5. applications of double integrals, volume and first

. . . . . .

.Chapter 14 Multiple Integrals..

......

...1 Double Integrals, Iterated Integrals, Cross-sections

...2 Double Integrals over more general regions, Definition,Evaluation of Double Integrals, Properties of Double Integrals

...3 Area and Volume by Double Integration, Volume by IteratedIntegrals, Volume between Two surfaces

...4 Double Integrals in Polar Coordinates, More general Regions

...5 Applications of Double Integrals, Volume and First Theorem ofPappus, Surface Area and Second Theorem of Pappus,Moments of Inertia

...6 Triple Integrals, Iterated Triple Integrals

Prepared by Dr. I.T. Leong for Tutorial 5 of Math 200

. . . . . .

.

......

Definition. The definite integral of a function f (x) of one variabledefined on an interval [a, b], is given by

∫ b

af (x) dx = lim

max ∆xk→0

n

∑k=1

f (x∗k )∆xk = limn→+∞

n

∑k=1

f (x∗k )∆xk.

This concept arose from the problem of finding areas under curves.Now we have similar problem if we replace the function of twovariables..

......

Volume Problem. Given a function f (x, y) of two variables that iscontinuous and nonnegative on a region R in the xy-plane, find thevolume of the solid enclosed between the surface z = f (x, y) and theregion R.


. . . . . .

.

......

Definition. The definite integral of a function f (x) of one variabledefined on an interval [a, b], is given by∫ b

af (x) dx = lim

max ∆xk→0

n

∑k=1


n

∑k=1

f (x∗k )∆xk.


......



. . . . . .

.

......


af (x) dx = lim

max ∆xk→0

n

∑k=1


n

∑k=1

f (x∗k )∆xk.

This concept arose from the problem of finding areas under curves.Now we have similar problem if we replace the function of twovariables.

.

......



. . . . . .

.

......


af (x) dx = lim

max ∆xk→0

n

∑k=1


n

∑k=1

f (x∗k )∆xk.


......



. . . . . .

.Volume and Double Integrals..

......

Let f (x, y) be a function of two variables defined over a rectangleR = [a, b]× [c, d]. We would like to define the double integral of f (x, y)over R as the algebraic volume of the solid under the graph ofz = f (x, y) over R.

The idea is similar to the case of integral∫ b

a f (x)dx in one variablecase, in which we subdivide the interval into smaller subintervals withuniform width ∆x = b−a

n , and then choose arbitrary points xi in thesubinterval Ii. Then we have the approximate Riemann sum

n

∑i=1

f (xi)∆x.

Of course, this sum depends on the choices of xi and the subinterval.In fact, this idea can be implemented in 2-dimensional cases as well.


. . . . . .


......



a f (x)dx in one variablecase, in which

we subdivide the interval into smaller subintervals withuniform width ∆x = b−a


n

∑i=1

f (xi)∆x.



. . . . . .


......




n , and then choose arbitrary points xi in thesubinterval Ii.

Then we have the approximate Riemann sumn

∑i=1

f (xi)∆x.



. . . . . .


......





n

∑i=1

f (xi)∆x.



. . . . . .

Let R = [a, b]× [c, d]. and f (x, y) be a function defined on R. We firstsubdivide the rectangle R into mn small rectangles Rij, each havingarea ∆A, where i = 1, · · · , m and j = 1, · · · , n. For each pair (i, j), pickan arbitrary point (xij, yij) inside Rij. Use the value f (xij, yij) as theheight of a rectangular solid erected over Rij. Thus its volume isf (xij, yij)∆A.

The sum of the volume of all these small rectangular solidsapproximates the volume of the solid under the graph of z = f (x, y)

over R. This summ

∑i=1

n

∑j=1

f (xij, yij)∆A is called Riemann sum of f .

.

......

Definition The double integral of f over R is∫∫R

f (x, y) dA = limm,n→∞

m

∑i=1

n

∑j=1

f (xij, yij)∆A, if this limit exists.


. . . . . .

Remarks. In general, it is very difficult to prove that the limit ofRiemann sum converges, because of the choices of the heightf (xij, yih) involved. The usual method is replace the height either themaximum and the minimum values of f within each smallerrectangles, and hence we obtain the upper and lower Riemann sumsrespectively..

......

Theorem. If f (x, y) is continuous on a domain containing the

rectangle R, then the double integral∫∫

Rf (x, y) dA always exists.


. . . . . .

.

......

Example. Approximate the value of the integral∫∫

R(4x3 + 6xy2) dA

over the rectangle R = [1, 3]× [−2, 1], by means of the Riemannsums, with ∆xi = 1, and ∆yj = 1.

Solution. Partition the rectangle R into six 1 × 1squares Ri with area ∆Ai = 1 (i = 1, · · · , 6).Choose the center points (x∗i , y∗i ) for each squareas shown on the right.

The desired Riemann sum is6

∑i=1

f (x∗i , y∗i )∆Ai

= f ( 32 ,− 3

2 )× 1 + f ( 52 ,− 3

2 )× 1 + f ( 32 ,− 1

2 )× 1+f ( 5

2 ,− 12 )× 1 + f ( 3

2 , 12 )× 1 + f ( 5

2 , 12 )× 1

=135

4+

3854

+634

+2654

+634

+2654

= 294,which is called the midpoint approximation of the integral∫∫

R(4x3 + 6xy2) dA.


. . . . . .

.

......


