Double Integrals in Polar Coordinates

Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.

Polar Rectangular Regions of Integration

When we defined the double integral for a continuous function in rectangular coordinates—say, g

over a region R

in the xy

-plane—we divided R

into subrectangles with sides parallel to the coordinate axes. These sides have either constant x

-values and/or constant y

-values. In polar coordinates, the shape we work with is a polar rectangle, whose sides have constant r

-values and/or constant θ

-values. This means we can describe a polar rectangle as in [link](a), with R={(r,θ)\|a≤r≤b,α≤θ≤β}.

In this section, we are looking to integrate over polar rectangles. Consider a function f(r,θ)

over a polar rectangle R.

We divide the interval [a,b]

into m

subintervals [ri−1,ri]

of length Δr=(b−a)/m

and divide the interval [α,β]

into n

subintervals [θi−1,θi]

of width Δθ=(β−α)/n.

This means that the circles r=ri

and rays θ=θi

for 1≤i≤m

and 1≤j≤n

divide the polar rectangle R

into smaller polar subrectangles Rij

([link](b)).

This figure consists of three figures labeled a, b, and c. In figure a, a sector of an annulus is shown in the polar coordinate plane with radii a and b and angles alpha and beta from the theta = 0 axis. In figure b, this sector of an annulus is cut up into subsectors in a manner similar to the way in which previous spaces were cut up into subrectangles. In figure c, one of these subsectors is shown with angle Delta theta, distance between inner and outer radii Delta r, and area Delta A = r* sub theta Delta r Delta theta, where the center point is given as (r* sub i j, theta* sub i j).

As before, we need to find the area ΔA

of the polar subrectangle Rij

and the “polar” volume of the thin box above Rij.

Recall that, in a circle of radius r,

the length s

of an arc subtended by a central angle of θ

radians is s=rθ.

Notice that the polar rectangle Rij

looks a lot like a trapezoid with parallel sides ri−1Δθ

and riΔθ

and with a width Δr.

Hence the area of the polar subrectangle Rij

is

ΔA=12Δr(ri−1Δθ+r1Δθ).

Simplifying and letting rij*=12(ri−1+ri),

we have ΔA=rij*ΔrΔθ.

Therefore, the polar volume of the thin box above Rij

([link]) is

f(rij*,θij*)ΔA=f(rij*,θij*)rij*ΔrΔθ.

In x y z space, there is a surface f (r, theta). On the x y plane, a series of subsectors of annuli are drawn as in the previous figure with radius between annuli Delta r and angle between subsectors Delta theta. A subsector from the surface f(r, theta) is projected down onto one of these subsectors. This subsector has center point marked (r* sub i j, theta* sub i j).

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as

∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.

As we have seen before, we obtain a better approximation to the polar volume of the solid above the region R

when we let m

and n

become larger. Hence, we define the polar volume as the limit of the double Riemann sum,

V=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.

This becomes the expression for the double integral.

Definition

The double integral of the function f(r,θ)

over the polar rectangular region R

in the rθ

-plane is defined as

∬Rf(r,θ)dA=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)ΔA=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.

Again, just as in Double Integrals over Rectangular Regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,

∬Rf(r,θ)dA=∬Rf(r,θ)rdrdθ=∫θ=αθ=β∫r=ar=bf(r,θ)rdrdθ.

Notice that the expression for dA

is replaced by rdrdθ

when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function f

is given in terms of x

and y,

using x=rcosθ,y=rsinθ,anddA=rdrdθ

changes it to

∬Rf(x,y)dA=∬Rf(rcosθ,rsinθ)rdrdθ.

Note that all the properties listed in Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.

Sketching a Polar Rectangular Region

Sketch the polar rectangular region R={(r,θ)\|1≤r≤3,0≤θ≤π}.

As we can see from [link], r=1

and r=3

are circles of radius 1and3

and 0≤θ≤π

covers the entire top half of the plane. Hence the region R

looks like a semicircular band.

Half an annulus R is drawn with inner radius 1 and outer radius 3. That is, the inner semicircle is given by x squared + y squared = 1, whereas the outer semicircle is given by x squared + y squared = 9.

Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.

Evaluating a Double Integral over a Polar Rectangular Region

Evaluate the integral ∬R3xdA

over the region R={(r,θ)\|1≤r≤2,0≤θ≤π}.

