LOCUS
1
Areas and volumes
This is the chapter where we will actually put our knowledge of (definite) integration integrati on to practical use. We will calculate areas under curves, areas area s bounded by curves, volumes under surfaces, volumes bounded by surfaces, etc. We We will mostly be concerned with areas; however, we will delve a little into calculating calculat ing volumes for the sake of completeness. Calculating areas and volumes volum es is totally about calculating calcul ating definite integrals i ntegrals and involves no new concepts. You are advised to read the chapter on “ Definite Integration” again if you are not very comfortable in evaluating integrals. integral s. We will now start with such calculations calculation s through actual examples.
Example – 1 Find the area bounded between the curves y = x2 and y = x. Solution: The bounded area is depicted in the figure below: y 2
y = x y = x
The shaded area represents represents the area A bounded between the two curves
1
x 1
Fig - 1
Thus, 1
A=
∫ ( x− x ) dx 2
0
1
x 2 x3 1 = − = 2 3 0 6
sq units
Example – 2 Find the area of the region bounded between x = 0, x = 2, y = 2 x and y = 2 x – x2.
Maths / Areas and Volumes
LOCUS
2
Solution: The bounded region has been sketched below. Verify Verify for yourself the validity of the figure drawn: y
x
y = 2
4
The shaded region represents the area A bounded between the given curves
1
0
1
x
2 y = 2x - x
2
Fig - 2
We have, have , 2
A=
∫ (2
x
− ( 2 x− x2 ))
dx
0
2
2 x 2 x3 = − x + l n 2 3 0 8
1
3
ln 2
3 4 = − sq. ln 3
units
=
4 ln 2
−4+ −
Example – 3 Compute the area of the region bounded between the curves y = exln x and y =
ln x ex
Solution: If you’ve read the chapter on “ Applications of Derivatives” properly, you should not have much of a problem plotting the graph of the two functions: y y =
1/e
1
ln x ex
x
The point of intersection is determined by equating the equation of the two curves, i.e, ln x ex ln x = ex ⇒
⇒
x = 1 or x =
1 e
The shaded area represents represents the area A between two curves
-1
Fig - 3 Maths / Areas and Volumes
2 2
ln x = 0 or e x =1
LOCUS
3
From the figure, it is clear that 1
∫ ex −
A=
ln x
exln x dx
1/ e 1
ln x
∫
=
ex
1/ e
1
dx − e
∫ x ln xdx
1/ e
1 1 1 0 x 2 1 = ∫ tdt − e ln x − ∫ x dx 2 1/ e e −1 2 1/ e
−1
1 1 1 − e 2 − 1 − 2 2e 2e 4 e 1 1 e e2 − 1 = − − + 2 2e 2e 4 e 2 1 e −1 =− + =
4e
e
e
=
2
−5
4e
sq. units
Example – 4 x) passes through the origin. Through any point ( x, y) on the curve, lines are drawn parallel to the A curve y = f ( x co-ordinate axis. If the curve divides the area formed form ed by these lines and the co-ordinate axes in the ratio m:n, find the equation of the curve. Solution: The situation described in the question is graphically depicted below: y = f(x)
It is given that A1
y
(x, y)
A2
=
m n
Not N otee that that A 1 + A 2 = xy x
and an d A2 =
A1 A2 x
Fig - 4
From the given constraint, we have x
xy−
dx ∫ y dx 0
x
∫ ydx
=
m n
0
x
⇒
∫ 0
Maths / Areas and Volumes
ydxx yd 0
n y=dx ( m+n
) xy
LOCUS
4
Differentiating both sides with respect to x, we obtain y =
xdy + y m + n dx n
⇒
my
= nx
dy
⇒
dy
=m
dx
⇒
y
dx
n x
ln y=
m
=
m
n n
ln x+ C ln x + ln k
where k is a constant
y= kxm
⇒
/ n
(Note that this passes through the origin)
Example – 5 Find the area of the region containing the points whose ( x, y) co-ordinates satisfy y
1
+ ≤ e− x 2
Solution: If you observe the given relation carefully, you will realise that whatever region we obtain will be symmetric about both the x-axis and y-axis.
