Line Integrals of Second Kind [608206]

CHAPTER 12
Line Integrals of Second Kind
12.1. Orientated curves
Recall that a curve means the class of all equivalent paths, in the sense of
Denition 10.6. The orientation on a curve is closely related to the monotoni-
city of the intermediate function φ;appearing in Denition 10.6. If we want
to dene the notion of an orientated curve, we will divide the class of all the
equivalent paths that belong to the curve into two subclasses: the paths for
which the equivalence relation is established through an increasing function
φ, and the paths for which the equivalence relation is established through a
decreasing function φ. Next we have to decide which of these two subclasses
represent the direct orientation (sense) and which represents the converse one.
Definition 12.1.Let
: [a;b]!R3be a path. When the parameter tgrows
fromatob, the point
(t)runs the image of the path, (
);in a sense called
the positive (direct )sense . Whentdecreases from btoa, the point
(t)
runs the image of the path, (
);in a sense called the negative (converse )
sense .
Definition 12.2.Anorientated path is every path for which a sense of
running is established.
Anorientated curve is every curve for which one species the sense of
running on a path that belongs to it.
So, a simple and smooth curve is considered to be positively (directly)
orientated , iff the sense of a parameterization that denes it is that of the
growing of the parameter , and negatively (conversely )orientated , by
contrary.
Thepositive (direct )sense on a closed, simple, and smooth path in R3
is the counterclockwise one.
A closed, simple, and smooth curve in R3is positively (directly) orientated
if a parameterization that denes it is counterclockwise traced. More speci-
cally, the contours in R3arepositively (directly )orientated iff the interior
of the domain remains on the left side, when running the contour.
237

238 12. LINE INTEGRALS OF SECOND KIND
12.2. Integrable functions with respect to the coordinates
Let
: [a;b]!R3be a smooth and orientated path, having the parametric
equations
8
<
:x=f(t)
y=g(t)
z=h(t); t2[a;b];
(
) be its image, and !r(t) be the position vector of the current point M(x;y;z )
from (
);i.e.
!r(t) =f(t) !i+g(t) !j+h(t) !k:
Let⃗V= (V1;V2;V3) : (
)!R3be a vector function. Let
(∆;) = (t0;t1;:::;tn;1;:::;n)
be a tagged division to [ a;b]. The real number
∆(
⃗V;)
: =n∑
i=1(
⃗V◦
)
(i) ( !ri !ri1)
=n∑
i=1[V1(
(i)) (f(ti)f(ti1))
+V2(
(i)) (g(ti)g(ti1))
+V3(
(i)) (h(ti)h(ti1))]
is called the Riemann sum associated to the vector function ⃗Vand the tagged
division (∆ ;).
Definition 12.3. !Vis called integrable with respect to the coordinates
along the path
iff there exists I2R, such that 8">0;9=(")>0;for
every tagged division (∆;) = (t0;t1;:::;tn;1;:::;n)to[a;b], with ∥∆∥< ;
one has∆( !V ;)
I<":
IfIdoes exists like in this Denition, then it is unique. We call it the
line integral with respect to the coordinates to the vector function !V
(orthe line integral of second kind ) along the path
and we denote it by
I=:∫

!V d !r=:∫

V1(x;y;z )dx+V2(x;y;z )dy+V3(x;y;z )dz;
where !r=x !i+y !j+z !kis the position vector of the current point belonging
to (
):

12.3. PROPERTIES OF THE LINE INTEGRALS OF SECOND KIND 239
In the case of a closed path, one prefers the notation
I=:I

!V d !r :
Theorem 12.1.Let
: [a;b]!R3be a smooth and positively orientated path
of parametric equations
8
<
:x=f(t)
y=g(t)
z=h(t); t2[a;b];
and !V= (V1;V2;V3) : (
)!R3be a continuous vector function. Then !Vis
integrable with respect to the coordinates along the path
and
∫

!V d !r=∫b
a[V1(f(t);g(t);h(t))f′(t)
+V2(f(t);g(t);h(t))g′(t)
+V3(f(t);g(t);h(t))h′(t)]dt:
We can easily deduce that if !Vis integrable with respect to the coordinates
along a path
;then it will be integrable with respect to the coordinates along
every path that is equivalent to
:We dene the line integral on a simple,
smooth, and orientated curve to be the line integral on a path that belongs to
it.
12.3. Properties of the line integrals of second kind
1. (Dependence on the orientation). Let
be a smooth and orientated
path and !V: (
)!R3be a vector function that is integrable with respect
to the coordinates along the path
. Then !Vis integrable with respect to the
coordinates along the path
;the opposite of the path
;and
∫

