Unit 3

UNIT 3

FURTHER APPLICATIONS OF INTEGRAL CALCULUS

Structure 3.1

Introduction Objectives

Length of a Plane Curve

32

3.2.1 Cartesian Form 3.2.2 Parametric Form 3.2.3 Polar Form

3.3 3.4 3.5 3.6

Volume of a Solid of Revolution Area of Surface ef Revolution Summary

Solutions and Answers

3.1

INTRODUCTION

In the last unit we have seen how definite integrals can be used to calculate areas. In fact, this application of definite integrals is not surprising. Because, as we have seen earlier, the problem of finding areas was the motivation behind the definition of integrals. In this unit we shall see that the length of an arc of a curve, the volume of a cone and other solids of revolution, the area of a sphere and other surfaces of revolution, can all be expressed as definite integrals. This unit also brings us to the end of this course on calculus.

Objectives After reading this unit you should be able to : find the length of an arc of a given curve whose equation is expressed in either the Cartesian or parametric or polar forms, find the volumes of some solids of revolution, find the areas of some surfaces of revolution.

3.2

LENGTH OF A PLANE CLTRVE

In this section we shall see how definite integrals can be used to find the lengths of plane curves whose equations are given in the Cartesian, polar or parametric form. A curve whose length can be found is called a rectifiable curve and the process of finding the length of a curve is called rectification. You will see here that to find the length of an arc of a curve, we shall have to integrate an expression which involves not only the given function, but also its derivative. Therefore, to ensure the existence of the integral which determines the arc length, we make an assumption that the function defining the curve is derivable, and is deriative is also continuous on the interval of integration. Let's first consider a curve whose equation is given in the Cartesian form.

3.2.1

Cartesian Form

Let y = f(x) be defined on the interval [a, b]. We assume that f is derivable and its derivative f is continuous. Let us consider a partition Pnof [a, b], given by

The ordinates x = a and x = b determine the extent of the arc AB of the curve y = f(x) Fig. 1 (a)]. Let Mi = 1,2, .........., n - 1, be the points in which the lines x = xi meet the curve. Join the successive points A, M,, M,, M3, ........,M,,, B by straight line segments. Here we have approximated the given curve by a series of line segments.

Appllcrtlons o f Calculus

't

If we can fmd the length of each line segment, the total length of this series will give us an approximation to the length of the curve. But how do we fmd the length of any of these line segments?Take M2,M,, for example. Fig. I(b) shows an enlargement of the encircled portion in Fig. I(a). Looking at it we fmd that

where A x, = M2Qis the length (x, - $),and AY, =M3Qtflx3)- fl$) =y3-y2. In this way we can fmd the lengths of the chords AMI, MlM2,........,M,,B, and take their sum

Sngives an approximation to thelength of the arc AB. When the number of division points is increased indefinitely, and the length of each segment tends to zero, we obtain the length of the arc AB as

provided this limit exists.

Our assumptions that f is derivable on [a,b], and that f is continuous, permit us to apply the mean value themtm meom 3, Block 21.

Thus,there exists a point P,' (xi', y,') between the points Mi_,and Mi on the curve, when the tangent to the curve is parallel to the chord M,, Mi. That is,

or

Ayi = f (xi*) Ax,

Hence we can write (1) as

-

, 2 J--~~~

o+IO

i-l

This is nothing but the definite integral h "

J(I*ll'Odx. ']

Further Appllcationr of Integral Calculus

Therefore,

Remark 1: It is sometimes convenient to express x as a single valued function of y. In this case we interchange the roles of x and y, and get the length

where the limits of integration are with respect to y. Note that the length of an arc of a curve is invariant since it does not depend on the choice of coordinates, that is, on the frame of reference. Our assumption that f is continuous on [a, b] ensures that the integrals in (2) and (3) exist, and their value L: is the length of the curve y = f(x) between the ordinates x = a and x = b. The following example illustratesthe use of the formulas given by (2) and (3). Example 1 :Suppose we want to find the length of the arc of the curve y = k x intercepted by the ordinates x = 1 and x = 2. We have drawn the curve y = lnx in Fig. 2.

Flg. 2

Using (2), the required length L,2 is given by

.

dx t If we put 1 + x2= t2, we get -= - ,and therefore,

dt

x

Applications of Calculus

We can also use (3) to solve this example. For this we write the equation y = lnx as x = ey. The limits x = 1 and x = 2, h n correspond to the limits y = 0 and y = 1112respectively. ,

Hence, using (3), we obtain

as we have seen earlier. This verifies our observation in Remark 1 that both (2) and (3) give us the same value of arc length.