This means that we only need to plot the region in the first quadrant. The regions in the other quadrants can then automatically be obtained by reflecting symmetrically the region in the first quadrant into all the other quadrants. So let us now consider just the first quadrant. quadrant . In this quadrant, both x, y > 0, so that the given relation can be simply written as y +
⇒
1 2
≤ e − x
y ≤ e − x
−
1
2 We now plot this region r egion for x > 0, y > 0: y The shaded region 1/2
-x
represents y < e –
-x
y=e –
1 2 x ln 2
Fig - 5 Maths / Areas and Volumes
1 2
LOCUS
5
Our required region is therefore, y The shaded region
1/2
represents - x
y < e
- ln2
ln2
–
1 2
x
-1/2
Fig - 6
The required area is: ln 2
A =4
∫ 0
− x
e
1 − 2
dx
ln 2
= 4 −e− x − x 2 0
= 4 1 − 1 ln 2 2 2 units = 2 (1 − ln 2 ) sq. un
Example – 6
Find all the possible values of b > 0 so that the area of the bounded region enclosed between the parabolas y= x− bx and y= 2
x
2
b
is maximum.
Solution: The first parabola has its it s zeroes given by y = x− bx
2
⇒ x = 0, Maths / Areas and Volumes
1 b
=0
LOCUS
6
This is a downwards facing parabola. The other parabola has its vertex at the origin. y
2
y =
x b
The point of intersection P is given by: 2
x = x – bx b b x = 2 1+b 2
P 2
y = x – bx
⇒
0
y =
x
1/b
b 2 2 (1+b )
Fig - 7
The figure above tells us how to calculate the co-ordinates of the intersection point P:
b b , P≡ 2 2 2 1 + b (1 + b ) Now, the bounded area A is given by: b
A= ∫ x− 0 1+ b
2
bx − 2
x
2
b
dx b
x 1 1 3 = − b + x 2 3 b 0 2
1+ b 2
b
=
2
2 (1 + b b
=
2
2
)
2
)
1 b2 + 1
−
3
b
⋅
b
(1 + b
6 (1 + b
⇒
dA1 db
=
dA1 db
6A =
)
6 A ):
b
1 + b2
2
2
2
1− b
2
1+ b
2
= 0 i.e, b = 1
(b is > 0 so it cannot be –1)
Thus, the bounded area is maximum when b = 1
Maths / Areas and Volumes
3
2
(1 + b ) − 2b = (1 + b ) =
A1 is maximum when
2
2
We now maximize maximi ze A (or more conveniently, convenient ly, and equivalently, A1
3
LOCUS
7
Example – 7
Find the area of the region bounded by x =
1 2
, x = 2, y = ln x and y = 2 x.
Solution: The region of interest is very straightforward to plot:
y
The shaded area represents the region whose area we need to determine
1
1
½
x
2
Fig - 8
The required area is 2
A=
∫ (2
x
− ln )x
dx
1/ 2
2
2 x = − xln x + x ln 2 1/ 2 4− 2 5 3 = − ln 2 + sq. units l n 2 2 2 Example – 8 Lett Le
{
2
}
2 f ( x) = max x , (1 − x) , 2 x(1 − x) . Determine the area of the region bounded by the curve
y = f ( x) , x − axis, x = 0 and x = 1.