 !V d !r=∫

!V d !r :
So, we easily deduce that the line integral of a continuous vector function
on a smooth curve does not depend on the parameterization, up to the sign,
that is determined by the orientation.
2. (Linearity). Let
be a smooth, and orientated path and !V1; !V2: (
)!
R3be two vector functions that are integrable with respect to the coordinates
along the path
, and; 2R. Then !V1+ !V2is integrable with respect to
the coordinates along the path
and
∫

(
 !V1+ !V2)
d !r=∫

!V1d !r+∫

!V2d !r :

240 12. LINE INTEGRALS OF SECOND KIND
3. (Additivity with respect to the curve). Let
1and
2be two smooth
and orientated curves and !V: (
)!R3be a vector function that is integrable
with respect to the coordinates along the path
1and
2. Then !Vis integrable
with respect to the coordinates along the path
1[
2and∫

1[
2 !V d !r=∫

1 !V d !r+∫

2 !V d !r :
Remark 12.1.The work done by the force !Vthat acts on a material point,
displacing it along the image of the path
;is
W=∫

!V d !r :
Remark 12.2.We can easily adapt Denitions 12.1-12.3 to the case of paths

inR2and vector functions !V= (V1;V2) : (
)!R2:
We denote
I=:∫

!V d !r=:∫

V1(x;y)dx+V2(x;y)dy:
where !r=x !i+y !jis the position vector of the current point belonging to
(
):And we can state a similiar result for this case.
Theorem 12.2.Let
: [a;b]!R2be a simple, smooth, and positively orien-
tated path of parametric equations{
x=f(t)
y=g(t); t2[a;b];
and !V= (V1;V2) : (
)!R3be a continuous vector function. Then !Vis
integrable with respect to the coordinates along the path
and
∫

!V d !r=∫b
a[V1(f(t);g(t))f′(t) +V2(f(t);g(t))g′(t)]dt:
12.4. Independence on the path of the line integral of second kind
We are interested to see weither the line inegral does not depend on the
path but only on the endpoints of the path. As we will remark, this question
is closely related to the work of a force and it will be reseached by using the
notion of conservative vector eld.
Definition 12.4.LetDR3be arbitrary. A vector eld of components V1;
V2; V3is a vector function !V:D!R3,
!V(x;y;z ) = (V1(x;y;z );V2(x;y;z );V3(x;y;z )):

12.4. INDEPENDENCE ON THE PATH OF THE LINE INTEGRAL OF SECOND KIND 241
To dene a vector eld on a set Dmeans to associate to each point (x;y;z )2D
a vector
!V(x;y;z ) =V1(x;y;z ) !i+V2(x;y;z ) !j+V3(x;y;z ) !k:
Similarly we can introduce the notion of vector eld in R2.
Definition 12.5.A vector eld !Vof classC1is called potential vector
eld orgradient vector eld on the setDR3iff there exist a scalar eld
U:D!R, such that
!V=gradU;
i.e.
!V(x;y;z ) =@U
@x(x;y;z ) !i+@U
@y(x;y;z ) !j+@U
@z(x;y;z ) !k;
for all (x;y;z )2D: U is called potential scalar eld of the vector eld !V
onD.
Remark 12.3.Every gradient vector eld admits an innitely many potential
scalar elds, all being equal up to a real constant.
Theorem 12.3.(Leibniz-Newton Formula for the Line Integral of Gradient
Vector Fields). Let !Vbe a gradient vector eld on the set DR3(orR2)
andU:D!Rbe a potential scalar eld of !VonD:Let
: [a;b]!Dbe a
smooth and positively orientated path. Then∫

!V d !r=U(
(b))U(
(a)) =:Uj
(b)

(a)
(the last equality represents a notation).
Theorem 12.4.LetDR3(orR2)be an open and connected set and !Va
vector eld of class C1(D):Then the following assertions are equivalent:
a) !Vis a gradient vector eld on D;
b) for every
: [a;b]!Da smooth, closed, and orientated path, we have∫

!V d !r= 0;
c) for every two smooth and orientated paths
1;
2: [a;b]!D, with

1(a) =
2(a);
1(b) =
2(b);we have∫

1 !V d !r=∫

2 !V d !r :