Now, here are some exercises for you to solve.

E E I)

Find the length of the line x = 3y between the points (3,l) and (6,2). Verify your answer by using the distance formula.

E E2)

Find the length of the curve y = In sec x between x = 0 and x = d2.

E E3)

Find the length of the arc of the catenary y = C cosh (x/c) measured fkom the vertex (0, c) to any point (x, y) on the catenary.

E E4)

Find the length of the semi-cubical parabola ay2= x3 fkom the vertex to the point

Farther A p p ! i e a W of Integral Caleu!us

Applications o f c a l c u ~ u s

E ES)

-

- .

Show that the length of the arc of the parabola y2 = 4ax cut off by the line 3y = 8x isa(ln2 + 15116).

In the next sub-section we shall consider curves whose equations are expressed in the panmetric f o m

Parametric Form

3.2.2

Sometimesthe equation of a curve cannot be written either in the form y = f(x) or in the form x = g(y). A common example is a circle x2+ $ = a2.In such cases,we try to write the equation of the curve in the parametric form. For example, the above circle can be represented by the pair of equations x = a cos t, y = a sin t. Here, we shall derive a formula to find the length of a curve given by a pair of panmetric equations. Let x = Nt), y = ~ ( t )a, 5 t S f3 be the equation of a curve in parametric f o m As in the previous sub-section, we assume that the functions $ and q~ are both derivable and have continuous derivatives $' and y/ on the interval [a,PI. We have dx = dt

I$'

Hence,

d~ (t), and - = y' (t). dt dy

W"t) =0 , ,and

- [$' (t)12 + [w' (t)12

4' (t)

Now, using (3) we obtain the length

(we assume that $' (t) # 0).

The following example ihows that sometunes ~tis more convenient to express the equation of a given c w e in the parametric form in order to find its length.

Example 2: Let us find the whole length of the c w e

By substitution, you can easily check that x =a cos3t,y = b sin3t is the parametric form of the given c w e . The m e lies between the lines x = f a and y = f b since -1 5 cos t 5 1, and -1 5 sin t S 1. The curve is symmetrical about both the axes since its equation remains unchanged if we change the signs of x and y. The value t = 0 corresponds to the point (a, 0) and t = d 2 corresponds to the point (0, b). By applying the m e tracing methods discussed in Unit 9 we can draw this c w e (see Fig. 3).

Fig. 3

Since the c w e is symmetricalabout both axes, the total length of the c w e is four times its length in the first quadrant.

($1 ($1 )'

+

= 9 sin%0x4 (a2C O S + ~ b2 sin%)

Hence, the length of the c w e is

L

= 4

7 '7

3 sin t cos t Ja2 cos2t + b2 sin2t dt

0

-

12

sin t cos t Ja2 cos2 t + b2 sin2t dt

0

Putting u2 = a2cos%+ b2sin2t,we obtain 2u = (2b2- 2a2)sin t cos t

dt du

and the limits t = 0, t = d 2 correspond to u = a, u = b, respectively.

Further Applications of Integral Calculus

Appllcatlons af.Caleulus

Thus, we have

12 b3 - a3 b2-a2 3

= --=

4(a2 + b2 + ab) a+b

Now you can apply equation (4) to solve these exercises. \

E E 6) Fmd the length of the cycloid x=a(8-sine) ;y=a(l-cos8)

E E7) Show that the length of the arc of the w

e

x=ets~t,y=e'costftomt=Opt=d2isf i (e*-1).

3.2.3

Pobr Form

In this sub-section we shall consider the case of a curve whose equation is given in the polar form, Let r = f(0) determine a curve as 8 varies from 9 = a to 8 = b, i.e.,-the function f is defined in the interval [a, @] (see Fig. 4). As before, we assume that the function f is derivable and its derivative f is continuous on [a,b]. This assumption ensures that the c w e represented by r = f(8) is rectifiable.

Furtbct Applieatfonr o f lntcgral Calcolur

Transforming the given equation into Cartesian coordinates by taking x = r cos 0, y = r sin 0, we obtain x = f(0) cos 0, y = f(0) sin 0. Now we proceed as in the case of parametric equations, and get,

Hence, the length of the arc of the curve r = q0) from 0 = a to 0 = $ is given by

changing the variable x to 8.

