)=(−1,−1)
(x4,y4)=(3,0)
(x5,y5)=(0,1)
Hint+
r is allowed to be negative.
Answer+
(a) (r1=2,θ1=nπ,n odd integer ) or
(r1=−2,θ1=nπ,n even integer ).
In particular, (r1=−2,θ1=0) has r1<0
and 0≤θ1<2π.
(b) (r2=2,θ2=π/4+2nπ) or
(r2=−2,θ2=5π/4+2nπ),
with n integer. In particular, (r2=−2,θ2=5π/4) has r2<0
and 0≤θ2<2π.
(c) (r3=2,θ3=5π/4+2nπ) or
(r3=−2,θ3=π/4+2nπ),
with n integer. In particular, (r3=−2,θ3=π/4) has r3<0
and 0≤θ3<2π.
(d) (r4=3,θ4=0+2nπ) or
(r4=−3,θ4=π+2nπ),
with n integer. In particular, (r4=−3,θ4=π) has r4<0 and 0≤θ4<2π.
(e) (r5=1,θ5=π/2+2nπ) or
(r5=−1,θ5=3π/2+2nπ),
with n integer. In particular, (r5=−1,θ5=3π/2) has r5<0
and 0≤θ5<2π.
Full solution+
In this solution, we'll supress the subscripts. That is, we'll write r in place of ri and θ in place of θi.
Note that the distance from the point (rcosθ,rsinθ)
to the origin is
r2cos2θ+r2sin2θ=r2=∣r∣
Thus r can be either the distance to the origin or minus the distance to the
origin.
(a)
The distance from (−2,0) to the origin is 2. So either r=2 or r=−2.
(−2,0)=(−2cosθ,−2sinθ)⟺sinθ=0,cosθ=1⟺θ=nπ,n integer ,cosθ=1⟺θ=nπ,n even integer
In particular, (r=−2,θ=0) has r<0
and 0≤θ<2π.
In the figure on the left below, the blue half-line is the set of all points
with polar coordinates θ=π, r>0 and the orange half-line is the set
of all points with polar coordinates θ=π, r<0.
In the figure on the right below, the blue half-line is the set of all points
with polar coordinates θ=0, r>0 and the orange half-line is the set
of all points with polar coordinates θ=0, r<0.
(b)
The distance from (1,1) to the origin is 2.
So either r=2 or r=−2.
In particular, (r=−2,θ=5π/4) has r<0
and 0≤θ<2π.
In the figure on the left below, the blue half-line is the set of all points
with polar coordinates θ=4π, r>0 and the orange half-line
is the set of all points with polar coordinates θ=4π, r<0.
In the figure on the right below, the blue half-line is the set of all points
with polar coordinates θ=45π, r>0 and the orange
half-line is the set of all points with polar coordinates
θ=45π, r<0.
(c)
The distance from (−1,−1) to the origin is 2.
So either r=2 or r=−2.
In the figure on the left below, the blue half-line is the set of all points
with polar coordinates θ=45π, r>0 and the orange half-line
is the set of all points with polar coordinates
θ=45π, r<0.
In the figure on the right below, the blue half-line is the set of all points
with polar coordinates θ=4π, r>0 and the orange
half-line is the set of all points with polar coordinates
θ=4π, r<0.
(d)
The distance from (3,0) to the origin is 3.
So either r=3 or r=−3.
In the figure on the left below, the blue half-line is the set of all points
with polar coordinates θ=π, r>0 and the orange half-line is the set
of all points with polar coordinates θ=π, r<0.
In the figure on the right below, the blue half-line is the set of all points
with polar coordinates θ=0, r>0 and the orange half-line is the set
of all points with polar coordinates θ=0, r<0.
(e)
The distance from (0,1) to the origin is 1.
So either r=1 or r=−1.
In the figure on the left below, the blue half-line is the set of all points
with polar coordinates θ=23π, r>0 and the orange half-line is the set
of all points with polar coordinates θ=23π, r<0.