Maths / Areas and Volumes
LOCUS
8
Solution: The technique to plot the curve for f ( x x) has been outlined in i n the unit on Functions. Functio ns. We We plot all the three thr ee 2
curves x2 , (1 − x) and 2 x(1 − x) on the same axes, scan the x-axis from left to right and at every point, pick out that graph which lies l ies uppermost of all the three graphs. In I n the figure below, the heavyset x): curve is the curve for f ( x y y = x
2
2
y = (1 – x)
1 ½ ¼ 0
A
B
x
1
½
The intersection point A is given by 2 (1 - x) x) = 2x (1 - x) 1 3 The intersection point B is given by 2 x = 2x 2x (1 (1 - x) 2 3
y = x (1 – x) Fig - 9
We can evaluate the required area, as is clear from fr om the figure above, by dividing the integration int egration interval [0, 1] into three sub-intervals: 1/ 3
∫ (1 − x)
2
A=
2/3
dx+
0
= = =
19 18
+
13 81
1
∫ 2 x(1 − x) dx+ ∫ x dx 2
1/ 3
+
19 81
2/3
(verify)
51 81 17 27
Example – 9 A square has its vertices at (1, 1), (1, –1), (–1, –1) and (–1, 1). Four circles of radius 2 are drawn, one centred at each vertex of the square. Find the area common to these four circles.
Maths / Areas and Volumes
LOCUS
9
y
Solution:
(-1, 1)
(1, 1) O Each circle has R
been partially drawn.
P
x
The area common to there four circles has bee shaded.
S (-1, -1)
(1, -1)
Fig - 10
Let us evaluate the area in the first quadrant. quadr ant. For that, we need to evaluate the x-co-ordinate of P. Notice that the circle centred at (–1, –1) is the one which intersects the x-axis at P. The equation of this circle is given by: 2
2
( x + 1) + ( y + 1) = 4
...(1)
When y = 0, x = 3 − 1
( 3 − 1, 0)
P≡
Thus,
The area in the first quadrant is now: 3−1 st
1
A=
∫ (
4−(
)
+x1)2 − 1
0
=
1 − x+ 2 ( x+ 1)
= − π
= − 3
3 + 1+
3 2
3
Therefore, the total bounded area is A = 4 × A1st π = 4 − 3
Maths / Areas and Volumes
2
4 − ( x+ 1)
+ 2 sin −1
3 sq. units
we used used (1) to write write down down the the equa equatio tion n of the circle in an explicit form
dx
3 2
+
4 2
−
sin 3 2
−1
x + 1 2 0 1
+ 2 sin −1 2
3 −1
(verify this step)
LOCUS
10
Example – 10 Find the area of the region bounded by x + 1 = 0, y = 0, y = x2 + x + 1 and the tangent to y = x2 + x + 1 at x = 1 Solution: Let us first determine the equation of the said tangent: dy dx x =1
= ( 2 x + 1) x=1 =3
Also, when x = 1, y = 3 The required equation is
y − 3 = 3 ( x − 1) ⇒ y = 3 x . y
3 The shaded area represents the bounded region whose area we wish to determine
The curve 2 y=x y =x +x+ +x+1 1
1
¾
-1
0
-½
x
1
The tangent y = 3x
Fig - 11
From the figure above, it should be clear that the area can be calculated as described below: 0
A=
1
∫ ( x + x+ 1) dx+ ∫ {( x + x+ 1) − 3 x} dx 2
2
−1
=
0
3
x
3
+
2
x
2
0
+ x + −1
3
x
3
1
− x + x 0 2
1
1 1 1 = − − + − 1 + −1 + 1 3 2 3 0
= Maths / Areas and Volumes
7 6
unitss sq. unit
LOCUS
11
Example – 11 Find the area of the region bounded by the curves y = x2 , y = 2 − x2 and y = 2,
which lies to the right of the line x = 1. Solution: The required area is sketched below y 2
2
y = |2-x |2-x |
