Calculus with Algebra and Trigonometry II Lecture 18 Arc ...frooney/M217_18_ArcSA.pdf · Calculus...

Calculus with Algebra and Trigonometry IILecture 18

Arc length and surface area

Apr 7, 2015

Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 1 / 21

Parametric equations revisited

Recall from earlier in the semester a curve is in parametric form if x and yare given as functions of another variable t, the parameter.

x = x(t) y = y(t)

If the curve is given in the form y = f (x) it can be considered as a specialcase with x as the parameter

For example a circle of radius r can be expressed in parametric form as

x = r cos t y = r sin t 0 ≤ t ≤ 2π

sincex2 + y2 = (r cos t)2 + (r sin t)2 = r2


Arc length

To find the length of a curve, s, consider the picture below. For smallvalues of ∆x and ∆y , the arc length is approximately the hypotenuse ofthe triangle.

If the curve is given parametrically

∆x ≈ x ′(t)∆t ∆y ≈ y ′(t)∆t

(∆s)2 ≈ (x ′(t)∆t)2 + (y ′(t)∆t)2 = ((x ′(t))2 + (y ′(t))2)(∆t)2


(∆s)2

(∆t)2≈ (x ′(t))2 + (y ′(t))2

then in the limit as ∆t → 0(ds

dt

)2

=

(dx

dt

)2

+

(dy

dt

)2

ds

dt=√

x ′2 + y ′2

So to calculate the arc length, use the fundamental theorem of calculus

s =

∫ √x ′2 + y ′2 dt

or if the curve is given in explicit form (y = f (x)) then

s =

∫ √1 + (f ′(x))2 dx


Example 1

As an easy first example we will use the arc length formula to calculate thecircumference of a circle. The circle of radius r is given parametrically as

x = r cos t y = r sin t ⇒ x ′ = −r sin t y ′ = r cos t

then √x ′2 + y ′2 =

√(−r sin t)2 + (r cos t)2 = r

and the circumference is

s =

∫ 2π

0r dt = 2πr

as it should be.


Example 2

Find the length of the arc of the curve y = x3/2 from x = 0 to x = 2.

f ′(x) =3

2x1/2 ⇒

√1 + (f ′(x))2 =

√1 +

9

4x


s =

∫ 2

0

√1 +

9

4x dx

Use u substitution. Let

u = 1 +9

4x ⇒ du =

9

4dx ⇒ dx =

4

9du

x = 0 ⇒ u = 1 x = 2 ⇒ u =11

2

s =

∫ 11/2

1

√u

(4

9du

)=

4

9

[2

3u3/2

]11/21

=8

27

((11

2

)3/2

− 1

)


Example 3

Find the length of the (catenary) curve

y =1

2

(ex + e−x

)− 1 ≤ x ≤ 1

y ′ = f ′(x) =1

2

(ex − e−x

)1 + y ′2 = 1 +

(1

2

(ex − e−x

))2


√1 + y ′2 =

√1 +

1

4(e2x − 2 + e−2x)

=

√1

4(e2x + 2 + e−2x)

=1

2

(ex + e−x

)and the arc length is

s =

∫ 1

−1

1

2

(ex + e−x

)dx

=1

2

[ex − e−x

]1−1

= e − e−1


Example 4

Find the length of the curve

x = et cos t y = et sin t 0 ≤ t ≤ π

x ′ = et(cos t − sin t) y ′ = et(sin t + cos t)

x ′2 + y ′2 = e2t((cos t − sin t)2 + (sin t + cos t)2

)Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 10 / 21

x ′2 + y ′2 = e2t(cos2 t − 2 cos t sin t + sin2 t

)+e2t

(sin2 t + 2 sin t cos t + cos2 t

)= 2e2t

Then √x ′2 + y ′2 =

√2et

and the arc length is

s =

∫ π

0

√2et dt

=√

2(eπ − 1)


Example 5

Find the length of the curve

x = cos3 t y = sin3 t 0 ≤ t ≤ 2π

By the symmetry of the curve we will calculate the arc length in the firstquadrant and multiply by 4.

