roxy-
16
The ABC’s of Calculus
case2: y = −x ⇒ x2 − x2 + x2 = 16 or x2 = 16.
x = ±4 But y = −x ⇒ y = ±4 too!
So we get four critical points:
4
,
x= √
3
4
x = −√
,
3
x=4 ,
x = −4 ,
x2 + y 2
32
3
y=4
2 + y2
x
32
3
4
y= √
3
4
y = −√
3
y = −4
32
3
32
√3
√32
32
32
32
check critical points for max/min.
Note that maximum is at (4, −4) and (−4, 4) and the value at this point is
Note that minimum is at (± √43 , ± √43 ) and the value here is
1.5
√
32
32
3
Double and Iterated Integrals
Let domain f = R and write z = f (x, y) continuous over a finite region R of
the x, y plane. We divide R into n subregions area ∆A1 , ∆A2 , ......, ∆An in any
fashion what so ever.
Figure 4
Next we pick a point (x1 , y1 ), (x2 , y2 ), ......, (xn , yn ) in each one of these subregions and form the sum:
n
f (xi , yi )∆Ai = f (x1 , y1 )∆A1 + ...... + f (xn , yn )∆An
i=1
Think about f (xi , yi )∆Ai : This is the signed volume of a parallelepipeds of
base area ∆Ai and height f (xi , yi ).
Functions and their properties
17
If this sum approaches a limit as n → ∞ and at the same time every subregion
shrinks to a point (∆Ai → 0) the limit (if it is unique) is called the double
integral of f over R and denoted by:
f dA or
R
f dA or
f (x, y)dA
R
R
by definition:
=
n
lim
n→+∞; ∆Ai →0
f (xi , yi )∆Ai
i=1
Remark:
1) Let f (x, y) ≥ 0 all (x, y) in R
⇒
R f dA = volume under that point of s whose projection is R
2) If f (x, y) = 1 for (x, y) in R then:
f dA =
dA = volume under the part of the plane z = 1 whose projection is in R.
R
R
So we claim that
dA = area of R (numerically)
R
The double integral (2D) has same properties as (1D) integral, namely:
1)
R
2)
R
(c · f )dA = c
f dA c is constant
R
(f ± g)dA = (
f dA) ± (
gdA)
R
R
3) If
R = R1 ∪ R2 ∪ ....... ∪ Rn where Ri ∩ Rj = 0
then
f dA =
R
f dA +
R1
f dA + .......... +
R2
f dA
Rn
In order to evaluate a double integral we need to set up an ”iterated integral”
in cartesian (rectangular) coordinates, such an integral takes the form:
f (x, y)dxdy
R1
or
f (x, y)dydx
R2
where
R1 = {(x, y) : f1 (y) ≤ x ≤ f2 (y),
a ≤ y ≤ b}
18
The ABC’s of Calculus
and
R2 = {(x, y) : g1 (x) ≤ y ≤ g2 (x),
c ≤ x ≤ d}
R1 &R2 are simply descriptions of the given region R in rectangular coordinates.
Figure 5
Describing Regions in R2
Example 15 Describe the circle of radius 4 centered at (0, 0) in two different
ways = {x2 + y 2 = 4}
1) Vertical slices:
Figure 6
= {(x, y) : −2 ≤ x ≤ 2,
−
4 − x2 ≤ y ≤ + 4 − x2
2) Horizontal slices:
= {(x, y) : − 4 − y 2 ≤ x ≤ + 4 − y 2 , −2 ≤ y ≤ 2}
Example 16 Figure 7
R = {(x, y) : 0 ≤ y ≤ (1 − x), 0 ≤ x ≤ 1}
Example 17 Figure 8
R = {(x, y) : x2 ≤ y ≤ x, 0 ≤ x ≤ 1}
NOTE: R1 and R2 are really the same region R described in two different
ways.
The iterated integral takes the form:
f (x, y)dxdy =
R1
f dA
(theorem......)
R
f (x, y)dxdy
means
b
f2 (y)
=
f (x, y)dxdy
a
f1 (y)
Functions and their properties
19
a, b, f1 (y), f2 (y):come from the description of R1 by horizontal slices.
b
f2 (y)
f dA =
f (x, y)dx
R
a
f1 (y)
dy
integrate this first,keep ”y” fixed.
Then integrate the resulting function of y given inside the box with respect to
”y”between a&b.
The other technique for evaluation involves taking vertical slices.
d
g2 (x)
f dA =
f (x, y)dy
R
c
g1 (x)
dx
integrate this first, keep ”x” fixed.
The double integral of (piece wise) continuous function f defined on R ⊂ Rn
exists.
We say f is integrable over R if
R ⇒ Fubini’s theorem says that:
| f | dA is finite. If f is integrable over
R
f (x, y)dxdy =
f (x, y)dydx =
R1
R2
f dA
R
Where R1 , R2 are the two (different) representations of region R. We can
reverse/interchange the order of integration.
Example 18
1
2
dxdy =
0
2
1
f dA.