R(4x3 + 6xy2) dA




∑i=1


= f ( 32 ,− 3

2 )× 1 + f ( 52 ,− 3

2 )× 1 + f ( 32 ,− 1

2 )× 1+f ( 5

2 ,− 12 )× 1 + f ( 3

2 , 12 )× 1 + f ( 5

2 , 12 )× 1

=135

4+

3854

+634

+2654

+634

+2654

= 294,

which is called the midpoint approximation of the integral∫∫R(4x3 + 6xy2) dA.


. . . . . .

.

......


R(4x3 + 6xy2) dA




∑i=1


= f ( 32 ,− 3

2 )× 1 + f ( 52 ,− 3

2 )× 1 + f ( 32 ,− 1

2 )× 1+f ( 5

2 ,− 12 )× 1 + f ( 3

2 , 12 )× 1 + f ( 5

2 , 12 )× 1

=135

4+

3854

+634

+2654

+634

+2654

= 294,which is called the midpoint approximation of the integral∫∫

R(4x3 + 6xy2) dA.


. . . . . .

.Iterated Integrals..

......

Let f (x, y) be a function defined on R = [a, b]× [c, d]. We write∫ d

cf (x, y) dy to mean that x is regarded as a constant and f (x, y) is

integrated with respect to y from y = c to y = d.

.

......

Therefore, the value of the integral∫ d

cf (x, y) dy is a function of x, and

we can integrate it with respect to x from x = a to x = b. The resulting

integral∫ b

a

(∫ d

cf (x, y) dy

)dx is called an iterated integral.

kk


. . . . . .


......




.

......


cf (x, y) dy is a function of x,

and


integral∫ b

a

(∫ d

cf (x, y) dy


kk


. . . . . .


......




.

......




integral∫ b

a

(∫ d

cf (x, y) dy


kk


. . . . . .


......




.

......




integral∫ b

a

(∫ d

cf (x, y) dy


Similarly

one can define the iterated integral∫ d

c

(∫ b

af (x, y) dx

)dy.

Remark. We call the blue and red segments inside the region R thecross-sections of R cut by the line y = y0 and x = x0 respectively.


. . . . . .


......




.

......




integral∫ b

a

(∫ d

cf (x, y) dy

)dx is called an iterated integral. Similarly

one can define the iterated integral∫ d

c

(∫ b

af (x, y) dx

)dy.

Remark. We call the blue and red segments inside the region R thecross-sections of R cut by the line y = y0 and x = x0 respectively.


. . . . . .

.

......

Example. Evaluate the iterated integrals

(a)∫ 3

0

∫ 2

1x2y dy dx, (b)

∫ 2

1

∫ 3

0x2y dx dy.

Solution.

(a)∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0x2∫ 2

1y dydx = =∫ 3

0x2[

y2

2

]y=2

y=1dx =

∫ 3

0

3x2

2dx =

[x3

2

]x=3

x=0=

272

.

(b)∫ 2

1

∫ 3

0x2y dx dy =

∫ 2

1

[x3y3

]x=3

x=0dy =

∫ 2

19ydy =

[9y2

2

]y=2

y=1=

272

.


. . . . . .

.

......


(a)∫ 3

0

∫ 2

1x2y dy dx, (b)

∫ 2

1

∫ 3

0x2y dx dy.

Solution.

(a)∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0x2∫ 2

1y dydx =

=∫ 3

0x2[

y2

2

]y=2

y=1dx =

∫ 3

0

3x2

2dx =

[x3

2

]x=3

x=0=

272

.

(b)∫ 2

1

∫ 3

0x2y dx dy =

∫ 2

1

[x3y3

]x=3

x=0dy =

∫ 2

19ydy =

[9y2

2

]y=2

y=1=

272

.


. . . . . .

.

......


(a)∫ 3

0

∫ 2

1x2y dy dx, (b)

∫ 2

1

∫ 3

0x2y dx dy.

Solution.

(a)∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0x2∫ 2

1y dydx = =∫ 3

0x2[

y2

2

]y=2

y=1dx

=∫ 3

0

3x2

2dx =

[x3

2

]x=3

x=0=

272

.

(b)∫ 2

1

∫ 3

0x2y dx dy =

∫ 2

1

[x3y3

]x=3

x=0dy =

∫ 2

19ydy =

[9y2

2

]y=2

y=1=

272

.


. . . . . .

.

......


(a)∫ 3

0

∫ 2

1x2y dy dx, (b)

∫ 2

1

∫ 3

0x2y dx dy.

Solution.

(a)∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0

∫ 2

1x2y dydx =

∫ 3

0x2∫ 2

1y dydx = =∫ 3

0x2[

y2

2

]y=2

y=1dx =

∫ 3

0

3x2

2dx =

[x3

2

]x=3

x=0=

272

.

(b)∫ 2

1

∫ 3

0x2y dx dy =

∫ 2

1

[x3y3

]x=3

x=0dy =

∫ 2

19ydy =

[9y2

2

]y=2

y=1=

272

.


. . . . . .

.Fubini’s Theorem for Rectangle case..

......

Thoerem. If f (x, y) is continuous on R = [a, b]× [c, d], then∫ b

a

∫ d

cf (x, y) dy dx =

∫∫R

f (x, y)dA =∫ d

c

∫ b

af (x, y) dx dy.

.

......

Example. f (x, y) is a positive function defined on a rectangleR = [a, b]× [c, d]. The volume V of the solid under the graph ofz = f (x, y) over R, is given by either one of the iterated integrals:∫ b

a

∫ d

cf (x, y) dy dx, or

∫ d

c

∫ b

af (x, y) dx dy.


. . . . . .

.Fubini theorem for non-rectangular region..