1
y = x
2
0
1
x
2
2
Fig - 12
The required area can be evaluated as follows: 2
A=
∫ ( x − (2 − x )) 2
2
1
2
2
dx+
∫ (2 − ( x − 2)) 2
2
2 x3 x3 = − 2 x + 4 x − 3 3 1 =
20 − 12 2 3
sq.units
dx
2
2
(verify the calculation calculations) s)
Example – 12 A curve y = f ( x) passes through the point P (1, 1) . The normal to the curve at P is a ( y − 1) + ( x − 1) = 0 . If the slope of the tangent at any point on the curve is proportional to the ordinate of that point, determine the equation of the curve. Hence obtain the area bounded by the y-axis, the curve and the normal to the curve at P. Solution: The slope of the given normal is obvious from the expression: y − 1 x − 1
Maths / Areas and Volumes
=−
1 a
LOCUS
12
⇒
m N = −
⇒
mT = a
1 a
⇒ dy = a dx P It is given that Since
dy dx
∝
dy dx (1.1)
⇒y
dy
a
⇒
=
dx
=
ky
=a
k
Thus, dy dx
= ay dy
⇒ ⇒
y
= adx
ln y= ax+ ln C (we took the constant of integration as ln C instead
of C so that the final expression for y is simpler)
⇒
y= Ceax
Since this curve passes through (1, 1), 1 = Ce a.1
⇒
C
= e− a
Thus, the equation of the curve is y = e
a ( x −1)
Let us now proceed to evaluate the bounded area, which is i s sketched below: y
The curve y=e
a(x-1)
p
1 -a
e
The normal to the curve at P x
1
Fig - 13
The equation of the normal has already been provided: y = 1 + Maths / Areas and Volumes
1 a
(1 − x )
LOCUS
13
Thus, the required area is:
1 − + − x 1 ( 1 ) ∫ 0 a 1
= A
x x2 = x + − − a 2a
a ( x −1)
e
a
a ( x −1)
e
dx
1
0
1 1 1 e− a = 1 + − − − − a 2a a a 1 1 = 1 − + a sq. units 2 a ae
Example – 13 Let f ( x) be a continuous function given by
2 x 2 x + ax+ b
f ( x) =
≤ 1 x > 1
, x ,
Find the area of the region in the third t hird quadrant bounded by the curves x = −2 y 2 and y = f ( x) lying to the left of the line 8 x + 1 = 0 . Solution: We can use the fact that f ( x) is continuous to determine the constants a and b. The critical points are x = –1 and x = 1. x = 1
lim
x →− 1–
f ( x) = f ( − 1) 1)
⇒ 1 − a + b = −2 ⇒ a −b = 3
...(1)
x = −1
lim f ( x) = f ( 1)
x →1+
⇒ 1+ a + b = 2 ⇒ a +b =1 From (1) and (2), a = 2 and b = −1 .
...(2)
Thus,
2 x 2 x + 2 x− 1
f ( x ) =
Maths / Areas and Volumes
≤ 1 x > 1
, x ,
LOCUS
14
We now proceed to sketch the bounded region that has been described in the question. y x= -1/8
Observe all the
y = 2x
intersection points very
2
2y ==-xx
carefully and how they are obtained. The function f(x) -2
-1 -1/8
0
changes its definition
x
at x = -1 so we have to split
-¼
the area we need to evaluate -1
accordingly.
(-2, -1) -2 2
y=x y =x +2x +2x-1 -1
Fig - 14
The required area is:
− x 2 2 1 A= ∫ − x + x − ) 2 ( −2 −1
= −
3
= − =− =−
2 3 2 3
1 48
= 4− =
x3 3/ 2 (− x) − + 3
2
(1 − 2
3/ 2
dx+
−1
x
− 2 2 x − x + (− −2 3
dx
−1 / 8
3/ 2 2 x) − x −1
5 − 10 + − 2 1 − 1 − 1 − 1 3 3 3 83/ 2 64
)−
4
5
1
3
3
48
+ + − −
x ∫ −1 − 2 − 2
−1 / 8
+
2 3
−
1 64
+1
1 64
7 192
761 192
sq. units
Example – 14 In what ratio does the x-axis divide the area of the region re gion bounded by the parabola y = 4 x − x2 and y = x2 − x ?