First we sketch a figure similar to [link] but with outer radius 2.

From the figure we can see that we have

∬R3xdA=∫θ=0θ=π∫r=1r=23rcosθrdrdθUse an iterated integral with correct limitsof integration.=∫θ=0θ=πcosθ[r3\|r=1r=2]dθIntegrate first with respect tor.=∫θ=0θ=π7cosθdθ=7sinθ\|θ=0θ=π=0.

Sketch the region R={(r,θ)\|1≤r≤2,−π2≤θ≤π2},

and evaluate ∬RxdA.

143
Hint

Follow the steps in [link].

Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral ∬R(1−x2−y2)dA

where R

is the unit circle on the xy

-plane.

The region R

is a unit circle, so we can describe it as R={(r,θ)\|0≤r≤1,0≤θ≤2π}.

Using the conversion x=rcosθ,y=rsinθ,

and dA=rdrdθ,

we have

∬R(1−x2−y2)dA=∫02π∫01(1−r2)rdrdθ=∫02π∫01(r−r3)drdθ=∫02π[r22−r44]01dθ=∫02π14dθ=π2.
Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral ∬R(x+y)dA

where R={(x,y)\|1≤x2+y2≤4,x≤0}.

We can see that R

is an annular region that can be converted to polar coordinates and described as R={(r,θ)\|1≤r≤2,π2≤θ≤3π2}

(see the following graph).

Two semicircles are drawn in the second and third quadrants, with equations x squared + y squared = 1 and x squared + y squared = 2.

Hence, using the conversion x=rcosθ,y=rsinθ,

and dA=rdrdθ,

we have

∬R(x+y)dA=∫θ=π/2θ=3π/2∫r=1r=2(rcosθ+rsinθ)rdrdθ=(∫r=1r=2r2dr)(∫π/23π/2(cosθ+sinθ)dθ)=[r33]12[sinθ−cosθ]\|π/23π/2=−143.

Evaluate the integral ∬R(4−x2−y2)dA

where R

is the circle of radius 2

on the xy

-plane.

8π
Hint

Follow the steps in the previous example.

General Polar Regions of Integration

To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in Double Integrals over General Regions. It is more common to write polar equations as r=f(θ)

than θ=f(r),

so we describe a general polar region as R={(r,θ)\|α≤θ≤β,h1(θ)≤r≤h2(θ)}

(see the following figure).

A region D is shown in polar coordinates with edges given by theta = alpha, theta = beta, r = h2(theta), and r = h1(theta).

Double Integrals over General Polar Regions

If f(r,θ)

is continuous on a general polar region D

as described above, then

∬Df(r,θ)rdrdθ=∫θ=αθ=β∫r=h1(θ)r=h2(θ)f(r,θ)rdrdθ
Evaluating a Double Integral over a General Polar Region

Evaluate the integral ∬Dr2sinθrdrdθ

where D

is the region bounded by the polar axis and the upper half of the cardioid r=1+cosθ.

We can describe the region D

as {(r,θ)\|0≤θ≤π,0≤r≤1+cosθ}

as shown in the following figure.

A region D is given as the top half of a cardioid with equation r = 1 + cos theta.

Hence, we have

∬Dr2sinθrdrdθ=∫θ=0θ=π∫r=0r=1+cosθ(r2sinθ)rdrdθ=14∫θ=0θ=π[r4]r=0r=1+cosθsinθdθ=14∫θ=0θ=π(1+cosθ)4sinθdθ=−14[(1+cosθ)55]0π=85.

Evaluate the integral

∬Dr2sin22θrdrdθwhereD={(r,θ)\|0≤θ≤π,0≤r≤2cos2θ}.
π/8
Hint

Graph the region and follow the steps in the previous example.

Polar Areas and Volumes

As in rectangular coordinates, if a solid S

is bounded by the surface z=f(r,θ),

as well as by the surfaces r=a,r=b,θ=α,

and θ=β,

we can find the volume V

of S

by double integration, as

V=∬Rf(r,θ)rdrdθ=∫θ=αθ=β∫r=ar=bf(r,θ)rdrdθ.