......

Let R be a region in the xy-plane, and suppose there exist twocontinuous function ymin(x), ymax(x) defined on the interval [a, b]such that R = { (x, y) | a ≤ x ≤ b, ymin(x) ≤ y ≤ ymax(x) }, then∫∫

Rf (x, y) dA =

∫ b

a

∫ ymax(x)

ymin(x)f (x, y) dy dx.

.Using Vertical Cross-sections..

......

In evaluating∫∫

R f (x, y)dA, one can ideally use the iterated integral

∫∫R

f (x, y)dA =∫ b

a

(∫ ∗

?f (x, y) dy

)dx,

the difficulties lies in determining upper and lower limits ∗, ? in theiterated integral.


. . . . . .

.

......

Example. Let R be the region in first quadrant bounded by the twocurves x2 + y2 = 1 and x + y = 1.

...1 Sketch and label the bounding curves, and determine the regionR of integration in the double integral.

...2 Project the region R onto one the coordinate axes, so that itsshadow is an interval [a, b] or union of intervals on the coordinateaxis.


. . . . . .

.

......


Usually the intersection point of curves C1 : x + y = 1 andC2 : x2 + y2 = 1 gives some important information.1 = x2 + y2 = x2 + (1 − x)2 = 2x2 − 2x + 1, i.e. 0 = x(x − 1), andhence we know (x, y) = (1, 0) and (0, 1) are the common intersection.

Want to see (by mathematical means) the relative position of thecurves y = 1 − x and y =

√1 − x2 when x varies in the interval [0, 1].

For 0 ≤ x ≤ 1, we have 0 ≤ 1 − x ≤ 1 + x, so

(1 − x)2 ≤ (1 + x)(1 − x) = 1 − x2,

it follows that 1 − x ≤√

1 − x2 for 0 ≤ x ≤ 1, i.e. the line segment C2is below the circle C2.


. . . . . .

.

......


Usually the intersection point of curves C1 : x + y = 1 andC2 : x2 + y2 = 1 gives some important information.1 = x2 + y2 = x2 + (1 − x)2 = 2x2 − 2x + 1, i.e. 0 = x(x − 1), andhence we know (x, y) = (1, 0) and (0, 1) are the common intersection.Want to see (by mathematical means) the relative position of thecurves y = 1 − x and y =

√1 − x2 when x varies in the interval [0, 1].

For 0 ≤ x ≤ 1, we have 0 ≤ 1 − x ≤ 1 + x, so

(1 − x)2 ≤ (1 + x)(1 − x) = 1 − x2,

it follows that 1 − x ≤√

1 − x2 for 0 ≤ x ≤ 1, i.e. the line segment C2is below the circle C2.


. . . . . .

.

......

Example. Let R be the region in first quadrant bounded by the twocurves x2 + y2 = 1 and x + y = 1. Rewrite the double integral∫∫

Rf (x, y)dA in iterated integrals.

...1 Choose any arbitrary point P(x, 0) or P(0, y) in the shadow, drawa line ℓ through P perpendicular to the axis with shadow.

...2 Ideally the line ℓ meets the boundary R at only two points(x, ymax) and (x, ymin). These two y’s depends on x, and henceare functions of x, i.e. the ones determined by the boundarycurves of R. Then the regionR = { (x, y) ∈ R2 | a ≤ x ≤ b, ymin(x) ≤ y ≤ ymax(x) }.

Answer:∫∫

Rf (x, y)dA

=∫ 1

0

∫ √1−x2

1−xf (x, y) dy dx


. . . . . .

.

......

Example. Let a > 0. Express the triangular region R with vertices A, Band C on xy-plane: (i) A(0, 0), B(a, 0) and C(a, 0);(ii) A(0, 0), B(0, a) and C(a, a); (iii) A(0, 0), B(−a, 0) and C(a, a).Hint: Draw the triangle ABC first.

Solution. (i) Note the the equation of BC is given by x + y = a, whichis above the x-axis AB defined by y = 0. So the region can bedescribed by vertical section as follows:R = { (x, y) | 0 ≤ x ≤ a, 0 ≤ y ≤ a − x }.

(ii) Note the the equation of BC is given by y = a, which is above thethe segment AC defined by the equation y = x. So the region can bedescribed by vertical section as follows:R = { (x, y) | 0 ≤ x ≤ a, x ≤ y ≤ a }.(iii) Note it would be better to describe the region by means ofsections parallel to x-axis, i.e. y-section. In this case the segment ABis below the segment AC. So the region can be described by verticalsection as follows: R = { (x, y) | 0 ≤ y ≤ a, − y ≤ x ≤ y }.


. . . . . .

.

......


Solution. (i) Note the the equation of BC is given by x + y = a, whichis above the x-axis AB defined by y = 0. So the region can bedescribed by vertical section as follows:R = { (x, y) | 0 ≤ x ≤ a, 0 ≤ y ≤ a − x }.(ii) Note the the equation of BC is given by y = a, which is above thethe segment AC defined by the equation y = x. So the region can bedescribed by vertical section as follows:R = { (x, y) | 0 ≤ x ≤ a, x ≤ y ≤ a }.

(iii) Note it would be better to describe the region by means ofsections parallel to x-axis, i.e. y-section. In this case the segment ABis below the segment AC. So the region can be described by verticalsection as follows: R = { (x, y) | 0 ≤ y ≤ a, − y ≤ x ≤ y }.


. . . . . .

.

......


