MA 105: Calculus Lecture 10 . Prof. B.V. Limaye IIT Bombay - - PowerPoint PPT Presentation

. .

MA 105: Calculus Lecture 10

• Prof. B.V. Limaye

IIT Bombay Friday, 3 February 2017

B.V. Limaye, IITB MA 105: Lec-10

. . Area under a Curve

Let f : [a, b] → R be a bounded function. Suppose f ≥ 0 on [a, b], and let Rf := {(x, y) ∈ R2 : a ≤ x ≤ b and 0 ≤ y ≤ f (x)}. We say that Rf has an area if f is Riemann integrable, and then we define Area (Rf ) := ∫ b

a

f (x)dx. If f : [a, b] → R is any function, then f = f + − f −, where f + = |f | + f 2 and f − = |f | − f 2 . Note that f + ≥ 0 and f − ≥ 0.

B.V. Limaye, IITB MA 105: Lec-10

. . Positive and Negative Parts of a Function

In fact, for x ∈ [a, b], f +(x) = max{f (x), 0} and f −(x) = − min{f (x), 0}. The functions f + and f − are known as the positive part of f and the negative part of f , respectively. Clearly, f is bounded if and only if f + and f − are both bounded. Also, f is integrable if and only if f + and f − are both integrable, and then ∫ b

a

f (x)dx = ∫ b

a

f +(x)dx − ∫ b

a

f −(x)dx = Area (Rf +) − Area (Rf −), which can be called the ‘signed area’ delineated by the curve y = f (x), x ∈ [a, b].

B.V. Limaye, IITB MA 105: Lec-10

. . Area between Curves

Let f1, f2 : [a, b] → R be integrable functions such that f1 ≤ f2. Let R := {(x, y) ∈ R2 : a ≤ x ≤ b and f1(x) ≤ y ≤ f2(x)} be the region between the curves y = f1(x) and y = f2(x). Define Area (R) := Area (Rf2−f1) = ∫ b

a

( f2(x) − f1(x) ) dx. Letg1, g2 : [c, d]→R be integrable functions such that g1 ≤ g2. Let R := {(x, y) ∈ R2 : c ≤ y ≤ d and g1(y) ≤ x ≤ g2(y)} be the region between the curves x = g1(y) and x = g2(y). Define Area (R) := ∫ d

c

( g2(y) − g1(y) ) dy. If two curves cross each other, then we must find areas of several regions between them separately, and add them up.

B.V. Limaye, IITB MA 105: Lec-10

a b y = f2(x) y = f1(x) x y

d x = g1(y) x = g2(y) c x y

b b

x y x = g1(y) x = g2(y) c d

B.V. Limaye, IITB MA 105: Lec-10

. . Examples

(i) Let R denote the region enclosed by the loop of the curve y 2 = x(1 − x)2, that is, the region bounded by the curves y = −√x(1 − x) and y = √x(1 − x). Now √x(1 − x) = −√x(1 − x) ⇐ ⇒ x = 0 or 1. Hence Area (R) = ∫ 1 (√x(1 − x) − (−√x(1 − x)) ) dx = 8 15. (ii) Let R denote the region bounded by the curves x = −2y 2 and x = 1 − 3y 2. Now −2y 2 = 1 − 3y 2 ⇐ ⇒ y = ±1, and −2y 2 ≤ 1 − 3y 2 if y ∈ [−1, 1]. Hence Area (R) = ∫ 1

−1

( 1 − 3y 2 − (−2y 2) ) dy = ∫ 1

−1

(1 − y 2) dy = 4 3.

B.V. Limaye, IITB MA 105: Lec-10

. . Polar coordinates

Review: The function cos−1 : [−1, 1] → [0, π] is one-one and onto. Let P := (x, y) ̸= (0, 0). There are unique r, θ ∈ R such that r > 0, θ ∈ (−π, π], x = r cos θ and y = r sin θ. In fact, r := √ x2 + y 2 and θ := { cos−1(x/r) if y ≥ 0, − cos−1(x/r) if y < 0. (If y < 0, then |x/r| < 1, and − cos−1(x/r) ∈ (−π,0).) The pair (r, θ) is defined as the polar coordinates of P.

B.V. Limaye, IITB MA 105: Lec-10

. . Examples

Example: Let 0 < ϕ ≤ π/2, and let R denote the sector of a disc of radius a, marked by the points (0, 0), (a, 0) and (a cos ϕ, a sin ϕ), that is, the region bounded by the curves x = (cot ϕ)y and x = √ a2 − y 2 for y ∈ [0, a sin ϕ].

b b b

• t

• s

• y

Then Area (R) = ∫ a sin ϕ (√ a2 − y 2 − (cot ϕ)y ) dy = a2ϕ 2 . = a2θ.