If the base of the solid can be described as D={(r,θ)\|α≤θ≤β,h1(θ)≤r≤h2(θ)},

then the double integral for the volume becomes

V=∬Df(r,θ)rdrdθ=∫θ=αθ=β∫r=h1(θ)r=h2(θ)f(r,θ)rdrdθ.

We illustrate this idea with some examples.

Finding a Volume Using a Double Integral

Find the volume of the solid that lies under the paraboloid z=1−x2−y2

and above the unit circle on the xy

-plane (see the following figure).

The paraboloid z = 1 minus x squared minus y squared is shown, which in this graph looks like a sheet with the middle gently puffed up and the corners anchored.

By the method of double integration, we can see that the volume is the iterated integral of the form ∬R(1−x2−y2)dA

where R={(r,θ)\|0≤r≤1,0≤θ≤2π}.

This integration was shown before in [link], so the volume is π2

cubic units.

Finding a Volume Using Double Integration

Find the volume of the solid that lies under the paraboloid z=4−x2−y2

and above the disk (x−1)2+y2=1

on the xy

-plane. See the paraboloid in [link] intersecting the cylinder (x−1)2+y2=1

above the xy

-plane.

A paraboloid with equation z = 4 minus x squared minus y squared is intersected by a cylinder with equation (x minus 1) squared + y squared = 1.

First change the disk (x−1)2+y2=1

to polar coordinates. Expanding the square term, we have x2−2x+1+y2=1.

Then simplify to get x2+y2=2x,

which in polar coordinates becomes r2=2rcosθ

and then either r=0

or r=2cosθ.

Similarly, the equation of the paraboloid changes to z=4−r2.

Therefore we can describe the disk (x−1)2+y2=1

on the xy

-plane as the region

D={(r,θ)\|0≤θ≤π,0≤r≤2cosθ}.

Hence the volume of the solid bounded above by the paraboloid z=4−x2−y2

and below by r=2cosθ

is

V=∬Df(r,θ)rdrdθ=∫θ=0θ=π∫r=0r=2cosθ(4−r2)rdrdθ=∫θ=0θ=π[4r22−r44\|02cosθ]dθ=∫0π[8cos2θ−4cos2θ]dθ=[52θ+52sinθcosθ−sinθcos3θ]0π=52π.

Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if f

has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.

Finding a Volume Using a Double Integral

Find the volume of the region that lies under the paraboloid z=x2+y2

and above the triangle enclosed by the lines y=x,x=0,

and x+y=2

in the xy

-plane ([link]).

First examine the region over which we need to set up the double integral and the accompanying paraboloid.

This figure consists of three figures. The first is simply a paraboloid that opens up. The second shows the region D bounded by x = 0, y = x, and x + y = 2 with a vertical double-sided arrow within the region. The second shows the same region but in polar coordinates, so the lines bounding D are theta = pi/2, r = 2/(cos theta + sin theta), and theta = pi/4, with a double-sided arrow that has one side pointed at the origin.

The region D

is {(x,y)\|0≤x≤1,x≤y≤2−x}.

Converting the lines y=x,x=0,

and x+y=2

in the xy

-plane to functions of r

and θ,

we have θ=π/4,

θ=π/2,

and r=2/(cosθ+sinθ),

respectively. Graphing the region on the xy

-plane, we see that it looks like D={(r,θ)\|π/4≤θ≤π/2,0≤r≤2/(cosθ+sinθ)}.

Now converting the equation of the surface gives z=x2+y2=r2.

Therefore, the volume of the solid is given by the double integral

V=∬Df(r,θ)rdrdθ=∫θ=π/4θ=π/2∫r=0r=2/(cosθ+sinθ)r2rdrdθ=∫π/4π/2[r44]02/(cosθ+sinθ)dθ=14∫π/4π/2(2cosθ+sinθ)4dθ=164∫π/4π/2(1cosθ+sinθ)4dθ=4∫π/4π/2(1cosθ+sinθ)4dθ.

As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as

V=∫01∫x2−x(x2+y2)dydx.

Evaluating gives

V=∫01∫x2−x(x2+y2)dydx=∫01[x2y+y33]\|x2−xdx=∫0183−4x+4x2−8x33dx=[8x3−2x2+4x33−2x43]\|01=43.

To answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.

Finding a Volume Using a Double Integral

Use polar coordinates to find the volume inside the cone z=2−x2+y2

and above the xy-plane.