Solution. (i) Note the the equation of BC is given by x + y = a, whichis above the x-axis AB defined by y = 0. So the region can bedescribed by vertical section as follows:R = { (x, y) | 0 ≤ x ≤ a, 0 ≤ y ≤ a − x }.(ii) Note the the equation of BC is given by y = a, which is above thethe segment AC defined by the equation y = x. So the region can bedescribed by vertical section as follows:R = { (x, y) | 0 ≤ x ≤ a, x ≤ y ≤ a }.(iii) Note it would be better to describe the region by means ofsections parallel to x-axis, i.e. y-section. In this case the segment ABis below the segment AC. So the region can be described by verticalsection as follows: R = { (x, y) | 0 ≤ y ≤ a, − y ≤ x ≤ y }.


. . . . . .

.

......

Example. Let a > 0, express the triangular region Rwith vertices A, B and C on xy-plane: A(0, 0), B(−a, a)and C(2a, 2a);

Solution. Suppose that the segment BC meets the y-axis at a point D.The equation of segment AC is y = x (0 ≤ x ≤ 2a), the equation ofsegment AB is y = −x (−a ≤ x ≤ 0), and the equation of BC is givenby y = 1

3 (x + a) + a = x+4a3 . No matter which direction you project the

region R to, the region can not be described so easily as we did in theprevious example. In this case, we divide the region R into 2 regions:R1 = △ACD =

{(x, y) | 0 ≤ x ≤ a, x ≤ y ≤ x+4a

3

}, and

R2 = △ACB ={(x, y) | − a ≤ x ≤ 0, − x ≤ y ≤ x+4a

3

}.


. . . . . .

.

......Example. Evaluate the integral

∫ √π/6

0

∫ x

0cos(x2) dydx.

Solution.∫ √

π/6

0

∫ x

0cos(x2) dydx =

∫ √π/6

0x cos(x2)dx =

12

∫ √π/6

0cos(x2)d(x2) =

12

[sin(x2)

]√π/6

x=0=

12

sin(π

6) =

14

.


. . . . . .

.

......

Example. Let 0 < a < b , and function f be continuous on the closedinterval [a, b]. Prove that∫ b

a

(∫ x

a(x − y)n−2f (y) dy

)dx =

∫ b

a

(b − y)n−1

n − 1f (y) dy for n = 1.

Hint: Interchange the order of integration.

Solution. Let R = { (x, y) | a ≤ x ≤ b, a ≤ y ≤ x } be the triangle withvertices A(a, a), C(b, a) and B(b, b). As the first definition is given bymeans of vertical section (line segment parallel to the y-axes).By using the horizontal section (line segment parallel to the x-axes),one has R = { (x, y) | a ≤ y ≤ b, y ≤ x ≤ b }. Hence∫ b

a

(∫ y=x

y=a(x − y)n−2f (y) dy

)dx

=∫ b

a

(∫ x=b

x=y(x − y)n−2 dx

)f (y) dy =

∫ b

a

[(x − y)n−1

n − 1

]x=b

x=yf (y)dy

=∫ b

a(b−y)n−1

n−1 f (y)dy.Remark. This kind of trick of rewriting the triangle ABC is verycommon, and student should know for the test.


. . . . . .

.

......Example. Evaluate the iterated integral

∫ 1

0

∫ 1

√y

√1 + x3 dxdy.

Solution. Let R = { (x, y) | 0 ≤ y ≤ 1,√

y ≤ x ≤ 1 }. Hence R isbounded by the lines x = 1, y = 0 and x =

√y, i.e. y = x2.

As we want to interchange the order of integration, soR = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 } and∫ 1

0

∫ 1

√y

√1 + x3 dxdy =

∫ 1

0

∫ x2

0

√1 + x3 dydx =

∫ 1

0x2√

1 + x3 dx =

13

[(1 + x3)1+1/2

1 + 1/2

]1

0

=29(2√

2 − 1).


. . . . . .

.

......

Example. For the following integral:∫ 2

0

∫ 4

y2

y3ex2

xdxdy, sketch the

region of integration, reverse the order of integration, and evaluatethe resulting integral.

Solution. The region D of integration is shown below.

D = { (x, y) | 0 ≤ y ≤ 2, y2 ≤ x ≤ 4 }= { (x, y) | 0 ≤ x ≤ 4, 0 ≤ y ≤

√x }.

Reversing the order of integration and evaluating the integral we

have:∫ 2

0

∫ 4

y2

y3ex2

xdxdy =

∫ 4

0

∫ √x

0

y3ex2

xdydx =

∫ 4

0

[y4ex2

4x

]√x

0

dx =

∫ 4

0

xex2

4dx =

[ex2

8

]4

0

=e16 − 1

8.


. . . . . .

.. Properties of Double Integrals...1∫∫

D( f (x, y) + g(x, y) )dA =

∫∫D

f (x, y) dA +∫∫

Dg(x, y) dA.

...2∫∫

Dcf (x, y) dA = c

∫∫D

f (x, y) dA....3 If f (x, y) ≥ g(x, y) for all (x, y) ∈ D, then∫∫

Df (x, y) dA ≥

∫∫D

g(x, y) dA.

...4∫∫

Df (x, y) dA =

∫∫D1

f (x, y) dA +∫∫

D2

f (x, y) dA, where

D = D1 ∪ D2, and D1 and D2 donot overlap except perhaps on theirboundary.

...5∫∫

DdA =

∫∫D

1 dA = Area of D = A(D)....6 If m ≤ f (x, y) ≤ M for all (x, y) ∈ D, then

mA(D) ≤∫∫

Df (x, y) dA ≤ MA(D).


. . . . . .