In the figure on the right below, the blue half-line is the set of all points
with polar coordinates θ=2π, r>0 and the orange half-line is the set
of all points with polar coordinates θ=2π, r<0.
Determine, for each angle θ, the lengths of the vectors
e^r(θ) and e^θ(θ) and the angle between
the vectors e^r(θ) and e^θ(θ). Compute
e^r(θ)×e^θ(θ) (viewing e^r(θ) and e^θ(θ) as vectors in three dimensions with zero k^
components).
For each 1≤i≤5, sketch, in the xy-plane, the point
(xi,yi)=(ricosθi,risinθi)
and the vectors e^r(θi) and e^θ(θi). In your
sketch of the vectors, place the tails of the vectors
e^r(θi) and e^θ(θi) at (xi,yi).
Hint+
Compute, for each angle θ, the dot product e^r(θ)⋅e^θ(θ).
Answer+
(a) Both e^r(θ) and e^θ(θ) have length 1.
The angle between them is 2π. The cross product is
e^r(θ)×e^θ(θ)=k^.
(b)
Here is a sketch of (xi,yi), e^r(θi),
e^θ(θi) for i=1,3,5 (the points on the axes)
and here is a sketch (to a different scale) of (xi,yi), e^r(θi),
e^θ(θi) for i=2,4 (the points off the axes).
(b) Note that for θ determined by x=rcosθ, y=rsinθ,
the vector e^r(θ) is a unit vector in the same direction as the
vector from (0,0) to (x,y) and
the vector e^θ(θ) is a unit vector that is perpendicular
to e^r(θ).
The y-component of e^θ(θ) has the same sign
as the x-component of e^r(θ).
The x-component of e^θ(θ) has opposite sign to that of the y-component of e^r(θ).
Here is a sketch of (xi,yi), e^r(θi),
e^θ(θi) for i=1,3,5 (the points on the axes)
and here is a sketch (to a different scale) of (xi,yi), e^r(θi),
e^θ(θi) for i=2,4 (the points off the axes).
the line y=x becomes rsinθ=rcosθ or tanθ=1
or θ=4π.
Thus the domain of integration is
R={(rcosθ,rsinθ)0≤r≤2,0≤θ≤4π}
On this domain,
θ runs from 0 to 4π.
For each fixed θ in that range, r runs from 0 to 2,
as in the figure on the left below.
In polar coordinates dxdy=rdrdθ, so that
∬Rf(x,y)dxdy=∫0π/4dθ∫02drrf(rcosθ,rsinθ)
Alternatively, on R,
r runs from 0 to 2.
For each fixed r in that range, θ runs from 0 to 4π,
as in the figure on the right above.
So
∬Rf(x,y)dxdy=∫02dr∫0π/4dθrf(rcosθ,rsinθ)
(b) The region
R={(x,y)1≤x2+y2≤4,x≥0,y≥0}
In polar coordinates,
the circle x2+y2=1 becomes r2=1 or r=1 and
the circle x2+y2=4 becomes r2=4 or r=2 and
the positive x-axis, x≥0, y=0, becomes θ=0 and
the positive y-axis, x=0, y≥0, becomes θ=2π.
Thus the domain of integration is
R={(rcosθ,rsinθ)1≤r≤2,0≤θ≤2π}
On this domain,
θ runs from 0 to 2π.
For each fixed θ in that range, r runs from 1 to 2,
as in the figure on the left below.
In polar coordinates dxdy=rdrdθ, so that
∬Rf(x,y)dxdy=∫0π/2dθ∫12drrf(rcosθ,rsinθ)
Alternatively, on R,
r runs from 1 to 2.
For each fixed r in that range, θ runs from 0 to 2π,
as in the figure on the right above.
So
∬Rf(x,y)dxdy=∫12dr∫0π/2dθrf(rcosθ,rsinθ)
(c) The region
R={(x,y)(x−1)2+y2≤1,y≥0}
In polar coordinates, the circle (x−1)2+y2=1, or x2−2x+y2=0,
is r2−2rcosθ=0 or r=2cosθ. Note that, on r=2cosθ,
when θ=0, r=2 and
as θ increases from 0 towards π/2, r decreases but remains strictly bigger than 0 (look at the figure below), until
when θ=2π, r=0.