Maths / Areas and Volumes
LOCUS
15
Solution: The two parabolas and the region they bound have been sketched below: y
2
x -x 4 15/4
We need to A evaluate 1 A2 The intersection of the curves can be evaluated by equating the equation of the curves, i.e
A1
0
A 2
1
x
4
2 5/2
2
2
4x - x = x - x 5 2
2
4x-x
Fig - 15
The total area A1 + A2 can be evaluated as the area of the region bounded between the two curves: 5/2
A1 + A2
= ∫ {(4 x−
x )−( x 2
2
0 5/ 2
=
∫ (5 x− 2 x ) dx 2
0
5/2
5 x 2 2 x3 = − 2 3 0 = =
125 8 125 24
−
125 12
sq. units
The area A2 is 1
A2
= ∫ ( x2 − x) dx 0
1
x3 x 2 = − 3 2 0 =
1
A1
=
125
A1
= 121
6
sq. units
The area A1 is therefore
Maths / Areas and Volumes
1
24 6 121 = sq. units 24
Thus, the required ratio is A2
−
4
− x)} dx
LOCUS
16
Example – 15 Let C1 , C2 and C 3 be the graphs of the functions y = x2 , y = 2 x and y = f ( x) fo f or x ∈ [0, 1] where f (0) = 0 . For a point P on C 1 , let the lines through P parallel to the axes meet C2 and C 3 at Q and R respectively as shown in the figure fi gure below. below. If for every position P on C 1 , the area of the shaded regions OPQ and ORP are equal, determine the function f ( x) . y
The area ORP can be evaluated simply as:
C2 : y = 2x 2x C1 : y = x
2
x 2
(x -f( -f(x)) x))dx dx 0
P(1, 1)
1
But to evaluated area OPQ, we will have to divide the integration interval into two parts OS and SX, i.e., the area OPQ would be:
2
P(t, P( t, t )
Q
z
t 2
2
2
(2x - x ) dx + (t - x ) dx 0
X
O
where z is the x- coordinate 2 of S and equals t /2. A better better more straightforward approach is described in the solution
x
R 1
S
z
C3 : f(x f(x))
Fig - 16
Soulution:
To evaluate area ar ea OPQ , we could follow the approach described above, that is by dividing the integration i ntegration interval way, OX into OS and SX . By this way, z
area OPQ =
t
∫ ( 2 x − x ) dx + ∫ (t − x ) dx 2
2
0
=
2
z
2
x
2
= =
2t
−
3
z
2 + t x − 0
x
3
3
t
z
t 2 where z = z
t 2 t 3 t 6 − + t t − − − 24 2 3 24
4
t
x
3
t
8
3
3
−
6
2
4
t
4
However this area can be more easily evaluated by just considering the inverse of the functions representing representi ng C1 and C 2 and integrating with respect to y. (This means that you view the figure above from left to right inside of
the usual top to bottom.) That is, we take y as the integration variable. y will vary from 0 to t 2 (the y -co-ordinate of P). The equations of the curves C1 and C 2 with respect to the y-axis will just be the inverse of the original equations i.e. C1 y
=
y and C 2 y
Maths / Areas and Volumes
=
y
2
.
LOCUS
17
Thus, t 2
∫
area OPQ =
− y dy 2
y
0
2
t
= 2 y − y 4 0 3 3/ 2
=
2t
3
4
t
−
3
2
4
We see that doing the calculation this way given us the result more quickly because we did not have to divide the integration interval i nterval into two sub-intervals sub- intervals as we had to do when we integrated integ rated w.r.t w.r.t x. The area ORP can be easily evaluated evaluat ed (w.r.t (w.r.t the variable variabl e x): t
area ORP =
∫ ( x
2
−
f ( x ) ) dx
0
=
= Since area OPQ = area ORP
∀
t
x 3
3
0
t
− ∫ f ( x) dx 0
t
3
t
− ∫ f ( x)
3
dx
0
t∈ [0, 1], we have
2t 3 3
−
t4
=
4
t
⇒
t 3
3 t
t
dx − ∫ f ( x) dx 0
4
3
t
dx = − ∫ f ( x) dx 4 3 0
Differentiating both sides w.r.t w.r.t t , we obtain our required function : f ( t) = t
3
− t2
Thus, the curve C 3 is given by f ( x) = x3 − x2 in [0, [0, 1] .