.. Double Integrals in Polar Coordinates..

......

Proposition. If f (x, y) is continuous on a region D in xy-plane, and Dcan be described in polar coordinates in the following form{ (r, θ) | α ≤ θ ≤ β, g1(θ) ≤ r ≤ g2(θ) }, then∫∫

Df (x, y) dA =

∫ β

α

∫ g2(θ)

g1(θ)f (r cos θ, r sin θ) r dr dθ.


. . . . . .

.

......

Example. Evaluate the double integral∫ 2

3/2

(∫ √4−x2

√4x−x2

f (√

x2 + y2)dy

)dx

Solution.∫ π/6

−π/6

(∫ 2 cos θ

3 sec θ/2f (r)r dr

)dθ.

.

......


0

(∫ √4−x2

√4x−x2

f (√

x2 + y2)dy

)dx

Solution.∫ π/6

−π/6

(∫ 2 cos θ

3 sec θ/2f (r)r dr

)dθ.


. . . . . .

.

......Example. Evaluate the double integral

∫∫x2+y2≤x+y

(x + y) dxdy.

Solution I. Let D = { (x, y ) | x2 + y2 ≤ x + y }. In terms of polarcoordinates, D can be described by D′ = { (r, θ ) | r ≤ cos θ + sin θ }.

So we have∫∫

D(x + y) dxdy =

∫∫D′

r(cos θ + sin θ) rdrdθ

=∫ 3π

4

− π4

∫ cos θ+sin θ

0r2(cos θ + sin θ) drdθ =

13

∫ 3π4

− π4

(cos θ + sin θ)4 dθ

=13

∫ 3π4

− π4

(1 + 2 sin 2θ + sin2 2θ) dθ =π

2.

Solution II. The region D is a circular disc with center at (1/2, 1/2),radius of 1/2. Define a translation asϕ(u, v) = (x(u, v), y(u, v)) = (u + 1/2, v + 1/2), then in terms ofuv-coordinates, D can be described byG = { (u, v ) | u2 + v2 ≤ 1/2 }. Here the jacobian ∂(x, y)/∂(u, v) = 1.

So we have I =∫∫

D(x + y)dxdy =

∫∫u2+v2≤1/2

(1 + u + v)dudv =∫∫u2+v2≤1/2

dudv =π

2. (Area of a circle)


. . . . . .

.

......


3/2

(∫ √4−x2

√4x−x2

f (√

x2 + y2)dy

)dx

Solution.∫ π/6

−π/6

(∫ 2 cos θ

3 sec θ/2f (r)r dr

)dθ.

.

......Example. Evaluate the double integral

∫ 1

0

(∫ √4−x2

√4x−x2

√x2 + y2dy

)dx

Solution.∫ π/6

−π/6

(∫ 2 cos θ

3 sec θ/2f (r2)r dr

)dθ.


. . . . . .

.

......

Example. Evaluate the integral∫ 1

0

∫ √2−y2

y3(x − y) dxdy by

converting into polar coordinates.

Solution. The region D = { (x, y) | 0 ≤ y ≤ 1, y ≤ x ≤√

2 − y2 }represents the sector with radius

√2 from θ = 0 to π/4. Hence∫ 1

0

∫ √2−y2

y3(x − y) dxdy =

∫ π/4

0

∫ √2

03r(cos θ − sin θ)rdrdθ

=

[33

r3]√2

0· [sin θ + cos θ]π/4

0 = 4 − 2√

2.


. . . . . .

.

......

Example. Let D be the circular region x2 + y2 ≤ R2, R > 0. Evaluate

A =∫∫

D|xy| dxdy, B =

∫∫D|x + y| dxdy.

Solution. Let D = { (x, y) | x2 + y2 ≤ R2 }, andD′ = { (x, y) | x2 + y2 ≤ R2, x ≥ 0, y ≥ 0 }.

By symmetry, we know that A = 4∫∫

D′xydxdy =

2∫ π/2

0

∫ R

0r3 sin(2θ) drdθ =

[R4

4sin 2θ

]π/2

0=

R4

2.

The second integral, in terms of polar coordinates, is

B =∫ 2π

0

∫ R

0r2| cos θ + sin θ| drdθ =

R3

3

∫ 2π

0

√2| sin(θ +

π

4)| dθ =

√2R3

3

∫ 2π

0| sin(θ)| dθ =

√2R3

34∫ π/2

0sin(θ) dθ =

4√

2R3

3.


. . . . . .

.Applications of Double Integrals..

......

Suppose a (planar) object, in region R, is made of different material inwhich the density (mass per unit area) is given by δ(x, y), dependingon the location (x, y). Then the total mass of R is given approximatelyby the Riemann sum ∑i δ(xi, yi)∆Ai, which will converges to the

double integral m =∫∫

Dδ(x, y) dA. We call it the mass of the object.

.

......

Similarly, one can define the center of mass (centroid) of the object by

(x, y) =(

1m

∫∫D

xδ(x, y) dA,1m

∫∫D

yδ(x, y) dA)

.


. . . . . .

.

......

Example. A lamina R is made of the part of circular disc of radius a in1st quadrant. Its density is proportional to the distance from theorigin. Determine the position of its centroid.

Solution. In terms of polar coordinates, D can be described as{ (r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ π/2 }. Let δ(x, y) = k

√x2 + y2 = kr.

Then the mass m =∫ π/2

0

∫ a

0kr · r dr dθ =

π

2· ka3

3=

kπa3

6.