R
1
1 2
1
( 1 dx)dy = 0 1 · dy = y |10 = 1
0
dx = (2 − 1) = 1 next
But this iterated integral by definition:
1
2
dxdy
0
=
dA
R
1
= area of the region R
= area of square
Example 19
1
x
xy 2 dydx
I=
0
x2
20
The ABC’s of Calculus
Figure 9
f (x, y) = xy 2
R = {(x, y) : x2 ≤ y ≤ x,
0 ≤ x ≤ 1}
I
1
=
y 2 dy)dx
x2
3
0
1
=
(x[
y y=x
|
2 ])dx
3 y=x
(x[
x6
x3
− ])dx
3
3
0
1
=
0
=
=
=
=
=
x
(x
1
x7
x4
− )dx
3
3
0
5
8
x
x
[ − ]10
15 24
1
1
−
15 24
1 1 1
·( − )
3 5 8
1
(
40
Example 20 HORIZONTAL SLICES:
Redescribe:
R = {(x, y) : 0 ≤ y ≤ 1, y ≤ x ≤
Figure 10
I
1
√
y
(xy 2 )dxdy
=
0
y
1
(y
=
2
0
1
(y 2 [
0
=
0
=
=
y
xdx)dy
y
=
√
1
x2 x=√y
]x=y )dy
2
y y2
y 2 ( − )dy
2
2
1 1 3
(y − y 4 )dy
2 0
1 y4
y5
[ − ]10
2 4
5
√
y}
Functions and their properties
=
=
=
21
1 1 1
[ − ]
2 4 5
1 1
( )
2 20
1
40
Example 21 Find the volume cut from 9x2 + 4y 2 + 36z = 36 by plane z = 0
1) Ensure surface is above xy-plane.
2) Give an idea of shape of region.
3) Set up an integral for volume.
4) Describe the ”projection” involved.
5) Use an appropriate coordinate system.
6) Use Fubini??
7) Write down iterated integrals
8) Evaluate
SOLUTION
1) Yes the surface lies above the xy-plane.
2) Done
3)
f dA = volume under s.
R
solve for z then 36z = 36 − 9x2 − 4y 2 ⇒ z = 1 −
x2
4
−
y2
9
The projection of s on the xy-plane (ie z = 0) is given by the use of
2
2
geometry z = 0 1 ≥ x4 + y9
4) volume =
R
y2
x2
− )dA
(1 −
4 9
=z
5) R is an ellipse,use elliptical coordinates.
x = ar cos Θ
y = br sin Θ
How to choose a,b?
x2
4
+
y2
9
=1
x2
a2
+
y2
b2
= r2
⇒ a=2,b=3
(Jacobian :abr)
Figure 11
22
The ABC’s of Calculus
2
2
6) R = {(x, y) : x4 + y9 ≤ 1} in rectangular coordinates.
R = {(r, Θ); 0 ≤ Θ < 2π, 0 ≤ r ≤ 1} in elliptical coordinates.
Volume =
R
(1 −
y2
x2
− )dA
4 9
=r 2
2π
1
a·b·r
6r drdΘ
(1 − r ) 2
=
0
0
2π
= 6·
0
2·3·r
1
( (r − r3 )dr)dΘ
0
1
= 12π ·
(r − r3 )dr
0
r4
r2
− ]10
2
4
1 1
= 12π[ − ]
2 4
= 3π
= 12π[
Example 22 Find the volume common to the cylinders x2 + y 2 = 16 and x2 +
z 2 = 16.
Idea: look at cylinder x2 + y 2 = 16 and its pojection onto the xy-plane namely:
D = {(x, y) : 0 < x2 + y 2 ≤ 16} =
{(x, y) : − 16 − x2 ≤ y ≤ 16 − x2 , −4 ≤ x ≤ 4}cartesian coordinates
√
The surface above D is given by x2 + z 2 = 16 or z 2 = 16 − x2 ⇒ z = ± 16 − x2
√
but z ≥ 0 ⇒ z = 16 − x2
1.6
Double and Iterated Integrals
Let domain f = R where f : R2 → R and let z = f (x, y) denote that part of
the surface S which lies above the dom f = R.
Figure 13
We divide R into m subregions of area ∆A1 , ∆A2 , . . . , ∆Am in any fashion.
Figure 14
Functions and their properties
23
Next we pick a point (x1 , y1 ), (x2 , y2 ), . . . , (xm , ym ) in each one of these subregions and form the sum:
m
f (xi , yi )∆Ai = f (x1 , y1 )∆A1 + f (x2 , y2 )δA2 + . . . + f (xm , ym )∆Am
i=1
If this sum approaches a limit as the number m → ∞ and at the same time
every subregion shinks to a point (ie ∆Ai → 0), the limit (if it is unique) is
called the double integral of f over R and is writeen as:
R
f dA =
R
or
=
f dA =
f (x, y)dA
R
or
m
lim
m→∞,∆Ai →0
f (xi , yi )∆Ai
i=1
If f (x, y) ≥ 0 then:
f dA = Volume under surface S lying directly above R
R
If f = 1, then
dA = area of R
R
Figure 15
The double integral(2 variables) has the following properties:
1)
R
2)
(c · f )dA = c
R
f dA c is any constant
R
(f ± g)dA = (
f dA) ± (
gdA)
R
R
3) If
R = R1 ∪R2 ∪. . .∪Rn where Ri ∩ Rj = φ (ie. these are pairwise disjoint)
then
f dA =
R
f dA +
f dA + . . . +
R1
R2
=
m i=1
Ri
f dA
f dA
Rn
24
The ABC’s of Calculus
In order to evaluate a double integral we set up a so-called iterated integral .