242 12. LINE INTEGRALS OF SECOND KIND
Definition 12.6.A smooth, closed, and orientated path is called contour and
the line integralI

!V d !r
is also called the circulation of the vector eld !Valong the path
:
So, Theorem 12.4 can be stated as follows.
Theorem 12.5.LetDR3(orR2)be an open and connected set and !Va
vector eld of class C1(D):Then the following assertions are equivalent:
a) !Vis a gradient vector eld on D;
b) the circulation of !Valong every contour in Dis zero;
c)∫

!V d !ris independent on the path
;it only depends on the endpoints
of the path.
Theorem 12.6.LetDR3be an open and connected set, !V= (V1;V2;V3) :
D!R3be a gradient vector eld,
!V(x;y;z ) =V1(x;y;z ) !i+V2(x;y;z ) !j+V3(x;y;z ) !k;
and(x0;y0;z0)2Dbe xed. Then U:D!R,
U(x;y;z ) =∫x
x0V1(t;y0;z0)dt+∫y
y0V2(x;t;z 0)dt+∫z
z0V3(x;y;t )dt
is a potential scalar eld to !VonD:
Remark 12.4.By choosing another point (x0;y0;z0)2Dwe get another po-
tential scalar eld, that is different by the rst one through an additive scalar
constant.
A similar result can be deduced in the case of vector elds !V= (V1;V2) :
D!R2:
Theorem 12.7.LetDR2be an open and connected set, !V= (V1;V2) :
D!R2be a gradient vector eld,
!V(x;y) =V1(x;y) !i+V2(x;y) !j ;
and(x0;y0)2Dbe xed. Then U:D!R,
U(x;y) =∫x
x0V1(t;y0)dt+∫y
y0V2(x;t)dt

12.4. INDEPENDENCE ON THE PATH OF THE LINE INTEGRAL OF SECOND KIND 243
is a potential scalar eld to !VonD:
Definition 12.7.A vector eld !V:DR3!R3of classC1(D)is called
conservative vector eld onDiff
curl( !V)
= !0;onD;
i.e.
⃗i ⃗j ⃗k
@
@x@
@y@
@z
V1(x;y;z )V2(x;y;z )V3(x;y;z )=⃗0;8(x;y;z )2D
or, equivalently,
@V2
@x(x;y;z ) =@V1
@y(x;y;z )
@V3
@y(x;y;z ) =@V2
@z(x;y;z )
@V1
@z(x;y;z ) =@V3
@x(x;y;z );8(x;y;z )2D
Remark 12.5.A vector eld !V= (V1;V2) :DR2!R2(V3= 0) of class
C1(D)is conservative on Dif and only if
@V2
@x(x;y) =@V1
@y(x;y);8(x;y)2D:
Remark 12.6.IfU:D!Ris a scalar eld of class C2(D), then
curl (gradU) = !0:
Hence, each gradient vector eld is a conservative vector eld on D. The
converse of this assertion does not hold true. Indeed, if one considers the
vector eld !V:D!R2,D=R2nf(0;0)g;
!V(x;y) =y
x2+y2 !i+x
x2+y2 !j ;
then !Vis conservative on D, since
@(
x
x2+y2)
@x=@(
y
x2+y2)
@y=y2x2
(x2+y2)2;8(x;y)2D;
and it is not a gradient vector eld on D, since on the circle
:x2+y2= 1;
that is traced once counterclockwise,
I

y
x2+y2dx+x
x2+y2dy= 2̸= 0:

244 12. LINE INTEGRALS OF SECOND KIND
Definition 12.8.A setDR3(
R2)
is called star-shaped iff there exists
a point (x0;y0;z0)2D, such that for all (x;y;z )2D, the line segment of
endpoints (x0;y0;z0)and(x;y;z )is enclosed into D:
Theorem 12.8.LetDR3be an open and star-shaped set and !V:D!R3
be a vector eld of class C1(D). Then !Vis gradient vector eld on Dif and
only if !Vis conservative vector eld on D:
So, since every star-shaped set is connected, we easily deduce the following
results.
Theorem 12.9.LetDR3be an open and star-shaped set and !V= (V1;V2;V3)
be a vector eld of class C1(D):Then the following assertions are equivalent:
a) !Vis a gradient vector eld on D;
b) the circulation of !Valong every contour in Dis zero;
c)∫