Thus the domain of integration is
R={(rcosθ,rsinθ)0≤θ≤2π,0≤r≤2cosθ}
On this domain,
θ runs from 0 to 2π.
For each fixed θ in that range, r runs from 0 to 2cosθ,
as in the figure on the left below.
In polar coordinates dxdy=rdrdθ, so that
∬Rf(x,y)dxdy=∫0π/2dθ∫02cosθdrrf(rcosθ,rsinθ)
Alternatively, on R,
r runs from 0 (at the point (0,0))
to 2 (at the point (2,0)).
For each fixed r in that range, θ runs
from 0 to arccos2r (which was gotten by solving
r=2cosθ for θ as a function of r), as in the figure
on the right above.
r runs from 0 (at the point (0,0))
to 22 (at the point (2,2)).
For each fixed r between 0 and 2, θ runs from 4π
to 2π, as in the central figure above.
For each fixed r between 2 and 22, θ runs
from 4π to arcsinr2 (which was gotten by solving
rsinθ=2 for θ as a function of r), as in the figure
on the right above.
(c) In polar coordinates, the triangle
with vertices (0,0),(1,0) and (1,1) has sides θ=0,
θ=4π and r=cosθ1 (which is the polar
coordinates version of x=1). The integral is
∬T(x2+y2)dxdy=∫04πdθ∫0cosθ1drr(r2)=∫04πdθ4r40cosθ1=41∫04πdθcos4θ1=41∫04πdθsec4θ=41∫04πdθsec2θ(1+tan2θ)=41∫01dt(1+t2) where t=tanθ=41[t+3t3]01=4134=31
(d) In polar coordinates, the domain of integration,
x2+y2≤1, becomes 0≤r≤1, 0≤θ≤2π. So
∬x2+y2≤1ln(x2+y2)dxdy=∫02πdθ∫01drrlnr2=2π∫01drrlnr2=π∫01dslns where s=r2=π[slns−s]01=−π
To be picky, lns tends to −∞ as s tends to 0.
So ∫01dslns is an improper integral. The careful way to
evaluate it is
Find the volume lying inside the sphere x2+y2+z2=2
and above the paraboloid z=x2+y2.
Answer+
π[342−67]≈2.26
Full solution+
The top surface x2+y2+z2=2 meets the bottom surface
z=x2+y2 when z obeys x2+y2=z=2−z2. That is, when
0=z2+z−2=(z−1)(z+2). The root z=−2 is inconsistent with
z=x2+y2≥0. So the top and bottom surfaces meet at the circle
z=1, x2+y2=1.
In polar coordinates, the top surface is z2=2−r2, or equivalently
z=2−r2, and the bottom surface is z=r2. So the height of the volume
above the point with polar coordinates (r,θ) is
2−r2−r2 and
Let a>0.
Find the volume lying inside the cylinder x2+(y−a)2=a2
and between the upper and lower halves of the cone z2=x2+y2.
Answer+
964a3
Full solution+
For this region x and y run over the interior of the
cylinder x2+(y−a)2=a2. For each (x,y) inside the
cylinder, z runs from −x2+y2 to x2+y2.
As x2+(y−a)2=a2 if and only if x2+y2−2ay=0, the cylinder
has equation r2=2arsinθ, or equivalently, r=2asinθ,
in polar coordinates.
Thus (r,θ) runs over 0≤θ≤π,0≤r≤2asinθ and for each (r,θ) in this region z runs from
−r to r. By symmetry under x→−x,
the volume is
Volume=2∫02πdθ∫02asinθdrr[r−(−r)]=4∫02πdθ∫02asinθdrr2=34∫02πdθ(2asinθ)3=332a3∫02πdθsinθ(1−cos2θ)=−332a3∫10dt(1−t2) where t=cosθ=−332a3[t−3t3]10=964a3
Let a>0.