The region D

for the integration is the base of the cone, which appears to be a circle on the xy-plane

(see the following figure).

A cone given by z = 2 minus the square root of (x squared plus y squared) and a circle given by x squared plus y squared = 4. The cone is above the circle in xyz space.

We find the equation of the circle by setting z=0:

0=2−x2+y22=x2+y2x2+y2=4.

This means the radius of the circle is 2,

so for the integration we have 0≤θ≤2π

and 0≤r≤2.

Substituting x=rcosθ

and y=rsinθ

in the equation z=2−x2+y2

we have z=2−r.

Therefore, the volume of the cone is

∫θ=0θ=2π∫r=0r=2(2−r)rdrdθ=2π43=8π3

cubic units.

Analysis

Note that if we were to find the volume of an arbitrary cone with radius a

units and height h

units, then the equation of the cone would be z=h−hax2+y2.

We can still use [link] and set up the integral as ∫θ=0θ=2π∫r=0r=a(h−har)rdrdθ.

Evaluating the integral, we get 13πa2h.

Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids z=x2+y2

and z=16−x2−y2.

V=∫02π∫022(16−2r2)rdrdθ=64π

cubic units

Hint

Sketching the graphs can help.

As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like

AreaA=∫αβ∫h1(θ)h2(θ)1rdrdθ.
Finding an Area Using a Double Integral in Polar Coordinates

Evaluate the area bounded by the curve r=cos4θ.

Sketching the graph of the function r=cos4θ

reveals that it is a polar rose with eight petals (see the following figure).

A rose with eight petals given by r = cos (4 theta).

Using symmetry, we can see that we need to find the area of one petal and then multiply it by 8.

Notice that the values of θ

for which the graph passes through the origin are the zeros of the function cos4θ,

and these are odd multiples of π/8.

Thus, one of the petals corresponds to the values of θ

in the interval [−π/8,π/8].

Therefore, the area bounded by the curve r=cos4θ

is

A=8∫θ=−π/8θ=π/8∫r=0r=cos4θ1rdrdθ=8∫−π/8π/8[12r2\|0cos4θ]dθ=8∫−π/8π/812cos24θdθ=8[14θ+116sin4θcos4θ\|−π/8π/8]=8[π16]=π2.
Finding Area Between Two Polar Curves

Find the area enclosed by the circle r=3cosθ

and the cardioid r=1+cosθ.

First and foremost, sketch the graphs of the region ([link]).

A cardioid with equation 1 + cos theta is shown overlapping a circle given by r = 3 cos theta, which is a circle of radius 3 with center (1.5, 0). The area bounded by the x axis, the cardioid, and the dashed line connecting the origin to the intersection of the cardioid and circle on the r = 2 line is shaded.

We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives

3cosθ=1+cosθ.

One of the points of intersection is θ=π/3.

The area above the polar axis consists of two parts, with one part defined by the cardioid from θ=0

to θ=π/3

and the other part defined by the circle from θ=π/3

to θ=π/2.

By symmetry, the total area is twice the area above the polar axis. Thus, we have

A=2[∫θ=0θ=π/3∫r=0r=1+cosθ1rdrdθ+∫θ=π/3θ=π/2∫r=0r=3cosθ1rdrdθ].

Evaluating each piece separately, we find that the area is

A=2(14π+9163+38π−9163)=2(58π)=54πsquare units.

Find the area enclosed inside the cardioid r=3−3sinθ

and outside the cardioid r=1+sinθ.

A=2∫−π/2π/6∫1+sinθ3−3sinθrdrdθ=8π+93
Hint

Sketch the graph, and solve for the points of intersection.

Evaluating an Improper Double Integral in Polar Coordinates

Evaluate the integral ∬R2e−10(x2+y2)dxdy.

This is an improper integral because we are integrating over an unbounded region R2.

In polar coordinates, the entire plane R2

can be seen as 0≤θ≤2π,

0≤r≤∞.

Using the changes of variables from rectangular coordinates to polar coordinates, we have

∬R2e−10(x2+y2)dxdy=∫θ=0θ=2π∫r=0r=∞e−10r2rdrdθ=∫θ=0θ=2π(lima→∞∫r=0r=ae−10r2rdr)dθ=(∫θ=0θ=2πdθ)(lima→∞∫r=0r=ae−10r2rdr)=2π(lima→∞∫r=0r=ae−10r2rdr)=2πlima→∞(−120)(e−10r2\|0a)=2π(−120)lima→∞(e−10a2−1)=π10.