Example – 16 Consider a square with vertices at (1, 1), ( −1, 1), ( −1, − 1) and (1 (1, − 1) . Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region S and find its area. Solution: The important part in this question is to correctly plot the required region. regi on. Once that is done, the area is more or less straight forward to evaluate. Consider the square ABCD described in the question and any point P inside it satisfying the given constraint.
Maths / Areas and Volumes
LOCUS
18
y B(-1, 1)
A(1, 1)
P satisfies the given constraint, i.e, d(P, O) < d(P, AB) d(P, O) < d(P, BC) d(P, O) < d(P, CD) d(P, O) < d(P, AD)
P x
O
C(-1, -1)
D(1, -1)
Fig - 17
Consider the point P and any one edge of the square, say AD. Suppose P was equidistant from O and AD (or its extended line). li ne). Where could it possible lie? li e? What is the locus of a point which moves so that its distance from a fixed point is equal to t o its distance from a fixed line? A parabola. Since d ( P, O ) < d ( P, AD ), the point P will lie not on the parabola but in one of the regions that the parabola divides the plane into: y Any point on this curve is equidistant from O and AD (extended). A(1, 1)
The parabola divides the plane into two regions. The region containing the origin is the is that of all points which are closer closer to O than to AD (or the extended line of AD). The equation of this parabola is (using the definition of a parabola):
Any point here is closer to O than to AD (extended)
O
½
1
x
2
2
x +y = (x (x - 1) ⇒
2
2
x +y = (x (x - 1)
2
2
This crosses the y-axis at y = ± 1 and the x-axis at x = ½ Any point here D(1, -1) -1 is closer to AD (extended) than O.
Fig - 18
We will now have to draw such a parabola parabol a for each edge of the square since we want that tha t each edge should satisfy the given constraint, that is any point in the required region should lie closer to the origin origi n than to any edge:
Maths / Areas and Volumes
LOCUS
19
y
(-1, -1)
(1, 1) C 3
X
M
The shaded region is the region in which any point x
N Y
O
is closer to the origin than to any edge
C 4 (-1, 1)
(1, -1) C 1
C 2 Fig - 19
Verify that the equations equat ions of the other three thr ee bounding parabolas (apart from f rom C 1 ) are : 2
= 2x + 1
C3 : x 2
= 2 y +1
C2 : y
C4 : x
= −2 y + 1
2
To calculate the required requi red area, let us calculate calculat e the shaded region in the first quadrant. quadr ant. We We can divide this area into two parts: area (region OXMN ) and area (region NMY ). ). We first need the intersection inter section point M and its x-co-ordinate N . This can be found out by equating the equations of the curves C1 and C 4 . However, we may directly observe that the point M lies symmetrically symmetrical ly w.r.t w.r.t the x-axis and the y-axis, and therefore, it will have equal x and y co-ordinates. co-ordi nates. We may now use this fact in either of the two equations C1 or C 4 : C4 : x
Thus, the point M is
2
+ 2 y −1 = 0
⇒ x 2 + 2 x − 1 = 0
(∵ x = y for M)
⇒ x =
( we want x > 0)
( 2 − 1,
2 −1
)
2 −1 .