B.V. Limaye, IITB MA 105: Lec-10

. . Curves given by Polar Equations

Let R denote the region bounded by the curve r = p(θ) and the 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 is approximated by the area of a sector of a disc of radius ri := p(γi), that is, by p(γi)2 (θi − θi−1) 2 . The sum of areas of these sectors is

n

∑

i=1

p(γi)2 (θi − θi−1) 2 .

B.V. Limaye, IITB MA 105: Lec-10

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 the circle r = a is equal to 1/2 ∫ π

−π a2dθ = πa2.

(ii) Let a > 0, and let R denote the region enclosed by the cardioid r = a(1 + cos θ). Then Area (R) = 1 2 ∫ π

−π

a2(1 + cos θ)2dθ = 3a2π 2 . (iii) Let R denote the region that lies inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ, where θ ∈ [0, π]. Now 3 sin θ = 1 + sin θ ⇐ ⇒ θ = π/6 or 5π/6, and 1 + sin θ ≤ 3 sin θ if θ ∈ [π/6, 5π/6]. Hence Area (R) = 1 2 ∫ 5π/6

π/6

( (3 sin θ)2 dθ − (1 + sin θ)2) dθ = π.

B.V. Limaye, IITB MA 105: Lec-10

. . Volume of a solid

Let D be a bounded subset of R3. A cross-section of D by a plane in R3 is called a slice of D. Let a < b, and suppose D lies between the planes x = a and x = 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 is approximated by the volume of a rectangular slab of width xi − xi−1 and base area A(si), that is, by A(si)(xi − xi−1). The sum of volumes of these slabs is

n

∑

i=1

A(si)(xi − xi−1).

B.V. Limaye, IITB MA 105: Lec-10

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 the cylinders x2 + y 2 = a2 and x2 + z2 = a2. Then D lies between the 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

−a

A(s)ds = 4 ∫ a

−a

( a2 − s2) ds = 16a3 3 .

B.V. Limaye, IITB MA 105: Lec-10

b b b

y z x y z √ a2 − s2 − √ a2 − s2 √ a2 − s2 − √ a2 − s2

b b b b

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

B.V. Limaye, IITB MA 105: Lec-10

. . Solids of Revolution

If a subset D of R3 is generated by revolving a planar region about an axis, then D known as a solid of revolution. Examples 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. The cylindrical solid {(x, y, z) ∈ R3 : x2 + z2 ≤ b2 and 0 ≤ y ≤ h} is obtained revolving the rectangle [0, b] × [0, h] about the y-axis.

B.V. Limaye, IITB MA 105: Lec-10

. . Volume of a Solid of Revolution: Washer Method

Let D be the solid obtained by revolving the region between the curves y = f1(x) and y = f2(x), and the lines x = 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 of radius f1(x) has been removed, and so the area of the slice is A(x) := π (f2(x)2 − f1(x)2). Suppose f1 and f2 are integrable on [a, b]. Then the area function 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.

B.V. Limaye, IITB MA 105: Lec-10

If the inner radius of a washer is equal to 0, then the washer is in fact a disk. If this is the case for every x ∈ [a, b], then the washer method is called the disk method.

Figure : Illustrations of a disk and of a washer

B.V. Limaye, IITB MA 105: Lec-10

Examples (i) Let D denote the solid obtained by rotating the region between the curves y = x and y = x2 about the x-axis. Let f1(x) := x2 and f2(x) := x for x ∈ [0, 1]. The curves y = 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 ( (x)2 − (x2)2) dx = 2π 15 .

B.V. Limaye, IITB MA 105: Lec-10

(ii) Let D denote the spherical ball with centre at (0, 0, 0), and radius a > 0. Then D is obtained by revolving the semidisc {(x, y) ∈ R2 : x2 + y 2 ≤ a2 and y ≥ 0} about the x-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 = −a and x = a, and 0 ≤ f1 ≤ f2 on [−a, a]. By the disc method, Vol (D) = π ∫ a

−a

((√ a2 − x2)2 − 02) dx = 4πa3 3 .

B.V. Limaye, IITB MA 105: Lec-10

Let D be the solid obtained by revolving the region between the curves x = g1(y) and x = g2(y), c ≤ y ≤ d, about the y-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 the first quadrant between the parabolas y = x2 and y = 2 − x2 about the y-axis. Now √y = √2 − y ⇐ ⇒ y = 1. By the disk method, Vol (D) = π ∫ 1 (√y)2dy + ∫ 2

1

(√ 2 − y )2 dy = π (1 2+1 2 ) = π.

B.V. Limaye, IITB MA 105: Lec-10