Evaluate the integral ∬R2e−4(x2+y2)dxdy.

π4
Hint

Convert to the polar coordinate system.

Key Concepts

Key Equations

In the following exercises, express the region D

in polar coordinates.

D

is the region of the disk of radius 2

centered at the origin that lies in the first quadrant.

D

is the region between the circles of radius 4

and radius 5

centered at the origin that lies in the second quadrant.

D={(r,θ)\|4≤r≤5,π2≤θ≤π}
D

is the region bounded by the y

-axis and x=1−y2.

D

is the region bounded by the x

-axis and y=2−x2.

D={(r,θ)\|0≤r≤2,0≤θ≤π}
D={(x,y)\|x2+y2≤4x}
D={(x,y)\|x2+y2≤4y}
D={(r,θ)\|0≤r≤4sinθ,0≤θ≤π}

In the following exercises, the graph of the polar rectangular region D

is given. Express D

in polar coordinates.

![Half an annulus D is drawn in the first and second quadrants with inner radius 3 and outer radius 5.](/calculus-book/resources/CNX_Calc_Figure_15_03_201.jpg)
![A sector of an annulus D is drawn between theta = pi/4 and theta = pi/2 with inner radius 3 and outer radius 5.](/calculus-book/resources/CNX_Calc_Figure_15_03_202.jpg)
D={(r,θ)\|3≤r≤5,π4≤θ≤π2}
![Half of an annulus D is drawn between theta = pi/4 and theta = 5 pi/4 with inner radius 3 and outer radius 5.](/calculus-book/resources/CNX_Calc_Figure_15_03_203.jpg)
![ A sector of an annulus D is drawn between theta = 3 pi/4 and theta = 5 pi/4 with inner radius 3 and outer radius 5.](/calculus-book/resources/CNX_Calc_Figure_15_03_204.jpg)
D={(r,θ)\|3≤r≤5,3π4≤θ≤5π4}

In the following graph, the region D

is situated below y=x

and is bounded by x=1,x=5,

and y=0.

A region D is given that is bounded by y = 0, x = 1, x = 5, and y = x, that is, a right triangle with a corner cut off.

In the following graph, the region D

is bounded by y=x

and y=x2.

A region D is drawn between y = x and y = x squared, which looks like a deformed lens, with the bulbous part below the straight part.

D={(r,θ)\|0≤r≤tanθsecθ,0≤θ≤π4}

In the following exercises, evaluate the double integral ∬Rf(x,y)dA

over the polar rectangular region D.

f(x,y)=x2+y2,D={(r,θ)\|3≤r≤5,0≤θ≤2π}
f(x,y)=x+y,D={(r,θ)\|3≤r≤5,0≤θ≤2π}
0
f(x,y)=x2+xy,D={(r,θ)\|1≤r≤2,π≤θ≤2π}
f(x,y)=x4+y4,D={(r,θ)\|1≤r≤2,3π2≤θ≤2π}
63π16
f(x,y)=x2+y23,

where D={(r,θ)\|0≤r≤1,π2≤θ≤π}.

f(x,y)=x4+2x2y2+y4,

where D={(r,θ)\|3≤r≤4,π3≤θ≤2π3}.

3367π18
f(x,y)=sin(arctanyx),

where D={(r,θ)\|1≤r≤2,π6≤θ≤π3}

f(x,y)=arctan(yx),

where D={(r,θ)\|2≤r≤3,π4≤θ≤π3}

35π2576
∬Dex2+y2[1+2arctan(yx)]dA,D={(r,θ)\|1≤r≤2,π6≤θ≤π3}
∬D(ex2+y2+x4+2x2y2+y4)arctan(yx)dA,D={(r,θ)\|1≤r≤2,π4≤θ≤π3}
7576π2(21−e+e4)

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

∫12∫0x(x2+y2)dydx=∫0π4∫secθ2secθr3drdθ
∫23∫0xxx2+y2dydx=∫0π/4∫0tanθsecθrcosθdrdθ
54ln(3+22)
∫01∫x2x1x2+y2dydx=∫0π/4∫0tanθsecθdrdθ
∫01∫x2xyx2+y2dydx=∫0π/4∫0tanθsecθrsinθdrdθ
16(2−2)

In the following exercises, convert the integrals to polar coordinates and evaluate them.