As all the conditions on the lamina R are symmetric in x and y, so

x = y =1m

∫ π/2

0

∫ a

0(r sin θ) · kr · r dr dθ =

km

(∫ π/2

0sin θdθ

)·(∫ a

0r3 dr

)=

kkπa3/6

× 1 × a4

4=

3a2π

.


. . . . . .

.First Theorem of Pappus: Volume of Revolution..

......

Suppose that a plane region R is revolved around an axis in its planegenerating a solid of revolution with volume V. Assume that the axisdoes not intersect the interior of R. Then the volume V of the solid is

V = A · d,

where A is the area of R and d is the distance traveled by the centroidof R.

Remark. Cut the region R into vertical strips, and each vertical stripafter rotating will form a ring, which contributes ∆Vi = 2πx∗i ∆Ai,where Ai is the area of the vertical strip. It follows from Riemann sumthat

V ≈n

∑i=1

∆Vi =n

∑i

2πx∗i ∆Ai.Prepared by Dr. I.T. Leong for Tutorial 5 of Math 200

. . . . . .

.Volume of a Sphere........Prove that the volume of a sphere is 4

3 πa3.

Solution. Let R be a region bounded by the up-per semi-circle x2 + y2 = a2, y ≥ 0 and x-axis.

In polar coordinates, R can be described as{ (r, θ) | 0 ≤ θ ≤ π, 0 ≤ r ≤ a }. Using polar coordinates, the area of

R is =∫ π

0

∫ a

0r dr dθ =

a2

2× π = πa2/2. Similarly, y of R is given by

1πa2/2

∫∫R

ydA =1

πa2/2

∫ π

0

∫ a

0r sin θ · r dr dθ

=1

πa2/2

(∫ π

0sin θ dθ

)( ∫ a

0r2 dr

)=

1πa2/2

× [− cos θ ]π0 ×[

r3

3

]a

0

=2

πa2 × 2 × a3

3=

4a3π

. By Pappus’s theorem, the volume of a sphere

of radius a = 2πy× Area of R = 2π · 4a3π

· πa2

2=

43

πa3.


. . . . . .


3 πa3.



R is =∫ π

0

∫ a

0r dr dθ =

a2

2× π = πa2/2.

Similarly, y of R is given by

1πa2/2

∫∫R

ydA =1

πa2/2

∫ π

0

∫ a


=1

πa2/2

(∫ π

0sin θ dθ

)( ∫ a

0r2 dr

)=

1πa2/2

× [− cos θ ]π0 ×[

r3

3

]a

0

=2

πa2 × 2 × a3

3=

4a3π



· πa2

2=

43

πa3.


. . . . . .


3 πa3.



R is =∫ π

0

∫ a

0r dr dθ =

a2


1πa2/2

∫∫R

ydA =1

πa2/2

∫ π

0

∫ a


=1

πa2/2

(∫ π

0sin θ dθ

)( ∫ a

0r2 dr

)=

1πa2/2

× [− cos θ ]π0 ×[

r3

3

]a

0

=2

πa2 × 2 × a3

3=

4a3π



· πa2

2=

43

πa3.


. . . . . .


3 πa3.



R is =∫ π

0

∫ a

0r dr dθ =

a2


1πa2/2

∫∫R

ydA =1

πa2/2

∫ π

0

∫ a


=1

πa2/2

(∫ π

0sin θ dθ

)( ∫ a

0r2 dr

)=

1πa2/2

× [− cos θ ]π0 ×[

r3

3

]a

0

=2

πa2 × 2 × a3

3=

4a3π

.

By Pappus’s theorem, the volume of a sphere


· πa2

2=

43

πa3.


. . . . . .


3 πa3.



R is =∫ π

0

∫ a

0r dr dθ =

a2


1πa2/2

∫∫R

ydA =1

πa2/2

∫ π

0

∫ a


=1

πa2/2

(∫ π

0sin θ dθ

)( ∫ a

0r2 dr

)=

1πa2/2

× [− cos θ ]π0 ×[

r3

3

]a

0

=2

πa2 × 2 × a3

3=

4a3π



· πa2

2=

43

πa3.


. . . . . .

.Second Theorem of Pappus: Surface Area of Revolution..

......

Suppose that a plane curve C is revolved around an axis in its planethat does not intersect the curve. Then the area A of the surface ofrevolution generated is

A = s · d,

where s is the arc-length of C and d is the distance traveled by thecentroid of C.


. . . . . .

.

......

Let R be a plane lamina and ℓ be a straight line that may or may notlie in xy-plane. Then the moment of inertia I of R around the axis ℓ is∫∫

Rp2δ(x, y)dA, where p = p(x, y) is the shortest distance from the

point (x, y) of R to the line ℓ, and δ(x, y) is the density of R at the point(x, y).

For the coordinate axii, we have.

......

Ix = Ix-axis =∫∫

R(y2)δ(x, y)dA, Iy = Iy-axis =

∫∫R(x2)δ(x, y)dA and

Iz = Iz-axis =∫∫

R(x2 + y2)δ(x, y)dA.

.

......

For any plane lamina R, Define the center (x, y, z) of gyration by

x =√

Ixm , y =

√Iym , z =

√Izm .


. . . . . .

.

......

Fubini’s Theorem for non-rectangular region. Let D be a solid regionin space.If D = { (x, y, z) ∈ R3 | (x, y) ∈ R, andzmin(x, y) ≤ z ≤ zmax(x, y) for all (x, y) ∈ R }, where zmax(x, y) andzmin(x, y) are continuous functions defined in the region D inxy-plane. Let f (x, y, z) be a scalar function defined in the region D,

then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)

zmin(x,y)f (x, y, z)dz

)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate. The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y) andthen exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.