In cartesian coordinates, such an integral is of the form:
f (x, y)dxdy or
f (x, y)dydx
R1
R2
where
R1
and
=
{(x, y) : f1 (y) ≤ x ≤ f2 (y),
R2
=
{(x, y) : c ≤ x ≤ d,
a ≤ y ≤ b}
g1 (x) ≤ y ≤ g2 (x)}
Now each one of R1 , R2 are a description of R in cartesian coordinates:
Figure 16
So, R1 and R2 are really the same region described in two different ways.
The iterated integral then becomes:
f (x, y)dxdy
b
f2 (y)
=
f (x, y)dxdy
R1
a
f1 (y)
Limits come from the description of R above
b
{
=
a
f2 (y)
f (x, y)dx
f1 (y)
}dy
integrate w.r.t. x and hold y constant
Then integrate w.r.t. y to get a number
So this id the inverse proceedure to partial differentiation or we integrate with
respect to x holding y constant in the inner-most integral, and then integrate
with respect to y.
The double integral of a continuous function f defined on R ⊂ R2 exists. We
say f is integrable over R.
Indeed if |f (x, y)| is integrable over R, the iterated integrals of f over R exist
and are equal (Fubini), ie
f (x, y)dxdy =
f (x, y)dydx
R1
or, we can reverse the order of integration.
R2
Functions and their properties
25
Example 23
1
2
dxdy = 1
0
1
Figure 17
Here R = {(x, y) : 1 ≤ x ≤ 2, 0 ≤ y ≤ 1}, so R is a square. The iterated
integral represents the area of this square, R.
Now
1
2
1
dxdy =
0
1
(
0
2
1
1
(2 − 1)dy =
dx)dy =
0
1
dy = 1
0
If we reverse the order (we need to change the limits!) of integration:
2 1
2 1
2
1 2
dxdy =
dydx =
(
dy)dx =
(1 − 0)dx = 1
0
1
1
0
1
Example 24
1
x
xy 2 dydx =
I=
0
x2
I
0
1
=
x6
x3
− ]dx
3
3
x[
y 3 y=x
]
2 dx
3 y=x
x[
x6
y3
− ]dx
3
3
1
0
1
=
0
=
=
=
1
40
x[
0
=
1
1 1 4
(x − x7 )dx
3 0
1 1 1
( − )
3 5 8
1
40
Here R = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x} is shown below:
Figure 18
To reverse the order of integration we need to describe R in the ”other” order,
ie. by taking a horizontal fibre.
For a given y we need to find the coordinates of the ”ends” of the fibre.
This gives: ( y , y) and (
y=x
√
y
, y)
√
y=x2 ⇒x= y
26
The ABC’s of Calculus
Therefore:
√
R = {(x, y) : y ≤ x leq y, 0 ≤ y ≤ 1}
So
I
1
√
y
xy 2 dxdy
=
0
y
1
0
1
0
=
Example 25
2π
y
xdx}dy
y
=
=
√
y {
=
=
2
y y2
y 2 ( − )dy
2
2
1 1 3
(y − y 4 )dy
2 0
1 1 1
( − )
2 4 5
1
40
1−cos Θ
ρ3 cos2 ΘdρdΘ =
0
0
49π
32
Here:
R = {(ρ, Θ) : 0 ≤ ρ ≤ 1 − cos Θ, 0 ≤ Θ < 2π}
R can also be described as:
R = {(ρ, Θ) : 0 ≤ Θ ≤ cos−1 (1 − ρ), 0 ≤ ρ ≤ 2}
Figure 19
So:
2
cos−1 (1−ρ)
ρ3 cos2 ΘdΘdρ
0
2π
1−cos Θ
ρ3 cos2 ΘdρdΘ
=
0
0
0
2π
=
cos2 ΘdΘ · [
0
0
=
ρ3 dρ]
0
2π
=
1
4
1−cos Θ
(1 − cos Θ)4
cos2 ΘdΘ
4
2π
(1 − cos Θ)4 cos2 ΘdΘ
0
Example 26 Evaluate using iterated integrals (and interchange) f (x, y) = x
over the region R bounded by x2 and x3 .