∫03∫09−y2(x2+y2)dxdy
∫02∫−4−y24−y2(x2+y2)2dxdy
∫0π∫02r5drdθ=32π3
∫01∫01−x2(x+y)dydx
∫04∫−16−x216−x2sin(x2+y2)dydx
∫−π/2π/2∫04rsin(r2)drdθ=πsin28

Evaluate the integral ∬DrdA

where D

is the region bounded by the polar axis and the upper half of the cardioid r=1+cosθ.

Find the area of the region D

bounded by the polar axis and the upper half of the cardioid r=1+cosθ.

3π4

Evaluate the integral ∬DrdA,

where D

is the region bounded by the part of the four-leaved rose r=sin2θ

situated in the first quadrant (see the following figure).

A region D is drawn in the first quadrant petal of the four petal rose given by r = sin (2 theta).

Find the total area of the region enclosed by the four-leaved rose r=sin2θ

(see the figure in the previous exercise).

π2

Find the area of the region D,

which is the region bounded by y=4−x2,

x=3, x=2,

and y=0.

Find the area of the region D,

which is the region inside the disk x2+y2≤4

and to the right of the line x=1.

13(4π−33)

Determine the average value of the function f(x,y)=x2+y2

over the region D

bounded by the polar curve r=cos2θ,

where −π4≤θ≤π4

(see the following graph).

The first/fourth-quadrant petal of the four-petal rose given by r = cos (2 theta) is shown.

Determine the average value of the function f(x,y)=x2+y2

over the region D

bounded by the polar curve r=3sin2θ,

where 0≤θ≤π2

(see the following graph).

The first-quadrant petal of the four-petal rose given by r = 3sin (2 theta) is shown.

163π

Find the volume of the solid situated in the first octant and bounded by the paraboloid z=1−4x2−4y2

and the planes x=0,y=0,

and z=0.

Find the volume of the solid bounded by the paraboloid z=2−9x2−9y2

and the plane z=1.

π18
  1. Find the volume of the solid S1

    bounded by the cylinder

    x2+y2=1

    and the planes

    z=0

    and

    z=1.
  2. Find the volume of the solid S2

    outside the double cone

    z2=x2+y2,

    inside the cylinder

    x2+y2=1,

    and above the plane

    z=0.
  3. Find the volume of the solid inside the cone z2=x2+y2

    and below the plane

    z=1

    by subtracting the volumes of the solids

    S1

    and

    S2.
  1. Find the volume of the solid S1

    inside the unit sphere

    x2+y2+z2=1

    and above the plane

    z=0.
  2. Find the volume of the solid S2

    inside the double cone

    (z−1)2=x2+y2

    and above the plane

    z=0.
  3. Find the volume of the solid outside the double cone (z−1)2=x2+y2

    and inside the sphere

    x2+y2+z2=1.

a. 2π3;

b. π2;

c. π6

For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure).

A spherical ring is shown, that is, a sphere with a cylindrical hole going all the way through it.

If the sphere has radius 4

and the cylinder has radius 2,

find the volume of the spherical ring.

A cylindrical hole of diameter 6

cm is bored through a sphere of radius 5

cm such that the axis of the cylinder passes through the center of the sphere. Find the volume of the resulting spherical ring.

256π3cm3

Find the volume of the solid that lies under the double cone z2=4x2+4y2,

inside the cylinder x2+y2=x,

and above the plane z=0.

Find the volume of the solid that lies under the paraboloid z=x2+y2,

inside the cylinder x2+y2=x,

and above the plane z=0.

3π32

Find the volume of the solid that lies under the plane x+y+z=10

and above the disk x2+y2=4x.

Find the volume of the solid that lies under the plane 2x+y+2z=8

and above the unit disk x2+y2=1.

4π

A radial function f

is a function whose value at each point depends only on the distance between that point and the origin of the system of coordinates; that is, f(x,y)=g(r),

where r=x2+y2.