We now proceed to evaluate the t he area. A
1st 1st quad quad
=
2 −1
∫ 0
1 1 − x 2 2 d+x ∫ 2 −1 ↑
Equation of C 4
Maths / Areas and Volumes
1− 2
x dx
↑ Equation of C 1
LOCUS
20 2 −1
1 x x3 1 3/ 2 x = − − ((1 − 2 ) ) 2 −1 3 2 6 0 3 ( 2 − 1) 1 1 = 2 −1 − − 0 − (3 − 2 2 3 3
(
= =
(
) (
2 −1
−
2
3
) (
2 −1 6
+
2
3/ 2
)
)
3
)
2 −1 3
( 2 − 1)(3 − 2 ) 3
Thus, the total required area is A =
4
( 2 − 1)(3 − 2 ) 3
Example – 17
(
)
1/ 3 be the vertices of a triangle. Let R be the region consisting of all those points Let O (0, 0), A(2, 0) and B 1, 1/ P inside
∆OAB which satisfy d ( P, OA) ≤ min {d ( P, OB ), d ( P,
AB )} , where d denotes the distance from the
point to the corresponding line. lin e. Sketch the area R and find its area. Solution: Like the last question, this question too is more about correctly plotting the required region rather than calculating area. (In fact, we need not even use definite integration here since we’ll see that the region R is bounded by straight lines).
Consider the triangle OAB as described in the question: y
B(1, 1
Consider a point P which is closer to OA than to either of OB or AB.
3)
P
x
0
A(2, 0)
Fig - 20
We need to find out the region r egion in which P can possibly lie.
Maths / Areas and Volumes
LOCUS
21
Suppose that P was equidistant from OA and OB. Then P would lie on the angle bisector of ∠ AOB. Since P is actually closer to OA, it lies ‘below’ the angle bisector, as shown in the next figure: y Any point in this region is closer to OB (or its extended line) than OA (extended)
B
Any point on this line is equidistant from OA (extended) and OB (extended). 30º 15º
x A Any point in this region is closer to OA (extended) than to OB(extended) . This is the region we are looking for.
0
Fig - 21
Since P also satisfies the constraint that it lies closer to OA than to AB, P will lie below the angle bisector of ∠ OAB too. Thus, P lies in the following region y
1 B(1, 3
Angle bisector of ∠OAB
) Angle bisector of ∠AOB
X
15º
P can lie anywhere inside the shaded region
15º
0
Fig - 22
x A(2, 0)
It is a matter of simple simpl e geometry now to evaluate the required requir ed area which is the area that this is
3 −1
∆ OAX . Verify
sq. units.
3 + 1
Example – 18 Let b ≠ 0 and for j = 0, 1, 2 ......, n, let S j be the area of the region bounded by the y-axis and the curve xeay
= sin by by,, jπ ≤ y ≤ ( j + 1)π . Show that S0, S1, S2 ......, S n are in G.P. b
b
Also, find their sum for a = –1 and b = π . Solution: The equation of the curve is − x = e ay sin by x versus y) will be If you recall our discussions on plotting plotti ng graphs, you will realise that the graph (of x
oscillatory in nature, bounded by an envelope given by e−ay . The zeroes of this function are given by. by. by = jπ j∈" jπ j ∈" ⇒ y = b Maths / Areas and Volumes
LOCUS
22
Thus, we can now plot x versus y as follows: x The shaded areas repr eprese esents nts S 0, S1, S2 and so on.(Thus S j is defined defined for jπ j + π y ) b b
The envelope of the function S
0
2π b
0
π
2
3π b
y
S
3
S
b
S
1
Fig - 23
Now we proceed to calculate S j: ( j +1)π / b
S j
∫
=
e− sin by dy ay
jπ / b
( j +1)π
−ay
e = 2 2 ( −a sin by − b cos by ) a + b j
b
π
b
we used integration by parts to evaluate the indefinite integral; verify this step
− ba ( j +1) − ba j e j +1 e j × ( −b ) × ( −1) − 2 2 × ( −b ) × ( −1) = 2 2 + a b a +b π
=
be a
−
2
jaπ / b
+b
2
π
(1 + e−
aπ / b
)
Thus, S j +1 S j
= e−a / b π
which is a constant
⇒ The series S0, S1, S2 ...... Sn is a G.P. For a = −1 and b = π , we have from (1): S j
=
π
1 + π
2
e j (1 + e )
+ e ) en+1 − 1 ⇒ ∑ S j = 2 π + 1 e − 1 j =0 n
Maths / Areas and Volumes
π (1
.... (1)
LOCUS
23
Example – 19 Find the lateral surface area and the volume of a cone of height H and base radius R. Solution: This question is an easy application question of definite integration. integr ation. Let us evaluate the surface area first.