!V d !ris independent on the path
;it only depends on the endpoints
of the path;
d) !Vis conservative vector eld on D;
e) The following equalities hold
@V2
@x(x;y;z ) =@V1
@y(x;y;z )
@V3
@y(x;y;z ) =@V2
@z(x;y;z )
@V1
@z(x;y;z ) =@V3
@x(x;y;z );8(x;y;z )2D:
A similar result can be easily deduced in the case of vector elds !V=
(V1;V2) :D!R2:
Theorem 12.10.LetDR2be an open and star-shaped set and !V= (V1;V2)
be a vector eld of class C1(D):Then the following assertions are equivalent:
a) !Vis a gradient vector eld on D;
b) the circulation of !Valong every contour in Dis zero;
c)∫

!V d !ris independent on the path
;it only depends on the endpoints
of the path;
d) !Vis conservative vector eld on D;
e) The following equality holds
@V2
@x(x;y) =@V1
@y(x;y);8(x;y)2D:
Remark 12.7.The most important conservative forces, for which the work is
independent on the path, are:

12.4. INDEPENDENCE ON THE PATH OF THE LINE INTEGRAL OF SECOND KIND 245
1. The weight force: !F=mg !k;
2. The force of attraction: !F=c
r3 !r(c>0is constant) ;
3. The elastic force: !F=c !r(c>0is constant),
where !r=x !i+y !j+z !kis the position vector of the curent point
M(x;y;z )andr=∥ !r∥=√
x2+y2+z2is its length :
Let us consider some examples.
1.Calculate∫

y2dx+x2dy;where
is the upper semi-ellipsex2
a2+y2
b2= 1;
that is traced once clockwise.
We have the parametric equations

:{
x=acost
y=bsint; t2[0;]
and
I=∫0
[
b2sin2t
(asint) +a2cos2t
(bcost)]
dt
=ab2∫0
sin3tdt+a2b∫0
cos3tdt=4
3ab2:
2.Calculate∫
C(yz)dx+(zx)dy+(xy)dz;where
is the rst loop
of the spiral
C:8
<
:x=acost
y=asint
z=bt; t2[0;2];
that is traced once positively.
We have
I=∫2
0[(asintbt) (asint) + (btcost) (acost)
+ (acostasint)b]dt
=2aba2a:
3.Prove that !V:R2!R2, !V(x;y) = (4x+ 2y;2x6y) is a conserva-
tive vector eld and determine a scalar potential of it.
SinceR2is an open and star-shaped set, V1(x;y) = 4x+ 2y; V 2(x;y) =
2x6y,
@V2
@x=@V1
@y= 2;

246 12. LINE INTEGRALS OF SECOND KIND
it follows that !Vis a conservative vector eld. A scalar potential eld is
U(x;y) =∫x
0P(t;0)dt+∫y
0Q(x;t)dt
= 2×2+ 2xy3y2:
12.5. The Green-Riemann Theorem
Theorem 12.11.(Green-Riemann Theorem). Let DR2be a compact do-
main that is simple with respect to the axis Oxand to the axis Oy; and
its boundary, @Dis a simple, closed, and piecewise smooth curve. If !V=
(V1;V2) :D!R2is a vector eld of class C1on an open set that includes D,
then,
∫∫
D[@V2
@x(x;y)@V1
@y(x;y)]
dxdy =I
@DV1(x;y)dx+V2(x;y)dy;
where@Dis traced once counterclockwise.
For example, let us calculate, by using the Green-Riemann Theorem, the
line integral of second kind,
I=I
Cx2ydx+xy2dy;
where
is the circle x2+y2=R2;that is traced once counterclockwise.
We haveV1(x;y) =x2y; V 2(x;y) =xy2and
@V2
@x=y2;@V1
@y=x2:
Then
I=∫∫
D(@V2
@x@V1
@y)
dxdy
=∫∫
D(
y2(
x2))
dxdy
=∫∫
D(
x2+y2)
dxdy:
By passing to polar coordinates, we obtain
I=∫2
0dφ∫R
03d=R4
2:
Remark 12.8.1) If@V2
@x=@V1
@yinD;then
I
@DV1(x;y)dx+V2(x;y)dy= 0:

12.6. EXERCISES 247
2) IfDR2is a compact domain that is simple with respect to the axis Ox
and to the axis Oy;then
Area (D) =1
2∫
@Dydx+xdy=∫
@Dxdy=∫
@Dydx:
12.6. Exercises
(1)Calculate
∫