=

hf'(0) cos 0 - f (0) sin 01'

+ If' (8) sin 8 + f(0) ms 81'

dB

We shall apply this formula to find the length of the curve in the following example. Example 3: To find the perimeter of the cardioid r = a (1 + cos 0) we note that the curve is symmetrical about the initial line (Fig. 5). Therefore its perimeter is double the length of the arc of the curve lying above the x-axis.

Now,

dr

= - a sin 0. Hence, we have a0

Appllcatlanr af Calculus

We know

In this section we have derived and applied the formulas for finding the length of a curve when its eqwtion is given in either of the three forms: Cartesian, parametric or polar. Let us rmnrmarise our discussion in the following table.

Table 1: Length of an arc of a curve

Using this table you will be able to solve these exercises now.

E E 8)

Find the length of the curve r = a cod (013).

E E9) Find the length of the circle of radius 2 which is given by the equations x=2cost+3,y=2sint+4,O~t~2x.

Further Appllc8tions of Integral Calculus

E E 10) Show that the arc of the upper half of the curve r = a (1 - cos 8) is bisected by 8 =2 d 3 .

3.3

VOLUME OF A SOLID OF REVOLUTION

Until now, in this course, we were concerned with only plane curves and regions. In this section we shall see how our knowledge of integration can be used to find the volume of certain solids. Look at the plane region in Fig. 6 (a). It is bounded by x = a, x = b, y = f(x) and the x-axis. If we rotate this plane region about the x-axis, we get a solid. See Fig. 6 (b). -- -

Such solids are called solids of nvoluhbn. Fig. 7 (a) and Fig. 7@) show two more examples of solids of revolution.

Fig. 7

The solid in Fig. 7(a) is obtained by revolving the region ABCO around the y-axis. The solid of revolution in Fig. 7(b) differs from the others in that its axis of rotation does not form a part of the boundary of the plane region PQRS which is rotated. We see many examples of solids of revolution in every day 1ife.The various kinds of pots made by a potter using his wheel are solids of revolution. See Fig. 8(a). Some objects manufactured . with the help of lathe machine are also solids of revolution. See Fig. 8 (b).

Now, let us try to find the volume of a solid of revolution. The method which we are going to use is d t e d the method of slicing. The reason for this will be clear in a few moments.

<x,< ......<x,,+x,,=

LetTn= {a=x,,<x, intervals.

Furtbw Appllcatloa~of Integral Calculus

b) bea~tionoftheinte~al[a,b]intonsub-

Fig. 9

Let A xi denote thc length of the ith sub-interval [xi_,, 31.Further, let P and Q be the points on the c w e , y = f(x) corresponding to the ordinates x = and x = xi, respectively. Then, as the c)we revolves about the x-axis, the sbaded strip PQNM (Fig. 9(a)) generates a disc of thickness A xi.In general, the ordinates PM and QN nay not be of equal length. Hence, the disc is ac ally the frustum of a cone with its volume A vi, lying between II PM2MN and n Q N 2 dm* is,between

$,

If we assume that f is a continuous function on [a, b], we can apply the intermediate value theoran (Theorem 7, ~ l & k1, Plso see margin remark), and exprtss thisvolume as A vi =~ ~ { f ( tA, )xi, )~ where ti is a suitable point in the interval [xc,, xi]. Now summingup over all the discs, we obtain .

0

Y . =

n

[f(ti11' A xi,

A vi = i -1

%,Sti dXi a~ mapproximation

I

to the volume of the solid of revolution. As we have observed earlier while defming a d e f ~ t e integral, the approximation ge$s bet* as the partition P, gets finer and finer and Axi tends to zero. Thus. we get the volume of the solid of revolution as n

v

=

lirn Vn = lirn

,

0-m

n [f(ti )I' A xi i-1

We shall use this formula to fmd the volume of the solia described in the following example.

Example 4: Let us fmd the volume of the solid of revolution formed when the arc of the parabola y2 = 4ax between the ordinates x = 0, and x = a is revolved about its axis. The solid of revolution is the parabolic cap in Fig. 10. The volume V of the cap is given by

IrPM2. MN is the volume of the disc with radius PM and thickness MN. rrQNIMNis the volume of the disc with radius QN and thickness MN. Iff is continuous og [a, b], fla)=candflb)=d,andz l i e between c and d, then 3x, €14b[ S.t. Rx,,) = Z.