. . . . . .

.

......


then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)


)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate.

The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y) andthen exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.


. . . . . .

.

......


then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)


)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate. The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y)

andthen exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.


. . . . . .

.

......


then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)


)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate. The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y) andthen exit the solid D when z reaches zmax(x, y).

(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.


. . . . . .

.

......


then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)


)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate. The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y) andthen exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.

(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.


. . . . . .

.

......


then∫∫∫

Df (x, y, z) dV =

∫∫R

(∫ zmax(x,y)


)dA.

Remark. (1). The region R in xy-plane is in fact the shadow of the Dunder the projection from R3 onto xy-plane, or by simply forgettingthe z-coordinate. The idea is to allow the point P(x, y, 0) varies withinthe region D, and then draw a line through P perpendicular toxy-plane, which will enter the solid D when z reaches zmin(x, y) andthen exit the solid D when z reaches zmax(x, y).(2). The triple integral has similar properties like the double integral,here we con’t repeat in stating these properties.(3). It is not necessary to project the solid D onto xy-plane, one canproject D onto xz-plane, or yz-plane, in these cases, we should usefunction y = y(x, z) or x = x(y, z) respectively.


. . . . . .

.

......

Sketch. Sketch the region D along with its "shadow" R (verticalprojection) in the xy-plane. Label the upper and lower boundingsurfaces of D and the upper and lower bounding curves of R.


. . . . . .

.

......

Find the z-limits of integration. Draw a line M passing through atypical point (x, y) in the shadow R parallel to the z-axis. As zincreases, M enters the solid region D at z = zmin(x, y) = f1(x, y) andleaves at z = zmax(x, y) = f2(x, y). These are the z-limits of integration


. . . . . .

.

......

Find the y-limits of integration. Draw a line L through (x, y) parallel tothe y-axis. As y increases, L enters R at y = ymin = g1(x) and leavesat y = ymax(x) = g2(x). These are the y-limits of integration.


. . . . . .

.

......

Find the x-limits of integration. Choose x-limits that include all linesthrough R parallel to the y-axis (x = a and x = b in the precedingfigure). These are the x-limits of integration. The integral is∫ x=b

x=a

∫ y=g2(x)

y=g1(x)

∫ z=f1(x,y)

z=f2(x,y)F(x, y, z) dz dy dx.

.

......

Follow similar procedures if you change the order of integration. The"shadow" of region D lies in the plane of the last two variables withrespect to which the iterated integration takes place.


. . . . . .

.

......

Example. Find the volume of thetetrahedron D bounded by the planesx + 2y + z = 2, x = 2y, x = 0, and z = 0.

Solution. Project the tetrahedron D onto xy-plane with shadow R = { (x, y) | 0 ≤ x ≤1, 2/x ≤ y ≤ 1 − x/2 }.

Then D = { (x, y, z) | 0 ≤ x ≤ 1, x/2 ≤ y ≤1 − x/2, 0 ≤ z ≤ 2 − x − 2y }.Volume of D=∫∫∫

DdV =

∫∫R(2 − x − 2y) dz dA

=∫ 1

0

∫ 1−x/2

x/2(2 − x − 2y) dydx

=∫ 1

0

[2y − xy − y2

]1−x/2

x/2dx

=∫ 1

0

[2 − x − x(1 − x

2)− (1 − x

2)2 − x +

x2

2+

x2

4

]dx =

13

.


. . . . . .

.

......

Example. Evaluate the triple integral

I =∫∫∫

D

dxdydz(1 + x + y + z)3 , where the solid

D is bounded by the planesx + y + z = 1, x = 0, y = 0, z = 0.

Solution. Let R = { (x, y ) | x + y ≤ 1, x ≥ 0, y ≥ 0} be theprojection image of D onto xy-plane. ThenD = { (x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y }.

I =∫∫∫

D

dxdydz(1 + x + y + z)3 =

∫∫R

(∫ 1−x−y

0

dz(1 + x + y + z)3

)dxdy

=∫∫

R

[−1

2(1 + x + y + z)2

]1−x−y

z=0dxdy

=12

∫∫R

[1

(1 + x + y)2 − 14

]dxdy

=12

∫ 1

0

∫ 1−y

0

dx(1 + x + y)2 dy − 1

16= · · · = 1

2ln 2 − 5

16.


. . . . . .

.

......

Example. Find the volume of the solid Dbounded by xy-plane, the cylinder x2 + y2 = 1and the plane z = y.

Solution. The shadow of D onto xy-plane is giv-en by in polar coordinates by{ (r, θ) | 0 ≤ r ≤ 1, − π/2 ≤ θ ≤ π/2 }.

D is then given in cylindrical coordinates by{ (r, θ, z) | 0 ≤ r ≤ 1, − π/2 ≤ θ ≤ π/2, 0 ≤ z ≤ r sin θ }.

The volume of the solid D is given by∫ π/2

−π/2

∫ 1

0

∫ r sin θ

0r dz dr dθ

=∫ π/2

−π/2

∫ 1

0r2 sin θ dr dθ =

13

∫ π/2

−π/2sin θ dθ =

[−cos θ

3

]π/2

−π/2=

23

.


. . . . . .

.

......

Example. Find the volume of the region D that lies inside both thesphere x2 + y2 + z2 = 4 and the cylinder x2 + y2 − 2x = 0.