(
x22xy)
dx+(
2xy+y2)
dy;
where
is the arc of the parabola y=x2that connects the startpoint
A(1;1) to the endpoint B(2;4):
A:We have the parametric equations
{
x=t
y=t2; t2[1;2]
and
I=∫2
1[((
t22t3)

1)
+(
2t3+t4)

(2t)]
dt
=∫2
1(
t22t3+ 4t4+ 2t5)
dt
=1219
30:
(2)Calculate
∫

(2ay)dx+xdy;
where
is the arc of the cycloid
{
x=a(tsint)
y=a(1cost); t2[0;2];
that is traced once in the sense of the growing of the parameter t.
A:2a2:
(3)Calculate
I

(x+y)dx(xy)dy
x2+y2;
where
is the circle x2+y2=a2;that is traced once counterclockwise.
A:By using the parametric equations,
{
x=acost
y=asint; t2[0;2]:

248 12. LINE INTEGRALS OF SECOND KIND
We obtain
I=∫2
0(acost+asint) (asint)(acostasint) (acost)
a2dt
=∫2
0(
sintcostsin2tcos2t+ sintcost)
dt
=2:
(4)Calculate∫

cosydxsinxdy;
where
is the line segment situated on the second bisectrix of the
axes of coordinates, y=xand it is traced from the point Ahaving
the abscissa 2, to the point Bof the ordinate 2 :
A:2 sin 2:
(5)Calculate
I

xy(ydxxdy)
x2+y2;
where
is the right loop of the Bernoulli's lemniscate
(
x2+y2)2=a2(
x2y2)
that is traced once counterclockwise.
A:0:
(6)Calculate∫

2xydxx2dy;
where
connects the startpoint O(0;0) to the endpoint A(2;1) as
follows (see Figure 12.1):
Figure 12.1
a) along the line OmA ;

12.6. EXERCISES 249
b) along the parabola OnA; whose axis of symmetry is Oy;
c) along the parabola OpA, whose axis of symmetry is Ox;
d) along the broken line OBA ;
e) along the broken line OCA:
A:a)4
3; b) 0; c)12
5; d)4; e) 4:
(7)Under the hypotheses of Exercise 6, calculate
∫

2xydx +x2dy:
A:4 for all the cases a)-e).
(8)Calculate,∫

y2dx+x2dy, where
is the upper semi-ellipse
{
x=acost;
y=bsint;
that is traced counterclockwise.
A:4ab2
3:
(9)Calculate
∫

xdx+ydy√
1 +x2+y2;
where
is the arc of the ellipse
x2
a2+y2
b2= 1
that is situated within the rst quarter of the plane and is traced
clockwise.
A:p
1 +a2p
1 +b2:
(10) Calculate
∫

(yz)dx+ (zx)dy+ (xy)dz;
where
is the rst arc of the spiral
8
<
:x=acost
y=asint
z=ct; t2[0;2];
that is traced is the sense of the growing of the parameter t:
A:2a(a+b):
(11) Calculate
∫

ydx+zdy+xdz;

250 12. LINE INTEGRALS OF SECOND KIND
where
is the circle
8
<
:x=Rcosacost
y=Rcosasint
z=Rsina; t2[0;2];
(ais constant) that is traced once is the sense of the growing of the
parametert
A:R2cos2a:
(12) Calculate
∫

xydx +yzdy +zxdz;
where
is arc of the curve
{
x2+y2+z2= 2Rx;
z=x;
that is traced once positively.
A:(
1
6+p
2
16)
R3:
(13) Prove that the following vector elds are conservative and determine
a scalar potential eld:
a)⃗F(x;y) = (2x+ 3y)⃗i+ (3x4y)⃗j;
b)⃗F(x;y) =(
3×22xy+y2)⃗i(
x2+ 2xy+ 3y2)⃗j;
c)⃗F(x;y) =exy(1 +x+y)⃗i+exy(1xy)⃗j;
d)⃗F(x;y) =1
x+y⃗i+1
x+y⃗j:
A:One proves that@V2
@x=@V1
@yand one deduces the following
scalar potential elds: a) x2+ 3xy2y2; b)x3x2y+xy2y3;
c)exy(x+y) ; d) ln jx+yj:
(14) Calculate the following line integrals of second kind:
a)
∫

xdy+ydx;
where
is a curve that connects the startpoint ( 1;2) to the endpoint
(2;3) ;
b)
∫

xdx+ydy;
where
is a curve that connects the startpoint (0 ;1) to the endpoint
(3;4) ;