Applicatloos of Calculus

Our next example illustratesa slight modification of Fonnula (6) to find the volume of a solid obtained by revolving a plane region about the y-axis. x2 y2 Example 5: Suppose the ellipse 2 + = I. (a >b) is revolved about the minor axis. AB

3

(see Fw. 11). La us tind the volume of the solid generated.

In this case the axis of rotatian is the yaxis. The am revolved about the y-axis is shown by the shaded region in Fig. 11. You will agree that we need to consider only the area to the right of the y-axis.

RI. 11 To fiad the v o h of this solid we intacbangc x a d y in (6) and get

We can also modify Formula (6) to apply to curves whose equations are given in the parametric or polar forms. Let us tackle these one by one. PIvruneMeForm

If a c w e is given by x = Nt), y = ~ ( t ) a , S t 5 @, then the volume of the solid of revolution about the x-axis can be found by substituting x and y in Formula (6) by Nt) and ~ ( t ) , respectively. Thus,

We'll now derive the formula for curves given by polar equations.

PdarForm Suppose a curve is given by r = f(8), 8 , I 8S 8,. The volume of the solid generated by rotating the area bounded by x = a, x =b, the x-axis and r = Q0) about the x-axis is d 1(r sin el2 (r cos 9) do d0

91

v

= n

er = II

1

If(@ sin 01' If' (9) cis 0 - f(0) sin 01dB

Let's use this formula to find the volume of the solid generated by a cardioid about its initial line. Example 6: The cardioid shown in Fig. 12is given by r = a (1 + cos 0).

Fig. I 2

The points A and 0 cornspond to 0 = 0 and 0 = IC, respectively. Here,again, we need to consider ody the part of the cardioid above the initial line. Thus, 0

=

1n (r sin 8)' d0 (r cos 9) d0

E

1

It

- na3

(I

0

+ con 0)'

sin30 (I + 2 sos 0) dB, since r = a (I + COS 0)

Further Applications of Integral Calculus

Appliertions of Calculus

= 256 na3

'r

sin3 ( cos9 I$ d( - 64na3

0

64 xa3 15

= ----

'j?

sin3 I$ cos7 ( d$, where ( = 9 12

0

8xa3 on applying a reduction formula from Unit 12.

5

In all the examples that we Save seen till now, the axis of rotation formed a boundary of the region which was rotated. Now we take an example in which the axis touches the region at only one point.

Example 7: Let us find the volume of the solid generated by revolving the region bounded by the parabolas y = x2 and y2 = 8x about the x-axis. We have shown the area rotated and the solid in Fig. 13 (a) and.(b), respectively.

Fig. 13

Here, the required volume will be the differencebetween the volume of the solid generated by the parabola y2 = 8x and that of the solid generated by the parabola y = x2.

Note the limits of integration. Here, we list some exercises which you can solve by applying the formulas derived in this section.

E,E12) Find the volume of the right circular cone of height h and radius of the circular base r. (Hint: The cone will be generated by rotating the triangle bounded by the x-axis and the h e y = (rlh) x).

ii

: t't'r $1,

5, -: .

:,*'I

b;

Further Applicatiolls o f Integral Calculus

i ,' a

154

i'ij

r j,i

1..? ,.# i t:t ii'i :*!

,.

8

:ST* .%.

1.i,!

it;; -,,

?&it E

> ~ ? .

Iij k.f f

ij/

i,:!l ,.'l

:Qi

It.!

::I.: s

ti >i :

i4;

i.i.1 [;I;

;:ej

_

X'S

'.

I

NI,~

1:;

il

E E 13) Show that the volume of the solid generated by revolving the curve x* + yU3= a2I3about the x-axis is 32m3/105.

i,?,! f!:!2

. 4:;I

:I" I

!.ij 5

i.I.; \.! .?! ;;,i

i ,,I i: t

,1

ill{

?;:I

b' ? i i') :i i j

i:i!i

$1p;,,

! 21

;iij ::.I

iii r ii!

:-$

~. .:;I ,._I

i"

l

i .'i ,>!,I

: $1

jj;l

[:-.1

[ iii

* i:i

i

;ii; .,.

I:/ i i:j i$4 ? .I1

;;!!

!

$1

;:$!

I!::? ;!.I

i,$j rI $81 73 6

$1

!:it!