Show that if f

is a continuous radial function, then ∬Df(x,y)dA=(θ2−θ1)[G(R2)−G(R1)],

where G′(r)=rg(r)

and (x,y)∈D={(r,θ)\|R1≤r≤R2,0≤θ≤2π},

with 0≤R1<R2

and 0≤θ1<θ2≤2π.

Use the information from the preceding exercise to calculate the integral ∬D(x2+y2)3dA,

where D

is the unit disk.

π4

Let f(x,y)=F′(r)r

be a continuous radial function defined on the annular region D={(r,θ)\|R1≤r≤R2,0≤θ≤2π},

where r=x2+y2,

0<R1<R2,

and F

is a differentiable function. Show that ∬Df(x,y)dA=2π[F(R2)−F(R1)].

Apply the preceding exercise to calculate the integral ∬Dex2+y2x2+y2dxdy,

where D

is the annular region between the circles of radii 1

and 2

situated in the third quadrant.

12πe(e−1)

Let f

be a continuous function that can be expressed in polar coordinates as a function of θ

only; that is, f(x,y)=h(θ),

where (x,y)∈D={(r,θ)\|R1≤r≤R2,θ1≤θ≤θ2},

with 0≤R1<R2

and 0≤θ1<θ2≤2π.

Show that ∬Df(x,y)dA=12(R22−R12)[H(θ2)−H(θ1)],

where H

is an antiderivative of h.

Apply the preceding exercise to calculate the integral ∬Dy2x2dA,

where D={(r,θ)\|1≤r≤2,π6≤θ≤π3}.

3−π4

Let f

be a continuous function that can be expressed in polar coordinates as a function of θ

only; that is, f(x,y)=g(r)h(θ),

where (x,y)∈D={(r,θ)\|R1≤r≤R2,θ1≤θ≤θ2}

with 0≤R1<R2

and 0≤θ1<θ2≤2π.

Show that ∬Df(x,y)dA=[G(R2)−G(R1)][H(θ2)−H(θ1)],

where G

and H

are antiderivatives of g

and h,

respectively.

Evaluate ∬Darctan(yx)x2+y2dA,

where D={(r,θ)\|2≤r≤3,π4≤θ≤π3}.

133π3864

A spherical cap is the region of a sphere that lies above or below a given plane.

  1. Show that the volume of the spherical cap in the figure below is 16πh(3a2+h2).

    A sphere of radius R has a circle inside of it h units from the top of the sphere. This circle has radius a, which is less than R.

  2. A spherical segment is the solid defined by intersecting a sphere with two parallel planes. If the distance between the planes is h,

    show that the volume of the spherical segment in the figure below is

    16πh(3a2+3b2+h2).

    A sphere has two parallel circles inside of it h units apart. The upper circle has radius b, and the lower circle has radius a. Note that a > b.

In statistics, the joint density for two independent, normally distributed events with a mean μ=0

and a standard distribution σ

is defined by p(x,y)=12πσ2e−x2+y22σ2.

Consider (X,Y),

the Cartesian coordinates of a ball in the resting position after it was released from a position on the z-axis toward the xy

-plane. Assume that the coordinates of the ball are independently normally distributed with a mean μ=0

and a standard deviation of σ

(in feet). The probability that the ball will stop no more than a

feet from the origin is given by P[X2+Y2≤a2]=∬Dp(x,y)dydx,

where D

is the disk of radius a centered at the origin. Show that P[X2+Y2≤a2]=1−e−a2/2σ2.

The double improper integral ∫−∞∞∫−∞∞e(−x2+y2/2)dydx

may be defined as the limit value of the double integrals ∬Dae(−x2+y2/2)dA

over disks Da

of radii a centered at the origin, as a increases without bound; that is, ∫−∞∞∫−∞∞e(−x2+y2/2)dydx=lima→∞∬Dae(−x2+y2/2)dA.

  1. Use polar coordinates to show that ∫−∞∞∫−∞∞e(−x2+y2/2)dydx=2π.
  2. Show that ∫−∞∞e−x2/2dx=2π,

    by using the relation

    ∫−∞∞∫−∞∞e(−x2+y2/2)dydx=(∫−∞∞e−x2/2dx)(∫−∞∞e−y2/2dy).

Glossary

polar rectangle
the region enclosed between the circles r=a

and

r=b

and the angles

θ=α

and

θ=β;

it is described as

R={(r,θ)\|a≤r≤b,α≤θ≤β}
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