Solution. The cylinder S is given by (x − 1)2 + y2 = 1, so any pointP(x, y, z) inside S is given by (x − 1)2 + y2 ≤ 1. In terms of cylindricalcoordinates, it can be described by (r, θ, z) byr2 cos2 θ + r2 sin2 θ − 2r cos θ ≤ 0, i.e. r ≤ 2 cos θ, it follows that{ (r, θ, z) | − π

2 ≤ θ ≤ π2 , and −

√4 − r2 ≤ z ≤

√4 − r2 }.

Volume of the region

D =∫∫∫

D1 dV =

∫ π/2

−π/2

∫ 2 cos θ

0

(∫ √4−r2

−√

4−r2dz

)r drdθ.


. . . . . .

.

......

Example. Find the volume of the solid region D bounded below by theparaboloid z = x2 + y2, and above by the plane z = 2x.

Solution. First we sketch the graphs of the paraboloid S1 and theplane S2, and we want to find the common intersection of S1 and S2.Let P(x, y, z) be any point of the common intersection, i.e. they satisfyboth 2x = z = x2 + y2, so (x − 1)2 + y2 = 1, which represents acylinder in space. In fact, the intersection is a curve obtained by theintersection of the plane and the cylinder above, which is a boundedclosed curve. Inside the cylinder, we have (x − 1)2 + y2 ≤ 1, i.e.x2 + y2 ≤ 2x, or equivalently the graph of S1 is below the graph of theplane S2. Hence the solid region D = { (x, y, z) | x2 + y2 ≤ 2x, andx2 + y2 ≤ z ≤ 2x }. Then one can switch to cylindrical coordinates todescribe D as { (r, θ, z) | 0 ≤ r ≤ 2 cos θ, − pi

2 ≤ θ ≤ pi2 and

r2 ≤ z ≤ 2r cos θ }. The volume of D =∫∫∫

D1 dV =∫∫

x2+y2≤2x

(∫ 2x

x2+y21 dz

)dA =

∫ π/2

−π/2

∫ 2 cos θ

0

(∫ 2r cos θ

r21 dz

)r drdθ.


. . . . . .

.

......

Example. Find the volume of the solid region D bounded above bythe spherical surface x2 + y2 + z2 = 2 and below by the paraboloidz = x2 + y2.

Solution. First determine the equation of curve of the intersection ofthese two surfaces, let P(x, y, z) be a point of the intersection,2 − z2 = x2 + y2 = z ≥ 0, so 0 = z2 + z − 2 = (z + 2)(z − 1), so itfollows that z ≥ 0 that we have z = 1. In other words, the intersectionlies on a plane defined by the equation z = 1, and hencex2 + y2 = z = 1, which represents a circle C on the plane z = 1. If(x, y, z1) and (x, y, z2) are points on spherical surface, and theparaboloid such that the point (x, y) lies inside the circle x2 + y2 = 1then x2 + y2 ≤ 1, andz2 =

√2 − x2 − y2 ≥

√2 − 12 = 1 ≥ x2 + y2 = z2, we have

D = { (x, y, z) | x2 + y2 ≤ 1 and x2 + y2 ≤ z ≤√

2 − x2 − y2 }. The

volume of D =∫∫∫

D1 dV =

∫∫x2+y2≤1

(∫ √2−x2−y2

x2+y21 dz

)dA =

∫ 1

0

(∫ 2π

0

∫ √2−r2

r21 dzdθ

)rdr = 2π

∫ 1

0(√

2 − r2 − r2)rdr = · · · .


. . . . . .

.

......

Example. Find the volume of the region D bounded by the planez = 1 and the cone z = r, (i) with cylindrical coordinates; (ii) withspherical coordinates.

Solution. (i) In cylindrical coordinates, D can be described as{ (r, θ, z) | 0 ≤ r ≤ 1 and r ≤ z ≤ 1 }, so the volume of D =∫∫∫

D1 dV =

∫∫x2+y2≤1

(∫ 1√

x2+y2dz)

dA =∫ 2π

0

∫ 1

0

(∫ 1

r1 dz

)r drdθ

(ii) As the cone is symmetric about z-axis, so one know that0 ≤ θ ≤ 2π, and Volume of

D =∫∫∫

D1 dV =

∫ 2π

0

(∫ π/4

0

∫ sec ϕ

0ρ2 sin ϕ dρdϕ

)dθ.


. . . . . .

.

......

Example. Let D be the (solid) region bounded by the following threesurfaces: paraboloid x =

√y − z2, the cylinder x =

√y/2, and the

plane y = 1. Then determine the volume of D.

Solution. For any fixed b, (0 ≤ b ≤ 2), the plane Sb : y = b, parallel tothe xz-coordinate plane, will intersect the given solid region D with anintersection D(b), with boundary given by the curvesC1 : x =

√b − z2, and C2 : x =

√b/2 on the plane Sb respectively.

The curve C1 represents part x =√

b − z2 of a circle defined byx2 + z2 = b2, and the curve C2 represents portion of the linex =

√b/2 on the plane Sb. With this description, we can sketch the

circle and the line with various values of b (0 ≤ b ≤ 2), and then findout the cross-section D(b) is given precisely by

D(b) = {(x, b, z) | x2 + z2 ≤ b2, x ≥√

b/2 } = {(x, b, z) | −√

32

b ≤

z ≤√

32

b,

√b

2≤ x ≤

√b − z2 }. Then the volume of D is given by∫ 2

0

(∫∫D(y)

dAxz

)dy =

∫ 2

0

∫ √3

2 y

−√

32 y

∫ √y−z2

√y/2

dxdzdy = · · ·


5. applications of double integrals, volume and first

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