Note: x3 < x2 if 0 < x < 1 Describe R in two different ways:
Figure 20
Functions and their properties
27
√
√
A horizontal fibre has the end- coordinates ( y, y), ( 3 y, y), therefore:
√
√
R = {(x, y) : y ≤ x ≤ 3 y, 0 ≤ y ≤ 1}
Therefore:
xdA
1
=
R
=
=
√
3 y
√
0
=
xdxdy
y
1
1 2
(y 3 − y)dy
0 2
y2
1 3 5
[ y 3 − ]y=1
2 5
2 y=0
1
20
Furthermore, a vertical fibre has end coordinates (x, x3 ) and (x, x2 ), So:
R = {(x, y) : 0 ≤ x ≤ 1, x3 ≤ y ≤ x2 }
Therefore
f dA =
R
xdA
R
1
x2
=
xdydx
0
0
x3
1
=
x(
x2
dy)dx
x3
1
x(x2 − x3 )dx
=
0
1
(x3 − x4 )dx
=
0
1 1
−
4 5
1
=
20
1 √
3 y
=
xdxdy as required by Fubini’s theorem
√
=
0
y
Example 27 Evaluate
1dA where R is the first quadrant region bounded
R
by 2y = x2 , y = 3x and x + y = 4
1) Sketch the region Find all points of intersection:
y
y
=
=
3x meets y = 4 − x at:
x = 1 (y = 3)
x2
meets y = 3x at
2
x2
= 3x ⇒ x = 0, x = 6
2
28
The ABC’s of Calculus
y
x2
meets y = 4 − x when:
2
x = −4 and x = 2
=
2) Describe R by vertical or horizontal fibres:
a) vertical Then R = R1 ∪ R2 where:
R1 = {(x, y) : 1 ≤ x ≤ 2, 4 − x ≤ y ≤ 3x}
x2
≤ y ≤ 3x}
2
R2 = {(x, y) : 2 ≤ x ≤ 6,
Figure 21
So:
R
1 · dA
1 · dA
dydx +
=
R1 ∪R2
=
R1
2 3x
=
dydx
R2
6 3x
dydx +
1
4−x
dydx
x2
2
2
2
(3x −
1
2
x2
)dx
2
x3
(x − 1) x=2
x
|x=1 + (3 − )|x=6
2
2
6 x=2
8
= 2 + [3 · 18 − 36 − 6 + ]
6
46
=
3
2
= 4·
Area of R
6
(3x − (4 − x))dx +
=
2
b) Horizontal: Here R = R3 ∪ R4 where:
y
≤ x ≤ 2y}
3
R4 = {(x, y) : 3 ≤ y ≤ 18,
and
R3 = {(x, y) : 2 ≤ y ≤ 3, 4 − y ≤ x ≤
2y}
Figure 22
So area of R :
dA =
R3 ∪R4
(
+
R3
3
√
2y
=
dxdy +
2
Area of R =
2
=
)dA
R4
4−y
3
3
18
√
2y
dxdy
y
3
( 2y − (4 − y))dy +
3
18
y
( 2y − dy
3
√ 2 3
√ 2 3
y2
y2
2
( 2 · y 2 − 4y + )|y=3
y=2 + ( 2 · y −
3
2
3
6
Functions and their properties
29
√
√
9 2 2 √ 3
2 2 3
· 3 2 − 12 + −
( 2) + 8 − 2]
[
3√
2 √3
2 2 32
182
2 2 23
18 −
−
6 + 6]
+[
3
6
3
46
3
=
=
Applications:
1) Finding the area of planar regions R, (as we have).
2) Evaluating complicated integrals by interchanging order of integration.
3) Centroids of planar regions:
If R has area A =
R dA then the relations:
My
=
x=
A
xdA
Mx
, y=
=
A
A
R
ydA
A
R
define the centroid (x, y) of R
4) finding the mass of planar regions having mass density, per unit area given
by δ(x, y). then:
mass R =
δ(x, y)dA
R
Example 28 POLAR:
∞
∞
I=
e
0
−x2 −y 2
dxdy =
0
∞
(e
−y 2
0
∞
0
Figure 23
{R = (x, y) : 0 ≤ x ≤ ∞, 0 ≤ y ≤ ∞}
{R = (x, y) : 0 ≤ y ≤ ∞, 0 ≤ y ≤ ∞}
when x2 + y 2 appearing in integrand use ”polar”.
Why? Because:
x2 + y 2
=
(r cos Θ)2 + (r sin Θ)2
=
=
r2 (cos2 Θ + sin2 Θ)
r2
Then R can be described easily in polar coordinates:
2
e−x dx)dy
30
The ABC’s of Calculus
R = {(r, Θ) : 0 < r < ∞, 0 ≤ Θ ≤
π
2}
But dA polar = rdrdΘ (= dxdy)
So:
R
π
2
=
0
R
f (r cos Θ, r sin Θ) dApolar
textpolar
∞
rdrdΘ
2
e−r rdrdΘ
0
π
2
=
∞
(
0
0
∞
0
=
=
=
=
)dΘ
π2
2
re−r dr) · (
1 · dΘ)
0
=π
2
=
2
re−r dr
not dependant on Θ ! like a constant
= (
=
y
x
f dA =
∞
2
π
·
re−r dr
2 0
π
−du
· eu · (
)
2
2
π −1
·(
eu du)
2
2
π −1 −r2 r=∞
·[
e ]r=0
2
2
−1
π
· [0 − (
)]
2
2
π
−r2 = u, 2rdr = du
4
Example 29 Evaluate:
∞
−∞
∞
e−(x
2
+y 2 )
dxdy
−∞
By changing to polar coordinates.
Here R = {(x, y) : −∞ < x < ∞,
becomes:
−∞ < y < +∞}. In polar coordinates this
R = {(r, φ) : 0 ≤ r < ∞, 0 ≤ φ < 2π}
Furthermore, the elements of area:
dxdy = rdrdφ
Also x2 + y 2 = r2 , thus:
earlier
Functions and their properties
∞
−∞
inf ty
e−(x
2
+y 2 )
31
dxdy
∞
2
e−r rdrdφ
=
−∞
0 0 R in polar
2π
=
(
0
∞
2
re−r dr)dφ
0
∞
2
re−r dr
= 2π
0
R
= 2π lim
R→∞
2
re−r dr
0
2
1 R
= 2π lim (−
(−2r)e−r dr)
R→∞
2 0
2
1
= 2π · ( lim (e−r |r=R
r=0 )
2 R→∞
2
= −π · lim [e−R − 1]
R→∞
=
REMARK Now
∞
∞
e
−∞
−(x2 +y 2 )
π
dxdy
∞
=
−∞
e
−∞
∞
=
(
−x2
dx ·
∞
2
e−y dy
−∞
2
e−x dx)2
−∞
=
∞
therefore
e
−x2
dx
=
π above
√
π
−∞
Plane areas by double integration
Recall the area of a planar region R is given by:
dA = area of R
R
where dA is the element of area in the particular coordinate system used.