Find the volume common to the cylinders x2+y2≤2ax and
z2≤2ax.
Answer+
15128a3
Full solution+
The figure below shows the top view of the specified solid.
(x,y) runs over the interior of the circle x2+y2=2ax. For each fixed
(x,y) in this disk, z runs from −2ax to +2ax. In polar
coordinates, the circle is r2=2arcosθ or r=2acosθ.
The solid
is symmetric under y→−y and z→−z, so we can restrict
to y≥0, z≥0 and multiply by 4. The volume is
Volume=4∫02πdθ∫02acosθdrr2arcosθ=4∫02πdθ2acosθ52r5/202acosθ=58∫02πdθ(2acosθ)3=564a3∫02πdθcosθ(1−sin2θ)=564a3∫01dt(1−t2) where t=sinθ=564a3[t−3t3]01=15128a3
Consider the region E in 3–dimensions specified by the inequalities
x2+y2≤2y and 0≤z≤x2+y2.
Draw a reasonably accurate picture of E in 3–dimensions.
Be sure to show the units on the coordinate axes.
Use polar coordinates to find the volume of E. Note that you will
be “using polar coordinates” if you solve this problem
by means of cylindrical coordinates.
Hint: ∫sinnudu=−n1sinn−1ucosu+nn−1∫sinn−2udu
Answer+
(a)
(b) 932
Full solution+
(a)
The equation x2+y2≤2y is equivalent to the
equation x2+(y−1)2=1, which is the equation of the cylinder whose
z=z0 cross–section is the horizontal circle of radius 1, centred on
x=0, y=1, z=z0. The part of this cylinder in the first octant is
sketched in the figure on the left below.
z≤x2+y2 is the equation of the cone with vertex (0,0,0),
and axis the positive z–axis, whose radius at height z=2 is 2.
The part of this cone in the first octant is sketched in the figure
on the right below.
The region E is the part of the cylinder that is above the
xy–plane (since z≥0) outside the cone (since z≤x2+y2).
The part of E that is in the first octant is outlined in red in the
figure below. Both x2+y2≤2y and 0≤z≤x2+y2
are invariant under x→−x. So E is also invariant
under x→−x. That is, E is symmetric about the yz–plane
and contains, in the octant x≤0, y≥0, z≥0, a mirror
image of the first octant part of E.
(b) In polar coordinates, x2+y2≤2y becomes
r2≤2rsinθ⟺r≤2sinθ
Let us denote by D the base region of the part of E in the first
octant (i.e. the shaded region in the figure above). Think of D as being
part of the xy–plane. In polar coordinates, on D
θ runs from 0 to 2π. (Recall that D is
contained in the first quadrant.)
For each θ in that range, r runs from 0 to 2sinθ.
Because
in polar coordinates dA=rdrdθ, and
the height of E above each point (x,y) in D is x2+y2,
or, in polar coordinates, r, and
the volume of E is twice the volume of the part of E in the first octant,
We'll use polar coordinates. The domain of integration is
R2={(rcosθ,rsinθ)0≤r<∞,0≤θ≤2π}
The given integral is improper, so we'll start by integrating r from 0
to an arbitrary R>0, and then we'll take the limit R→∞.
In polar coordinates, the integrand
(1+x2+y2)21=(1+r2)21, and
dA=rdrdθ, so
Evaluate the integral by transforming to polar coordinates.
Answer+
(a)
(b) 12π[5ln(5)−4]
Full solution+
(a) On the domain of integration
y runs from 0 to 1. In inequalities, 0≤y≤1.
For each fixed y in that range, x runs from 3y to
4−y2. In inequalities, that is 3y≤x≤4−y2.
Note that the inequalities x≤4−y2, x≥0 are equivalent to
x2+y2≤4, x≥0.
Note that the line x=3y and the circle x2+y2≤4
intersect when 3y2+y2=4, i.e. y=±1.
Here is a sketch.
(b)
In polar coordinates, the circle x2+y2=4 is r=2 and the
line x=3y, i.e. xy=31,
is tanθ=31 or θ=6π.