x Take a variable vari able x as depicted along the slant of the cone. H
We analyse an infinitesmally small strip of width dx along this slant now.
Magnify
R
Fig - 24 Note from the geometry of r
d x
the cone that r R = x L where L is the slant height of the cone. (L= R 2 + H 2 ) The surface area of this eleme el ementa ntall strip strip is is 2 πrd rdxx
Fig - 25
From the explanation given in the figure: L
=A ∫ 2π
Lateral surface area
rdx
0
=
2π R L
L
∫ xdx 0
= π RL To calculate the volume, we again use the elemental strip analysed in Fig -25. The thickness of this strip is not dx but dx cosθ where cos θ =
Maths / Areas and Volumes
H L
LOCUS
24
θ d x
Actual thickness = dx cos
θ
Fig 26 2 The volume of this strip is therefore π r dx cos θ . Thus, the total volume is L
V
= ∫ π r 2dx cosθ 0
π
R 2
= ⋅
2
L L
= =
2
0
H L3
3
3
⋅
L 3
∫ x dx
2
π R
1
L
π R
2
H
Example – 20 An ellipse
x
2
a2
+
y
2
b2
= 1 (where a > b) is rotated about its major axis. Find the volume generated.
Solution: An ellipse, when rotated about its major axis, will form an “egg-shaped” volume. (It won’t be exactly egg-shaped though! It will be symmetric about its centre plane (a plane through the line y = 0 perpendicular to the x - y plane) while an egg is not exactly symmetric in this way). y
The solid shape formed by rotating the ellipse about its major axis
b
a
Fig - 27
Maths / Areas and Volumes
x
LOCUS
25
We take an elemental strip stri p of thickness dx at a distance x from the origin. y dx
y x
x
Fig - 27
As is clear from the figure, the volume of this elemental strip is dV
= π y 2 dx
= π b
2
x2 1 − a 2 dx
Thus, the volume generated of the right half would be : a
Vhalf
a
= ∫ dV = ∫ π b 0
0
2
x2 1 − a 2 dx a
x 3 2 = π b x − 2 3a 0 a = π b2 a − 3
=
2π ab2 3
The total volume is therefore V
= 2 × V half =
4π ab2 3
Now,, try calculating the surface area of this volume generated. Now
Maths / Areas and Volumes
LOCUS
26
TRY YOURSELF - 1
Q. 1
Show that the area of the ellipse
x 2 a2
+
y2 b2
2π
Q. 2
If S1 =
= 1 is π ab . 2π
∫0 max (sin x, cos x ) dx and S = ∫ min(sin x, cos x) dx , show that S = S . 2
1
0
2
Q. 3
Prove that the area of the t he region represented by x− 2 y + x+ 2 y ≤ 8 and xy≥ 2 is 4(3 − ln 4) .
Q. 4
2 Prove that the area enclosed by the region R ≡ ( x, y) : y ≤ x ≤ y
Q. 5
{
Find the surface area of the volume generated by rotating the ellipse axis. Hence find the surface area of a sphere of radius r .
Maths / Areas and Volumes
} is x 2 a2
1 3
+
.
y2 b2
= 1 (a > b) about its major