Example 30 Find the area of the region R bounded by 3x+ 4y = 24, x = 0, y =
0 (by double integration).
1) sketch the region: Find all points of intersection of the curves making up
its boundary.
Figure 24
Here the points are: (0, 0), (0, 6) and (8, 0).
32
The ABC’s of Calculus
2) Choose a description of R -Either by horizontal or vertical fibres, say,
horizontal.
Figure 25
3) Find the coordinates of the end-points of the fibre. -Do this by finding ”inverse functions”
Figure 26
For fixed y, the set of points on a typical fibre is given by:
4
{(x, y) : 0 ≤ x ≤ 8 − y}
3
4) give a description of R. Thus:
4
R = {(x, y) : 0 ≤ x ≤ 8 − y, 0 ≤ y ≤ 6}
3
5) Set up an iterated integral for evaluating the double integral. Here:
6
8− 43 y
dA =
R
dxdy
0
(in rectangular coordinates)
0
6) Evaluate the iterated integral to find the area
6
4
(8 − y)dy
3
0
2 2 y=6
[(8y − y )]y=0
3
2
48 − · 36 = 48 − 24
3
24
Area of R =
=
=
=
(As a check: Area of ∆ = 12 (base) × (height) = 12 (8) × (6) = 24)
Remark Use of vertical fibres in step(3) gives the iterated integral:
8
Area of R =
6− 34 x
dydx (= 24)
0
0
Example 31 Area within ρ = 2(1 − cos Θ)
A typical ”fibre” here is Θ = constant which gives a ray from O to the boundary
ρ = 2(1 − cos θ)
Figure 27
Functions and their properties
33
So:
R = {(ρ, Θ) : 0 ≤ ρ ≤ 2(1 − cos Θ 0 ≤ Θ < 2π}
So the area:
dA
2π
2−2 cos Θ
=
R
0
element of area in polar coordinates
=
ρdρdΘ
0
2π
(1 − cos Θ)2 dθ
2
0
=
2π
(1 − 2 cos Θ + cos2 Θ)dΘ
2
0
=
2π
2 · 2π + 2
cos2 ΘdΘ
0
=
=
1
4π + 2( · 2π)
2
6π
Example 32 f (x, y) = x over R bounded by y = x2 and y = x3 . Graph the
region (if possible).
NOTE: x3 < x2 if x < 1.
Figure 28
setting x3 = x2 we get either x = 0 or x = 1 so points of intersection of
these curves are (0, 0), (1, 1). Describe R(in some coordinate system), stick with
cartesian coordinates, (x, y)...so:
R = {(x, y) :
√
√
y ≤ x ≤ 3 y, 0 ≤ y ≤ 1}
Figure 29
Therefore using a horizontal slice:
f dA =
x dxdy → notice the order
R
R
1
=
√
y
0
(
y
y3
− )dy
2
2
2
1
0
5
=
=
=
=
xdxdy
3 y
x2 x= √
| √ ]dy
2 x= y
0
=
y
[
1
=
√
3
1 y3
y2
· [ 5 − ]10
2
2
3
1 3 1
·( − )
2 5 2
1 6−5
·(
)
2
10
1
20
34
The ABC’s of Calculus
Vertical Slice:
R = {(x, y) : 0 ≤ x ≤ 1, x3 ≤ y ≤ x2 }
rdA
1
x2
=
R
x3
0
1
=
x2
dy)dx
x3
0
x(
x dydx → notice the order
1
x(x2 − x3 )dx
=
0
1
(x3 − x4 )dx
=
0
x5
x4
− ) |n=1
4
5 n=o
1 1
−
4 5
5−4
20
1
20
=
(
=
=
=
Example 33
R
1
dA
this is the area of R!
R is first quadrant region bounded by 2y = x2 , y = 3x and x + y = 4
1) sketch & find all points of intersection of these curves:
y = 3x meets y = 4 − x at x = 1 or y = 3
2
y = x2 meets y = 3x at x = o and x = 6 or y = 0 and y = 18
2
y = x2 meets y = 4 − x at
x = −4
and x = 2 or y = 8
not in 1st quadrant, so ignore it
and y = 2.
2) Describe R
Either way we need to divide R into two pieces.
a) vertical slice.
R = R1 ∪ R2
Where:
R1 = {(x, y) : 1 ≤ x ≤ 2, 4 − x ≤ y ≤ 3x}
2
R2 = {(x, y) : 2 ≤ x ≤ 6, x2 ≤ y ≤ 3x}
Functions and their properties
35
R
1 · dA =
1 · dA +
=
R2
6 3x
dydx +
1
4−x
(3x −
2
2
(x − 1)dx + (
4·
1
=
=
x2
2
6
(4x − 4)dx +
1
=
2
2
=
=
1 · dA
R1
2 3x
dydx
x2
)dx
2
x3
3x2
− ) |x=6
2
6 x=2
(x − 1)2 x=2
4
|x=1 +[(54 − 46) − (6 − )]
4·
2
3
4
2 + 18 − 6 +
3
46
3
Similarly we can show:
R = R3 ∪ R4
where:
√
R3 = {(x, y) : (4 − y) ≤ x√≤ 2y, 2 ≤ y ≤ 3}
R4 = {(x, y) : ( y3 ) ≤ x ≤ 2y, 3 ≤ y ≤ 18}
Example 34 Evaluate the integral
der of integration.