x ′ = 3 cos2 t(− sin t) y ′ = 3 sint(cos t)

x ′2 + y ′2 = 9 cos4 t sin2 t + 9 sin4 t cos2 t


x ′2 + y ′2 = 9 cos2 t sin2 t(cos2 t + sin2 t)

= 9 cos2 t sin2 t

The arc length will be

s = 4

∫ π2

0

√x ′2 + y ′2 dt =

∫ π2

03 cos t sin t dt


u = sin t ⇒ du = cos t dt ⇒ dt =du

cos t

t = 0 ⇒ u = 0 t =π

2⇒ u = 1

s = 4

∫ 1

03 cos t u

(du

cos t

)= 12

∫ 1

0u du = 12

[1

2u2]10

= 6


Surface Area

A surface of revolution is formed when a curve is rotated about an axis.To find its area consider a tiny piece of the curve with length ds

The piece sweeps out a ring with area

dA = 2π(radius) ds

So the total surface area is

Surface area = 2π

∫radius ds


Example 6

Find the area of the surface obtained when the curve

y2 = 12x 0 ≤ x ≤ 3

is rotated about the x axis.

The radius is y and

y =√

12x y ′ =√

12

(1

2x−1/2

)and

1 + y ′2 = 1 + 12

(1

4x

)= 1 +

3

x


So the surface area is

Surface area = 2π

∫ 3

0y√

1 + y ′2 dx

= 2π

∫ 3

0

√12x

√1 +

3

xdx

= 2π

∫ 3

0

√12x + 36 dx

= 2π

([1

12× 2

3(12x + 36)3/2

]30

)=

π

9

(723/2 − 363/2

)= 24π(23/2 − 1)


Example 7

Find the surface area of a sphere. Rotate a semicircle about the x axis

The radius is y and we will use the parametric form

x = r cos t y = r sin t ⇒ x ′ = −r sin t y ′ = r cos t

then√x ′2 + y ′2 = r and

Surface Area = 2π

∫ π

0r sin t(r dt) = 2πr2 [− cos t]π0 = 4πr2


Example 8

Find the area of the surface formed when the curve

y =1

2

(ex + e−x

)− 1 ≤ x ≤ 1

is rotated about the x axis. The radius is again y and from example 3 weknow √

1 + y ′2 =1

2

(ex + e−x

)Then

Surface area = 2π

∫ 1

−1

(1

2

(ex + e−x

))2

dx

=π

2

∫ 1

−1(e2x + 2 + e−2x) dx

=π

2

[1

2e2x + 2x − 1

2e−2x

]1−1

=π

2

(e2 + 4− e−2

)Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 18 / 21

Example 9

Find the surface area of a torus

We are going to rotate the circle shown above about the y axis. Theradius is x . Use the parametric form

x = R + r cos t y = r sin t ⇒ x ′ = −r sin t y ′ = r cos t

then√

x ′2 + y ′2 = r and

Surface area = 2π

∫ 2π

0(R + r cos t)r dt = 2πr [Rt + r sin t]2π0 = 4π2rR


Example 10

Find the area generated when the curve

x = cos3 t y = sin3 t 0 ≤ t ≤ π

is rotated about the x axis. The radius is x . From example 5 we know√x ′2 + y ′2 = 3 cos t sin t

So

Surface area = 4π

∫ π2

0(sin3 t)3 cos t sin t dt = 4π

∫ π2

0sin4 t cos tdt



u = sin t ⇒ du = cos t dt ⇒ dt =du

cos t

t = 0 ⇒ u = 0 t =π

2⇒ u = 1

Surface area = 4π

∫ 1

03 cos t u4

(du

cos t

)= 12π

∫ 1

0u4 du

= 12π

[1

5u5]10

=12π

5


Calculus with Algebra and Trigonometry II Lecture 18 Arc ...frooney/M217_18_ArcSA.pdf · Calculus...

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