12.6. EXERCISES 251
c)∫

(x+y)dx+ (x+y)dy;
where
is a curve that connects the startpoint (0 ;0) to the endpoint
(1;1) ;
d)∫

ydxxdy
y2;
where
is a curve that connects the startpoint (1 ;2) to the endpoint
(2;1) and does not intersects the axis Ox;
e)∫

dx+dy
x+y;
where
is a curve that connects the startpoint(1
2;1
2)
to the endpoint
(x;y) and does not intersects the line x+y= 0;
f)∫

(
x4+ 4xy3)
dx+(
6x2y25y4)
dy;
where
is a curve that connects the startpoint ( 2;1) to the end-
point (3;0):
A:a) 8; b) 12; c) 2; d)3
2; e) ln (x+y) ; f) 62:
(15) Calculate the following line integrals of second kind:
a)∫

xdx+ydyzdz;
where
is a curve that connects the startpoint (1 ;0;3) to the end-
point (6;4;8) ;
b)∫

yzdx +zxdy +xydz;
where
is a curve that connects the startpoint (1 ;1;1) to the endpoint
(a;b;c ) ;
c)∫

xdx+ydy+zdz√
x2+y2+z2;
where
is a curve that connects the startpoint (0 ;0;0) to the endpoint
(3;4;5) ;

252 12. LINE INTEGRALS OF SECOND KIND
d)
∫

yzdx +zxdy +xydz
xyz;
where
is a curve that connects the startpoint (1 ;1;1) to the endpoint(
x;y;1
xy)
and it is situated in the positive dihedron x > 0; y > 0;
z>0:
A:a)20; b)abc1; c) 5p
2; d) 0:
(16) Transform, by applying the Green-Riemann Theorem, the line integral
of second kind
I=I

√
x2+y2dx+y[
xy+ ln(
x+√
x2+y2)]
dy;
where
is the contour that bounds a compact and simple with respect
to each axis of coordinates domain DR2:
A:We have
V1(x;y) =√
x2+y2;
V2(x;y) =y[
xy+ ln(
x+√
x2+y2)]
and
@V2
@x=(
y√
x2+y2+ 1)y√
x2+y2;
@V1
@y=1√
x2+y2y:
So, by applying the Green-Riemann Theorem, we obtain
I=∫∫
D[(
y√
x2+y2+ 1)y√
x2+y21√
x2+y2y]
dxdy
=∫∫
Dy2dxdy:
(17) Calculate, by applying the Green-Riemann Theorem, the line integral
of second kind
I=I

2(
x2+y2)
dx+ (x+y)2dy;
where
is the contour of the triangle of vertices A(1;1),B(2;2);
C(1;3), that is traced once positively.
A:4
3:

12.6. EXERCISES 253
(18) Calculate, by applying the Green-Riemann Theorem, the line integral
of second kind
I=I

x2ydx+xy2dy;
where
is the circle x2+y2=R2that is traced once counterclockwise.
A:R4
2:
(19) Calculate directly and by applying the Green-Riemann Theorem, the
line integral of second kind
I=I
AmBnA(x+y)dx(xy)dy;
whereA(1;0); B(2;3); AnB is the segment line and AmB is the
parabola having as axis of symmetry Oy(see Figure 13.1).
Figure 12.2
A:4
3:
(20) Determine the area of the domains that are bounded by the following
curves:
a) the ellipse
{
x=acost
y=bsint; t2[0;2] ;
b) the astroid
{x=acos3t
y=asin3t; t2[0;2] ;
c) the cardioid
{
x=a(2 costcos 2t)
y=a(2 sintsin 2t); t2[0;2] ;
d) the loop of folium of Descartes x3+y3= 3axy; a> 0:

254 12. LINE INTEGRALS OF SECOND KIND
A:We can apply one of the following formulas that gives us the
area:
Area (D) =1
2∫
@Dydx+xdy=∫
@Dxdy=∫
@Dydx:
We deduce: a) ab; b)3a2
8; c) 6a2; d)3a2
2:Hint: By setting y=tx;
the parametric equations to the loop of the folium of Descartes are{
x=3at
1+t3
y=3at2
1+t3; t2[0;1):