11 √
1 + y 3 dydx by interchaging the or0
x
Figure 32
f dA
Here
f (x, y) =
1 + y3
R
R = {(x, y) :
√
x ≤ y ≤ 1, 0 ≤ x ≤ 1}
Figure 33
R = {(x, y) : 0 ≤ x ≤ y 2 ,
0 ≤ y ≤ 1}
So by Fubini, using horizontal slices:
0
1
1
√
x
1 + y 3 dydx =
Figure 34
0
1
0
y
2
1 + y 3 dxdy
1 + y3 = u
3y 2 dy = du
y 2 dy = du
3
y=0⇒u=1
y=1⇒u=2
36
The ABC’s of Calculus
1 + y3(
1
=
0
y2
dx)dy
0
1
y2
=
1 + y 3 dy
0
2
=
1
=
1
3
du √
· u
3
2
1
u 2 du
1
3
=
=
=
1 u2 2
·
|
3 32 1
3
2
· u 2 |21
9
√
2
· (2 2 − 1)
9
Example 35
x2 + y 2 + 2x − s
x2 + y 2 = 0 equation in polar?
x = r cos Θ
} ⇒ x2 + y 2 = r2
y = r sin Θ
r2 + 2 · r cos Θ − 2r = 0
Divided by r, (r = 0)
r = 2 · (1 − cos Θ)
r + 2 cos Θ − 2 = 0
Area of R =
R
dA in polar =
rdrdΘ
R
Describe R in polar:
R = {(r, Θ) : 0 ≤ r ≤ 2 · (1 − cos Θ), 0 ≤ Θ < 2π}
2π
2·(1−cos Θ)
rdrdΘ
Area =
0
0
2
2π
=
[
r 2·(1−cos Θ)
]
dΘ
2 0
(
4 · (1 − cos Θ)2
− 0)dΘ
2
0
2π
=
0
=
2π
2·
(1 − cos Θ)2 dΘ
0
Functions and their properties
37
2π
(1 − 2 cos Θ + cos2 Θ)dΘ
=
0
Example 36 Find the centroid of:
3x + 4y = 24, x = 0, y = 0
Centroid
coordinates are
(x, y) where (x =
My =
xdA,
M
=
ydA
x
R
R
My
A ,,
and (y =
Mx
A .
A is area and
Here A = 24 (found earlier) whereas:
My =
8
xdA =
R
0
6− 34 x
xdydx =
0
0
8
3
x(6 − x)dx = 64
4
and
Mx =
6
ydA =
R
8− 43 y
ydxdy =
0
0
4
+06 y(8 − y)dy = 48
3
ie
64 48
8
My Mx
,
) = ( , ) = ( , 2)
A A
24 24
3
Note Either description of R in the expression for the iterated integral could be
used.
(x, y) = (
Example 37 Find the centroid of the first quadrant area bounded by y 2 =
4x, x2 = 5 − 2y, x = 0.
xdA
ydA
R
(x, y) = ( , R
)
dA
dA
R
R
Idea: As above
Figure 36
The points of intersection are: (0, 0), 0, 52 ), and those for which x =
x = 5 − 2y ⇒
2
y4
16
y2
4
and
+ 2y − 5 = 0 or y = 2 ⇒ x = 1 This gives (1, 2).
Take a vertical fibre (horizontal ones lead to the sum of two or more complicated
integrals).
Then:
Figure 37
√
5 − x2
, 0 ≤ x ≤ 1}
R = {(x, y) : 2 x ≤ y ≤
2
38
The ABC’s of Calculus
area, A
=7
dA
R
1
=
√
2 x
0
1
=
0
5−x2
2
dydx
1
5 x2
− 2x 2 }dx
{ −
2
2
1
=
Also
1
xdA =
R
√
2 x
0
xdydx
1
1
5 x2
− 2x 2 )dx
x( −
2
2
1
3
x3
5
− 2x 2 }dx
{ x−
2
2
=
0
5−x2
2
=
0
13
=
24
And
1
yda =
R
√
2 x
0
1
{
=
0
=
=
=
=
=
1
2
1
2
5−x2
2
2
y 2 y= 5−x
|y=2√2x }dx
2
1
{(
0
1
{
0
ydydx
5 − x2 2
) − 4x}dx
2
25 − 10x2 + x4
− 4x}dx
4
1 1
{25 − 16x − 10x2 + x4 }dx
8 0
1 208
(
)
8 15
26
15
Therefore the centroid is:
(x, y) = (
13 26
, )
40 15
Functions and their properties
1.7
39
Volumes under a surface by double integration
We recall that if z = f (x, y) and f (x, y) ≥ 0 over its domain D, a subset of the
xy-plane (or z = 0), the volume of that part of the surface S which lies directly
above the domain D is given by:
zdA =
f (x, y)dA
Volume under S =
D
D
where dA is the element of area on D
Example 38 Find the volume cut from 9x2 +4y 2 +36z = 36 by the plane z = 0.
