MA 105 : Calculus Division 1, Lecture 11srg/courses/autumn2019/MA105-D1/MA105-20… · the y-axis....

MA 105 : Calculus

Division 1, Lecture 11

Prof. Sudhir R. GhorpadeIIT Bombay

Prof. Sudhir R. Ghorpade, IIT Bombay MA 105 Calculus: Division 1, Lecture 11

Recap of the previous lecture

The natural logarithm function and its properties

The exponential function and its properties

Real powers of a positive real number

Definition of the arctangent function via integration, andits properties

A self-contained definition of using the arctangentfunction

An outline of the development of trigonometric functionsbeginning with arctan

Applications of integration

Area of planar regions between two curves. ExamplesReview of Polar coordinatesArea of the sector of a disk


Area of a sector of a diskLet 0 ≤ ϕ ≤ π/2, and let R denote the sector of a disc ofradius a, marked by the points (0, 0), (a, 0) and(a cosϕ, a sinϕ), that is, the region bounded by the curvesx = (cotϕ)y and x =

√a2 − y 2 for y ∈ [0, a sinϕ], and by

the x-axis.

0

b

b

b

a

x

=

y

o

t

'

(a os'; a sin')

x =

p

a

2

� y

2

x

y

Then Area (R) =

∫ a sinϕ

0

(√a2 − y 2 − (cotϕ)y

)dy =

a2ϕ

2.

By symmetry, this result holds for ϕ ∈ (π/2, π] as well.Prof. Sudhir R. Ghorpade, IIT Bombay MA 105 Calculus: Division 1, Lecture 11

Curves given by Polar Equations

Let R denote the region bounded by the curve r = p(θ) andthe rays θ = α, θ = β, where −π ≤ α < β ≤ π. Thus

R := {(r cos θ, r sin θ) : α ≤ θ ≤ β and 0 ≤ r ≤ p(θ)}.Suppose p : [α, β]→ R is integrable. (If α = −π and β = π,then we suppose p(−π) = p(π).)

Partition [α, β] into α = θ0 < θ1 < · · · < θn = β.

Pick sample points γi ∈ [θi−1, θi ] for i = 1, . . . , n.

Area between the rays θ = θi−1 and θ = θi isapproximated by the area of a sector of a disc of radius

ri := p(γi), that is, byp(γi)

2 (θi − θi−1)

2.

The sum of areas of these sectors is a Riemann sum,

namelyn∑

i=1

p(γi)2 (θi − θi−1)

2.


We define

Area (R) :=1

2

∫ β

α

p(θ)2dθ =1

2

∫ β

α

r 2dθ.

Examples: (i) Let a > 0. Area of the disc enclosed by thecircle r = a is equal to 1/2

∫ π−π a

2dθ = πa2.

(ii) Let a > 0, and let R denote the region enclosed by thecardioid r = a(1 + cos θ). Then

Area (R) =1

2

∫ π

−πa2(1 + cos θ)2dθ =

3a2π

2.

(iii) Let R denote the region that lies inside the circler = 3 sin θ and outside the cardioid r = 1 + sin θ, whereθ ∈ [0, π]. Now 3 sin θ = 1 + sin θ ⇐⇒ θ ∈ {π/6, 5π/6}, and1 + sin θ ≤ 3 sin θ if θ ∈ [π/6, 5π/6]. Hence

Area (R) =1

2

∫ 5π/6

π/6

((3 sin θ)2 − (1 + sin θ)2

)dθ = π.


Volume of a solid

Let D be a bounded subset of R3. A cross-section of Dobtained by cutting D by a plane in R3 is called a slice of D.

Let a < b, and suppose D lies between the planes x = a andx = b, which are perpendicular to the x-axis. For s ∈ [a, b],consider the slice of D by the plane x = s, namely{(x , y , z) ∈ D : x = s}, and suppose it has an ‘area’ A(s).

To find the volume of D, we proceed as follows.

Partition [a, b] into a = x0 < x1 < · · · < xn = b.

Pick sample points si ∈ [xi−1, xi ] for i = 1, . . . , n.

Volume between the planes x = xi−1 and x = xi isapproximated by the volume of a rectangular slab of widthxi − xi−1 and base area A(si), that is, by A(si)(xi − xi−1).

The sum of volumes of these slabs isn∑

i=1

A(si)(xi − xi−1).


Slice Method: We define the volume of D by

Vol (D) :=

∫ b

a

A(x) dx ,

provided the ‘area function’ A : [a, b]→ R is integrable.

Examples(i) If D is a cylinder with cross-sectional area A and height h,then Vol (D) = Ah. (The ‘area function’ is the constant A.)

