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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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 2 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 3 / 21
(∆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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 4 / 21
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.
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 5 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 6 / 21
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
)
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 7 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 8 / 21
√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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 9 / 21
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)
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 11 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 12 / 21
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
Use u substitution. Let
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 13 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 14 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 15 / 21
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)
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 16 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 17 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 19 / 21
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
Calculus with Algebra and Trigonometry II Lecture 18Arc length and surface areaApr 7, 2015 20 / 21
Use u substitution. Let
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 18Arc length and surface areaApr 7, 2015 21 / 21