IDEA
1) Find the projection of this surface on one of the coordinate axes,
Figure 38
Here, say, z = 0. This gives the ellipse:
D = {(x, y) : 9x2 + 4y 2 = 36}
2) solve for z in terms of x and y
Here:
z =1−
y2
x2
−
= f (x, y)
4
9
Note that for z on D, z = 0 while for z in d, z ≥ 0
3) So volume under S:
(1 −
D
y2
x2
− )dA
4
9
4) need to describe D: We need to introduce elliptical (plane) coordinates centered at (0, 0)
x
y
x = ar cos Θ
; r2 = ( )2 + ( )2
a
b
y = br sin Θ
Figure 39
The Jacobian is: abr
Here a = 2, b = 3
Therefore 1 −
x2
4
−
y2
9
= 1 − r2
40
The ABC’s of Calculus
5) Volume
y2
x2
− )dA
(1 −
4
9
D
2π
1
(1 − r2 ) (1 · b · r) drdθ
=
0
0
2π
Jacobian
1
(1 − r2 ) · 6rdrdΘ
=
0
=
0
1
(1 − r2 )rdr
12π
0
1 1
12π · ( − )
2 4
3π
=
=
Remark: S can be ”reconstructed” via its ”projections” eg.
S = {(x, y, z) : 9x2 + 4y 2 + 36z = 36}
Projection of S onto:
xy-plane is:(set z = 0) ⇒ 4x2 + 9y 2 = 36
xz-plane is:(set y = 0) ⇒ 9x2 + 36z = 36
yz-plane is:(set x = 0) ⇒ 4y 2 + 36z = 36
Figure 40
The surface S is reconstructed (basically) by superimposing the three projections.
the more projections, the better the schematic representation of S.
Figure 41
Example 39 Find the volume in the first octant bounded by xy = 4z, y = x
and x = 4.
Here z =
xy
4
Volume =
= f (x, y) ≥ 0 for (x, y) in D.
D
zdA because x > 0 over D.
(
V ol =
D
xy
)dA
4
Stick with cartesian coordinates, because z = ( xy
4 ) has an ”interesting form”
and because D is a ”triangle” so:
D = {(x, y) : 0 ≤ y ≤ x,
0 ≤ x ≤ 4}
Functions and their properties
41
or
D = {(x, y) : y ≤ x ≤ 4, 0 ≤ y ≤ 4}
Using Vertical Slices:
vol
=
=
=
=
=
=
=
4
x
xy
dydx
4
0
0
x
1 4
x·(
ydy)dx
4 0
0
x2
1 4
x · [ − 0]dx
4 0
2
4
1
x3 dx
8 0
1 x4 4
[ ]
8 4 0
1 3
(4 )
8
8
Example 40 D is found by projection of S on xy-plane, z = 0.
The domain of f (x, y) =
xy
4
is clearly all (x, y) on the xy-plane.
We need to restrict D to that part of xy-plane which is bounded by:
y = x (projection of plane y = x on z = 0)
x = 4 (projection of plane x = 4 on z = 0)
y = 0 (part of projection of z = xy on z = 0)
Figure 45
Thus volume under S is:
(
D
xy
)dA
4
Figure 46
Now:
D = {(x, y) : 0 ≤ y ≤ x,
So the volume under S is:
0 ≤ x ≤ 4}
42
The ABC’s of Calculus
4
x
(
0
0
xy
)dydx
4
=
=
=
x
1 4
{
xydy}dx
4 0
0
... ...
1 4 3
x dx
8 0
=
8
or, using horizontal slices,
vol
=
=
=
=
=
=
4
4
xy
dxdy
4
0
y
4
1
y(
xdx)dy
4
y
1 4
y(16 − y 2 )dy
8 0
1
[128 − 64]
8
64
8
8
Example 41 Find the volume common to the cylinders x2 + y 2 = 16 and x2 +
z 2 = 16.
Idea: We look at the cylinder x2 +z 2 = 16 and consider the projection of x2 +y 2 =
16 on the xy-plane or z = 0.
Figure 47
the projection of x2 + y 2 = 16 on z = 0 is the circle x2 + y 2 = 16. The projection
of x2 + z 2 = 16 on z = 0 is the region bounded by the parallel lines x = ±4. So,
the projection of the solid of intersection is simply:
D
=
=
{(x, y) : 0 ≤ x2 + y 2 ≤ 16}
{(x, y) : − 16 − x2 ≤ y ≤ 16 − x2 , −4 ≤ x ≤ +4}
Hence the volume of the solid:
=2
16 − x2 dA
D
z
Figure 48
Functions and their properties
43
√
because there is a symmetric part of the solid below z = 0, (as z = ± 16 − x2 )
Thus:
16 − x2 dA =
2
2
− 16−x2
( 16 − x2 dydx
z
4
2
−4
4
=
√
+ 16−x2
√
−4
D
=
4
4
−4
4
2(16 − x2 )dx
(16 − x2 )dx
=
(16 − x2 )dx (because integrand is an even function)
8
0
=
8(16x −
=
8·
x3 4
)|
3 0
128
3
1024
3
=
Example 42 Find the volume inside ρ = 2 and outside the cone z 2 = ρ2 .(cylindrical
coordinates)
Figure 49
The projection of the solid on z = 0 is given by the inside of the circle ρ = 2.
There are four identical parts making up the required volume and so it suffices
to find the volume of that part of the solid which lies above the semi-circle D
shown in the figure.
Next,
volume of solid
zdA
=
D
=
4
(z 2 = ρ2 ⇒ z = ρ above D)
ρdA
D
Recall z 2 = ρ2 = x2 + y 2 (in cylindrical coordinates). since D is on the xy-plane
z = 0, it is best to pick a description of D in polar coordinates. Therefore:
D = {(ρ, Θ) : 0 ≤ ρ ≤ 2, 0 ≤ Θ ≤
π }
note this!