(ii) Let a > 0, and let D denote the solid enclosed by thecylinders x2 + y 2 = a2 and x2 + z2 = a2. Then D lies betweenthe planes x = −a and x = a. For s ∈ [−a, a], the slice{(x , y , z) ∈ D : x = s} is the square{

(s, y , z) ∈ R3 : |y | ≤√a2 − s2 and |z | ≤

√a2 − s2

},

and so A(s) =(2√a2 − s2

)2= 4 (a2 − s2). Hence

Vol (D) =

∫ a

−aA(s)ds = 4

∫ a

−a

(a2 − s2

)ds =

16a3

3.


b

bb

y z

x

y

z

√a2 − s2−

√a2 − s2

√a2 − s2

−√a2 − s2

b

bb

b

Figure: Solid enclosed by two cylinders and a slice resulting in asquare region


Solids of Revolution

If a subset D of R3 is generated by revolving a planar regionabout an axis, then D is known as a solid of revolution.

Examples

Let a > 0. The spherical ball

{(x , y , z) ∈ R3 : x2 + y 2 + z2 ≤ a2}is obtained by revolving the semidisc

{(x , y) ∈ R2 : x2 + y 2 ≤ a2 and y ≥ 0}about the x-axis.

Let b > 0. The cylindrical solid

{(x , y , z) ∈ R3 : x2 + z2 ≤ b2 and 0 ≤ y ≤ h}is obtained by revolving the rectangle [0, b]× [0, h] aboutthe y -axis.


Volume of a Solid of Revolution: Washer Method

Let D be the solid obtained by revolving the regionbetween the curves y = f1(x) and y = f2(x), and the linesx = a and x = b, about the x-axis, where 0 ≤ f1 ≤ f2.

The slice of D at x ∈ [a, b] looks like a circular washer,that is, a disc of radius f2(x) from which a smaller disc ofradius f1(x) has been removed, and so the area of theslice is A(x) := π (f2(x)2 − f1(x)2).

Suppose f1 and f2 are integrable on [a, b]. Then the areafunction A is integrable on [a, b], and by the slice method,

Vol (D) =

∫ b

a

A(x)dx = π

∫ b

a

(f2(x)2 − f1(x)2

)dx .

This special case of the slice method is called the washermethod.


If the inner radius of a washer is equal to 0, then the washer isin fact a disk. If this is the case for every x ∈ [a, b], then thewasher method is also known as the disk method.

y = f

2

(x)

y = f

1

(x)

x

y

z

a

b

Figure: Illustrations of a disk and of a washer


Examples

(i) Let D denote the solid obtained by rotating the regionbetween the curves y = x and y = x2 about the x-axis.

Let f1(x) := x2 and f2(x) := x for x ∈ [0, 1]. The curvesy = f1(x) and y = f2(x) intersect at x = 0 and x = 1,and 0 ≤ f1 ≤ f2 on [0, 1].

By the washer method,

Vol (D) = π

∫ 1

0

((x)2−(x2)2

)dx = π

∫ 1

0

(x2−x4)dx =2π

15.


(ii) Let D denote the spherical ball with centre at (0, 0, 0),and radius a > 0. Then D is obtained by revolving thesemidisc {(x , y) ∈ R2 : x2 + y 2 ≤ a2 and y ≥ 0} about thex-axis.

Let f1(x) := 0 and f2(x) :=√a2 − x2 for x ∈ [−a, a].

The curves y = f1(x) and y = f2(x) intersect at x = −aand x = a, and 0 ≤ f1 ≤ f2 on [−a, a].

By the disc method,

Vol (D)=π

∫ a

−a((√

a2 − x2)2−02)dx =π

∫ a

−a(a2−x2)dx =

4πa3

3.


Let D be the solid obtained by revolving the region betweenthe curves x = g1(y) and x = g2(y), c ≤ y ≤ d , about they -axis, where 0 ≤ g1 ≤ g2. Then, as before,

Vol (D) = π

∫ d

c

(g2(y)2 − g1(y)2

)dy .

Example

Let D denote the solid obtained by revolving the region in thefirst quadrant between the parabolas y = x2 and y = 2− x2

about the y -axis. Now√y =√

2− y ⇐⇒ y = 1. By thedisk method,

Vol (D) = π

∫ 1

0

(√y)2dy + π

∫ 2

1

(√2− y

)2dy = π

(1

2+

1

2

)= π.


Volume of a Solid of Revolution: Shell Method

Let D be a bounded subset of R3. A cross-section of Dobtained by piercing a cylinder throughD is called a sliver of D.