As dxdy=rdrdθ,
the domain of integration is
Let D be the shaded region in the diagram. Find the average
distance of points in D from the origin. You may use that
∫cosn(x)dx=ncosn−1(x)sin(x)+nn−1∫cosn−2(x)dx
for all natural numbers n≥2.
Answer+
2π+8π+44/9≈1.442
Full solution+
We'll use polar coordinates. In D
θ runs from 0 to 2π and
for each fixed θ between 0 and 2π, r runs from
1 to 1+cos(θ).
the condition x2+y2≤1 restricts G to the interior of the
circle of radius 1 centred on the origin, and
the conditions 0≤x≤2y restricts G to x≥0, y≥0,
i.e. to the first quadrant, and
the conditions x≤2y and y≤2x restrict 2x≤y≤2x.
So G lies below the (steep) line y=2x and lies above the (not steep)
line y=2x.
Here is a sketch of G
(b) Observe that the line y=2x crosses the circle x2+y2=1 at
a point (x,y) obeying
x2+(2x)2=x2+y2=1⟹5x2=1
and that the line x=2y crosses the circle x2+y2=1 at
a point (x,y) obeying
(2y)2+y2=x2+y2=1⟹5y2=1
So the intersection point of y=2x and x2+y2=1 in the first octant
is (51,52) and
the intersection point of x=2y and x2+y2=1 in the first octant
is (52,51).
We'll set up the iterated integral using horizontal strips as in the
sketch
Looking at that sketch, we see that, on G,
y runs from 0 to 52, and
for each fixed y between 0 and 51,
x runs from 2y to 2y, and
for each fixed y between 51 and 52x runs from 2y to 1−y2.
for each y in that range, x runs from y to 4−y2.
We can rewrite x=4−y2 in the more familiar form x2+y2=4,
x≥0.
The figure on the left below provides a sketch of the domain of integration.
It also shows the generic horizontal slice that was used to set up the given
iterated integral.
(b)
To reverse the order of integration observe, we use vertical, rather than
horizontal slices. From the figure on the
right above that, on the domain of integration,
x runs from 0 to 2 and
for each x in the range 0≤x≤2, y runs from 0
to x.
for each x in the range 2≤x≤2, y runs from 0
to 4−x2.
A symmetrical coffee percolator holds 24 cups when full. The interior has a circular cross-section which tapers from a radius of 3" at the centre to 2" at the base and top, which are 12" apart. The bounding surface is parabolic. Where should the mark indicating the 6 cup level be placed?
Answer+
About 3.5” above the bottom
Full solution+
Let r(z) be the radius of the urn at height z above its middle.
Because the bounding surface of the urn is parabolic, r(z) must be a
quadratic function of z that varies between 3 at z=0 and 2 at z=±6. That is, r(z) must be of the form r(z)=az2+bz+c.
The condition that r(0)=3 tells us that c=3. The conditions that
r(±6)=2 tells us that
62a+6b+362a−6b+3=2=2
So b=0 and 62a=−1 so that a=−621.
All together r(z)=3−(6z)2.
Slice
the urn into horzontal slices, with the slice at height z a disk of radius
r(z) and thickness dz and hence of volume πr(z)2dz. The volume
to height z0 is
We are told that the mark is to be at the 6 cup level and that the urn holds 24 cups. So the mark is to be at the height z0 for which the volume, V(z0),
is one quarter of the total volume, V(6).
That is, we are to choose z0 so that V(z0)=41V(6) or
Since [9z0−18z03+6480z05]z0=−2.495=−21.61
and [9z0−18z03+6480z05]z0=−2.490=−21.57,
there is a solution z0=−2.49 (to two decimal places). The mark should
be about 3.5” above the bottom.
The union of those two domains (as well as horizontal strips that were used in
setting up the two given integrals) is sketched in the figure on the left below.
To reverse the order of integration, we decompose the domain using vertical
strips as in the figure on the right above. As
x runs from 0 to 1 and
for each fixed x between 0 and 1, y runs from x to 2−x.