Therfore:
V olume =
ρ(ρdρdθ)
4
D
44
The ABC’s of Calculus
=
π
4
(
0
=
2
ρ2 dρ)dΘ
0
π
4
0
8
dΘ
3
32π
3
=
Example 43 Find the volume common to r2 + z 2 = a2 and r = a sin Θ (cylindrical coordinates!).
Note r2 +z 2 = a2 is a sphere centered at (0, 0, 0) of radius a (since r2 = x2 +y 2 ).
x2 + y 2 = a √
Furthermore, r = a sin
Θ ⇔
y
x2 +y 2
=a y
r
(polar coordinates). Or,
x2 + y 2 = ay, But x2 + y 2 − ay = 0 ⇒ x2 + (y − a2 )2 =
square).
a2
4
(completing the
Therefore:
a
a2
r = a sin Θ ⇔ (x − 0)2 + (y − )2 =
2
4
And this is a cylinder centered at (0, a2 of radius
a
2.
Figure 50
The resulting volume is shown in the diagram.
Therefore the volume of the solid:
=4
zdA
D
where D = {(r, Θ) : 0 ≤ r ≤ a sin Θ, 0 ≤ θ ≤ π2 } centered at (0, a2 ) in polar
coordinates. Note that θ ≤ π2 and not Θ ≤ 2π (Why?? else r < 0!).
Or
π
2
a sin Θ
V olume = 4
a2 − r2 (rdrdΘ)
0
0
(Why? because volume = 4 × (volume of a half-circular region))
π
2
=
4
=
4
(−2)
0
1
r(a2 − r2 ) 2 dr}dΘ
0
π
2
(−2)
0
π
2
a sin Θ
{
{
1
(−2r)(a2 − r2 ) 2 dr }dΘ
0
0
=
a sin Θ
{
{u=a2 −r 2 ; du=−2rdr}
2
2
a cos Θ
1
u 2 du}dΘ
a2
Functions and their properties
=
=
=
=
=
=
=
45
r = 0 → u = a2 r = a sin Θ → u = a2 cos2 Θ
π2
2 3 2 2
{ u 2 |aa2 cos Θ }dΘ
(−2)
3
0
π2
4(−1)
{a3 cos3 Θ − a3 }dΘ
3
0
π
−4a3 2
(1 − sin2 Θ) cos ΘdΘ
3
0
π2
−4a3 4a3
a3 π
+
sin2 Θ cos ΘdΘ + 2
3
3 0
3
3
3
3
3
π
4a
sin Θ 2
2a π
−4a
+
·[
| ]+
3
3
3 0
3
2a3 π
−4a3 4a3
+
+
3
9
3
2a3
(3π − 4)
9
Example 44 Find the volume cut from paraboloid 4x2 + y 2 = 4z by the plane
z − y = 2.
2
Note that paraboloid is z = x2 + y4 and plane z = y + 2. Therefore the curve of
intersection of the two surfaces is (equates z’s):
y + 2 = x2 +
y2
4
Collecting terms we obtain (after completing the square):
4x2 + (y − 2)2 − 12 = 0
Or:
x2
(y − 2)2
: √
+ √
=1
( 3)2
( 12)2
Figure 51
Which is an ellipse. So, the projection of this solid of intersection on the plane
z = 0 is given by This ellipse, whose center is at (0, 2).
The paraboloid is basically constructed by its projections:
xz-plane (y = 0): z = x2
2
yz-plane (x = 0): z = y4
Figure 52
To compute the volume of this solid we compute the height of a typical vertical
fibre which intersects the solid.
46
The ABC’s of Calculus
2
In our case (height of fibre) = (y + 2) − (x2 + y4 ) (=difference in z- coordinates)
So a volume element = ((y + 2) − ((x2 +
y2
))dA
4
Where dA = element of area on the projection D of the solid on z = 0,(or the
interior of ). Therefore:
y2
V olume =
((y + 2) − ((x2 + ))dA
4
D
We introduce elliptical coordinates centered at (0, 2)(Why? because our D is
centered there!).
y−2
x
: ( √ )2 + ( √ )2 = 1
3
12
So, let:
x =
y =
√
3r cos Θ
√
2 + 12r sin Θ
and the
Jacobian =
√ √
3 · 12r
Figure 53
√ )2 = r2 the basic equation of our occurs when r = 1)
(then ( √x3 )2 + ( y−2
12
So in our coordinate system
D = {(r, Θ) : 0 ≤ r ≤ 1, 0 ≤ Θ ≤ 2π}
and thus:
V olume− =
y2
((y+2)−(x + ))dA =
4
D
(Also y + 2 − x2 −
2π
1
2
3(1−r2 )·
0
y2
4
0
√ √
3 12rdrdΘ
area element in new coord. syst.
= . . . = 3(1 − r2 ) as defined above).
2π 1
√ √
3(1 − r2 ) · 6rdrdΘ ( 3 12 = 6)
V olume =
0
0
1
(1 − r2 ) · rdr
= 36π
0
2
r4
r
− )|r=1
2
4 r=0
1 1
= 36π( − )
2 4
1
= 36π ·
4
= 9π
= 36π(
Functions and their properties
47
0 2
Remark: for elliptical coordinates centered at (x0 , y0 ) based on the ellipse ( x−x
a ) +
y−y0 2
( b ) =1:
Figure 54
we set
x
y
Jacobian
= x0 + ar cos Θ
= y0 + br sin Θ
= abr
(as before) etc.