Let f1, f2 : [a, b]→ R be such that f1 ≤ f2, and suppose0 ≤ a < b. Let D denote the solid generated by revolvingthe region between the curves y = f1(x) and y = f2(x),and the lines x = a and x = b about the y-axis.For s ∈ [a, b], the sliver {(x , y , z) ∈ D : x2 + z2 = s2} ofD by the cylinder x2 + z2 = s2 is a right circular cylinderhaving height f2(s)− f1(s) and radius s, and so itssurface area is 2πs

(f2(s)− f1(s)

). (To be justified later)

Suppose f1 and f2 are integrable on [a, b]. Since D is‘made up’ of these cylindrical slivers (or shells), we define

Vol (D) := 2π

∫ b

a

x(f2(x)− f1(x)

)dx ,

given by the shell method.Prof. Sudhir R. Ghorpade, IIT Bombay MA 105 Calculus: Division 1, Lecture 11

y = f2(x)

y = f1(x)

a b

y

x

z

Figure: Illustration of the Shell Method

Vol (D) := 2π

∫ b

a

x(f2(x)− f1(x)

)dx .


Shell Method (continued)

Let g1, g2 : [c , d ]→ R be integrable functions such thatg1 ≤ g2, and suppose 0 ≤ c < d . Let D denote the solidobtained by revolving the region between the curves x = g1(y)and x = g2(y) and between the lines y = c and y = d aboutthe x-axis.

For t ∈ [c , d ], the sliver {(x , y , z) ∈ D : y 2 + z2 = t2} of D bythe cylinder y 2 + z2 = t2 is a right circular cylinder havingheight g2(t)− g1(t) and radius t, and so its surface area is2πt(g2(t)− g1(t)

).

Since D is ‘made up’ of these cylindrical slivers (or shells), wedefine

Vol (D) := 2π

∫ d

c

y(g2(y)− g1(y)

)dy .


Examples

(i) Let D denote the solid obtained by revolving the regionbounded by the curves y = 2x2 − x3 and y = 0 about they -axis. Now 2x2 − x3 = 0 ⇐⇒ x = 0 or x = 2. By the shellmethod,

Vol (D) = 2π

∫ 2

0

x(2x2 − x3 − 0)dx =56π

5.

(ii) Let D denote the solid obtained by revolving the rectangle[0, b]× [0, h] about the y -axis. By the shell method,

Vol (D) = 2π

∫ b

0

x (h − 0) dx = πhb2.

(iii) Let E denote the solid obtained by revolving the rectangle[0, b]× [0, h] about the x-axis. By the shell method,

Vol (E ) = 2π

∫ h

0

y (b − 0) dy = πbh2.


(iv) Let D denote the solid in R3 obtained by revolving theregion in the first quadrant bounded by the curve y = x3 andthe line y = 4x about the line x = −1.

Now in the first quadrant, x3 = 4x ⇐⇒ x = 0 or x = 2.

By the washer method,

Vol (D) = π

∫ 8

0

((y 1/3 + 1

)2 − ((y/4) + 1)2)

dy

= π

∫ 8

0

(y 2/3 + 2y 1/3 − (y 2/16)−(y/2)

)dy =

248π

15.

By the shell method,

Vol (D) = 2π

∫ 2

0

(x + 1)(4x − x3

)dx

= 2π

∫ 2

0

(4x + 4x2 − x3 − x4) dx =248π

15.


0−1bb b

b

b

b

x 2

b

b

b b by

8

y = x3

y = 4x


Remarks on Volume of a Solid of Revolution

Let D denote a bounded solid of revolution.

Remark 1. In the washer method, the slices (which look likewashers) are taken perpendicular to the axis of revolution. Onthe other hand, in the shell method, the slivers (which look likecylindrical shells) are taken parallel to the axis of revolution.

Remark 2. The basic expression in the washer method isπ(r 22 − r 21 ), where r2 and r1 are outer and inner radii of thewasher. The basic expression in the shell method is2πr(h2 − h1), where r is the radius of the sliver, while h2 − h1is the height of the sliver.

Remark 3. The volume of D found by the washer method andby the shell method must be the same! This result wouldfollow from a general definition of the volume of a solid in R3.This can serve as a check on your calculations.


Parametrized Curve

A parametrized curve or a path C in R2 is given by(x(t), y(t)), where x , y : [α, β]→ R are continuous functions.

Here [α, β] is called the parameter interval.

We wish to define the ‘length’ of C .

Basic assumption: The (Euclidean) length of a line segmentjoining points (x1, y1) and (x2, y2) in R2 is equal to√

(x2 − x1)2 + (y2 − y1)2.

We shall assume that C is smooth, that is, the functions x , yare continuously differentiable on [α, β]. This means thatx , y are differentiable on [α, β], and their derivatives x ′, y ′ arecontinuous on [α, β].


MA 105 : Calculus Division 1, Lecture 11srg/courses/autumn2019/MA105-D1/MA105-20… · the y-axis....

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