E E 14) The arc of the cycloid x = a (t - sin t), y = a (1 - cos t) in [O,~R]is rotated about the y-axis. Find the volume generated. (Caution:The rotation is about the y-axis.)

Applicrtlons of C ~ l c u l u s

E

E 15) Find the volume of the solid obtained by revolving the limacon r = a + b cos 8 about the initial line.

E E 16) The serpicirculariegion bounded by y - 2 = J9 - xZ and the line y = 2 is rotated about the x-axis. Find the volume of the solid generated.

3.4 AREA OF SURFACE OF REVOLUTION Instead of rotating a p l h e region, if we rotate a curve about an x-axis, we shall get a surface of revolution. In this section we shall fmd a formula for the area of such a surface. Let us start with the case when the equation of the curve is given in the Cartesian form.

Cartesian Form Suppose that the c w e y f(x) [Fig. 141 is rotated about the x-axis. To find the area of the area of the generated surfilce, we consider a partition P, of the interval [a, b], namely, P l l = { a = ~ < x , < x 2........ < <x,,<x,,=b)

I I

/

Fig. 14

Let the lines x = xi intersect the curve in points Mi, i = 1,2. ......n. If we revolve the chord Ma Mi about the x-axis, we shall get the surface of the frustum of a cone of thickness Ax: = x: xL,. Let As;be the area of the area of the surface of this frustum. Then the total surfwe area of all the frusta'is 8

8

r.

This Snis an approximation to the area of the surface of revolution. The area of the surface of revolution generated by the curve y = fix), is the limit of Sn(if it exist), as n +wand each Ax; +0.

To find the area A of the curved surface of a typical frustum, we use the formula A = IC (r, + r,) I. where I is the slant height of the frustumand r, and r. are the radii of its bases (F;~.1;). In the frustum under consideration the radii of the bases are the ordinates fixo) and fixi), I

1

while the slant height Ma Miis give* by ,/(Axi12 +'(A Y , ) ~,where Ayi = fixi)- fixk,).

-

We assume that f is derivable on [a, b] and f is continuous. Then by the mean value theorem (Theorem 3, ~ 1 & 2), k we obtain A yi f (ti) xi, for some ti E [xi_,,xi]. Therefore, -

I

=

2n

Yi

where (fixi_,)+ fix,))/2is the mean radius of revolution

Jw A x,

\zJ Fig. 15

Exunple 8: Let us find the area of the surface of revolution obtained by revolving the parabola $ = 4ax h m x =a to x = 3%about the x-axis.

The area of the surface of revolution.

Instead of revolving the given curve abour the x-axis, if we revolve it about the y-axis, we get another surface of revolution. The area of the surface of revolution gmmted by the curve x = g(y), c 5 y 5 d, as it revolves about the y-axis is given by,

Now let us look at curve rtprtscnted by parametric tquations.

Parametric Form Suppose a curve is given by the parametric equations x = Mt), y = yr (t), t E [a, PI. Then we know that

Substitutingthis in formula (lo), we get the area of the surface of~evolutiongenerated by the curve as it revolves about the x-axis, to be

Now we shall state the formula for the surface generated by a curve represented by a polar equation.

PoIar Form If I = h(8) is the polar equation of the curve, then the area of the surface of revolution generated by the arc of the curve for 8, 5 0 4 B2,as it revolves about the initial line, is

Study the following examples carefully before trying the exercises given at the end of this section. Example 9: Suppose the astroid x = a sin3t, y = a co$t, is revolved about the x-axis. Let us find the area of the surface of revolution. You will agree that we need to consider only the portion of the curve above the x-axis. For this portion y > 0, and thus t varies from - d 2 to.d2. dy

dx

= 3a sin2tcos t, - = - 3a cos2t sin t dt dt Therefore,

($1'

(z) 2

+

+ i

.

=9 '.

sin%cos2t

We therefore get, nl2

s

= 2 1 J'acos3t

I

J9a- T -7 sm T tcos t d t-

nl2

=

271

a cos3 t / 3a sin t cos t 1 dt

-nlZ

= 6na2

'f

c0s4 t I sin t ( dt

-zl2

Example 10: Suppose we want to find the area of the surface generated by revolving the cardioid r = a (1 + cos 8) about its initial line. Notice that the cardioid is symmetrical about the initial line, and extends above this line from 8 = 0 to 8 = n.The surface generatedby revolving the whole w e about the initial line is the same as that generated by the upper half of the curve. Hence,

Further Applicatio~~s of Integral Calc-ulus

Applications ~ ( C a l c u l u i

dr Simer=a(l +cosO),and - =-asin€), wehave de 0 =a2(l + c o s 0 ~ ~ + a ~ s i n ~ 0 = 4 a ~ ~ ~ 2

f + ($J2 Therefore,

-

4na2

J 4 sin -2 ms4 Bd0 2

0

-

sin ( ma4 ( d(, *hie ( = 8 1 2

32xa2 0

E- E 17) Find the area of the surface generated by revolving the circle r = a aboutthe x-axis, and thus verify that the surface area of a sphere of radius a is 4 d .

E E 18) The arc of the curve y = sin x, h m x =0 to x =n is revolved about the xlaxis. Find the area of the surface of the solid of revolution generated.

E E 19) The ellipse x21a2+ y'/b2= 1 revolves about the x-axis. Find the area of the surface of the solid of revolution thus obtained.

E E20) Prove that the surface of the solid generated by the rwolution about the x-axis of the ~ q o f t h e c u r v e x = ~y, =

(t -g)

is3n.

E E2-1)Find the surface area of the solid generated by revolving the cycloid x = a(8 - sin 8), '

y = a(1- cos 8), about the line y = 0.

Furtber Applientions of Integral Cnleulus

Appllcmtions of Cmkulus

Now let us quickly recall what we have covekd ih this unit.

In this unit we have seen how to find

1)

the lengths of curves

2)

volumes of solids of revolution and

3)

the areas of surfaces of revolution.

In each case we have derived formulas when the equation of the curve is given in either the Cartesian or parametric or polar form. w e give the results here in the form of the following tables.

Length of an arc of a curve

Volume of the solid of revolution

w=w

Volume

n

x = g(y) about y - axis

n

x = +(t), Y = ~ ( t ) about x - axis

n

r = h(0)

n

about the initial line

1

b

Y = f(x) about x -axis

y2

dx

a d

J x2 dx

C

1 1

[ul(t)12 +* (t) dl

a

[h(B) sin 01' [hl(B) cos 6 - h(6) sin 01 dB

01

-

Area ofthe surface of revolution

E
Arpa

b

Y = f(x) abut x -axis

I

x = g(y) about y - axis

c /

B

= Mt). y = W(t) about x -axis

2%

r = h(0) about the initial line .

3.6

J W (0 ,/[v (t)lZ+ [w' (t)lZ dl

a

01

SOLUTIONS AND ANSWERS

1

By distanceformula,

secx13+tanx/3 sec 0 + tan 0

Further Applicatloar ef Integral Calculur

Appllcatloas of Calculus

0

H-

(0,O) and

9a 3a (16 T ) are the points of intersection

Further Appllcrtlons of Integral Calculus

- cos 0)

= 2a2(1 = 4a2s&

- 8a

E9)

'I

(812)

sin + d + = 8 a

dy = 2 cos t x=-2 sin t dt dx

.:

L =2

2i 0

dt = 4x

Note that L = 2m since, here, r = 2.

:.

/r2

+

(:I2

0

= 2a sin -

2

The length of the curve in the upper half

=

2a sin

2,

The length from 8 = 0 to 6 = 2 d 3

The arc of the curve in the upper half is bisected by 8= 2d3.

dr Ell) r=a(@- l), d0 -2a8

Note &t the total volume generated is equal to twice the volume generated by the arc of the curve between x = 0 and x = a.

.:

v

n2] a2 (I-sintl2 asin tdt 0

(t2 sin t - 21 sin2 t sin3 t) dt

na3 0

E15)

dx r = a+bcose 3 -=

d(r cos 8)

de

I

de

(The other two integrals are equal to zero since cos (x - 8) = - cos 0.)

(using a reduction formula)

-

I

2na2 f, sine de 0

='

4na2 cos 8

":I

Further Applications of Integral Calculus

Applications 01 Calculus

dy E18) y = sinxa-=cosx

dx

:.

S E 2x

1

*. J l+cos x d x smx

0

E20) l%e loop is between t =

- & and

t=

J5 .Because ofsymmetry,it is enoughto

considerthccwebetwant=Oandt=

&.

2x

El) S = 271

I

a (I

- cos 0) ,/a2 (I - EOS

0

= 4na2

7 0

0 (I - cos 0) sin -dB 2

+ a2 sin2 0 dB

Further Applications o f Integral Calculus