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# Areas of Surface of Revolution and Arc Length Worksheet

Areas of Surface of Revolution and Arc Length Worksheet
• Page 1
1.
Find the arc length of the polar curve $r$ = cos θ, 0 ≤ θ ≤ 2$\pi$.
 a. 2 units b. 4$\pi$ units c. 1 unit d. 2$\pi$ units

#### Solution:

r = cos θ, 0 ≤ θ ≤ 2π
[Equation of the polar curve.]

drdθ = - sin θ
[Differentiate r with respect to θ.]

Arc length, L = 0(cos θ)2+(- sin θ)2 dθ
[Use L = abr2+(drdθ)2 dθ]

= 0cos2 θ+sin2 θ dθ

= 01
[Use cos2 θ + sin2 θ = 1.]

= [θ]0
[Integrate.]

= 2π - 0 = 2π
[Substitute the limits and simplify.]

Arc length of the given polar curve is 2π units.

Correct answer : (4)
2.
Find the arc length of the polar curve $r$ = $e$θ, 0 ≤ θ ≤ 1.
 a. units b. ($e$2 - 1) units c. $e$$\sqrt{2}$ units d. $\frac{{e}^{2}-2e}{\sqrt{2}}$ units e. $\sqrt{2}$($e$ - 1) units

#### Solution:

r = eθ, 0 ≤ θ ≤ 1
[Equation of the polar curve.]

drdθ = eθ
[Differentiate r with respect to θ.]

Arc length, L = 01(eθ)2+(eθ)2 dθ
[Use L = abr2+(drdθ)2 dθ]

= 01e2θ+e2θ dθ

= 012e2θ dθ

= 201 eθ

= 2[eθ]01
[Integrate.]

= 2[e1 - e0]
[Substitute the limits.]

= 2(e - 1)
[Simplify.]

Arc length of the given polar curve is 2(e - 1) units.

Correct answer : (5)
3.
Find the arc length of the polar curve $r$ = $e$- θ, 0 ≤ θ ≤ 1.
 a. $\sqrt{2}\left(\frac{1}{e}-1$) units b. $\frac{{e}^{2}-1}{2{e}^{2}}$ units c. units d. $\sqrt{2}\left(1-\frac{1}{e}$) units e. (1 - $\frac{1}{e}$) units

#### Solution:

r = e- θ, 0 ≤ θ ≤ 1
[Equation of the polar curve.]

drdθ = - e- θ
[Differentiate r with respect to θ.]

Arc length, L = 01(e- θ)2+((- e)- θ)2 dθ
[Use L = abr2+(drdθ)2 dθ]

= 01e- 2θ+e- 2θ dθ

= 012e- 2θ dθ

= 201 e- θ

= - 2[e- θ]01
[Integrate.]

= - 2[e- 1 - e0]
[Substitute the limits.]

= 2(1-1e)
[Simplify.]

Arc length of the given polar curve is 2(1-1e) units.

Correct answer : (4)
4.
Determine the arc length of the cardiod $r$ = 1 + cos θ between 0 and $\pi$.
 a. 2$\sqrt{2}$ units b. 1 unit c. 2 units d. $\pi$$\sqrt{2}$ units e. 4 units

#### Solution:

r = 1 + cos θ, 0 ≤ θ ≤ π
[Equation of the polar curve.]

drdθ = - sin θ
[Differentiate r with respect to θ.]

Arc length, L = 0π(1+cos θ)2+(- sin θ)2 dθ
[Use L = abr2+(drdθ)2 dθ.]

= 0πcos2 θ+1+2cos θ+sin2 θ  dθ

= 0π2+2cos θ dθ
[Use cos2 θ + sin2 θ = 1.]

= 0π21+2cos2θ2-1 dθ
[Use cos θ = 2cos2θ2 - 1.]

= 20πcos θ2 dθ

= 2[2sin θ2]0π
[Integrate.]

= 4[sin π2 - sin 0] = 4
[Substitute the limits and simplify.]

Arc length of the given polar curve is 4 units.

Correct answer : (5)
5.
Find the arc length of the curve $r$ = 1 + sin θ, 0 ≤ θ ≤ $\frac{\pi }{2}$.
 a. (4 - 2$\sqrt{2}$) units b. 2$\sqrt{2}$ units c. 4$\sqrt{2}$ units d. $\sqrt{2}$ units e. $\frac{1}{\sqrt{2}}$ units

#### Solution:

r = 1 + sin θ, 0 ≤ θ ≤ π2
[Equation of the polar curve.]

drdθ = cos θ
[Differentiate r with respect to θ.]

Arc length, L = 0π/2(1+sin θ)2+(cos θ)2 dθ
[Use L = abr2+(drdθ)2 dθ.]

= 0π/21+2sin θ+sin2 θ+cos2 θ  dθ

= 0π/22+2sin θ dθ
[Use cos2 θ + sin2 θ = 1.]

= 0π/22(1+sin θ) dθ

= 0π/221+2sinθ2cosθ2 dθ
[Use sin θ = 2sin θ2cosθ2.]

= 0π/22(sinθ2+cosθ2)2 dθ
[Use (sinθ2+cosθ2)2 = 1 + 2sin θ2cosθ2.]

= 0π/22(sinθ2+cosθ2) dθ

= [2[- 2cosθ2+2sinθ2]0π/2
[Integrate.]

= 22[- cosπ4+sinπ4 - (- cos 0 + sin 0)] = 22
[Substitute the limits and simplify.]

Arc length of the given polar curve is 22 units.

Correct answer : (2)
6.
Find the area of surface formed by revolving the circle $r$ = 2 cos$\theta$, 0 ≤ $\theta$$\frac{\pi }{2}$ about polar axis.
 a. $\pi$ b. $\frac{\pi }{2}$ c. 4$\pi$ d. $\frac{3\pi }{2}$ e. 8$\pi$

#### Solution:

r = f(θ) = 2 cosθ
[Write the polar equation.]

If the derivative of the function f(θ) is continuous on the interval α ≤ θ ≤ β, then the area of the surface formed by revolving the graph of f(θ) from θ = α to θ = β about the polar axis is
s = 2π αβ f(θ) · sinθ · (f(θ))2+(f(θ))2

f ′(θ) = - 2 sinθ is continuous on the interval [0, π2].

Surface area of revolution = 2π 0π/2 4 cosθ · sinθ · cos2θ+sin2θ dθ

= 4π 0π/2 sin 2θ dθ

= 4π[- cos 2θ]0π/2 = 4π

So, the area of surface formed by revolving the circle r = 2 cosθ is 4π.

Correct answer : (3)
7.
Find the area of surface formed by revolving the curve $r$ = sin$\theta$ 0 ≤ $\theta$$\frac{\pi }{2}$ about the line $\theta$ = $\frac{\pi }{2}$.
 a. $\frac{\pi }{2}$ b. 2$\pi$ c. 4$\pi$ d. $\pi$ e. $\frac{3\pi }{2}$

#### Solution:

r = f(θ) = sinθ
[Write the polar equation.]

If the derivative of the function f(θ) is continuous on the interval α ≤ θ ≤ β, then the area of the surface formed by revolving the graph of f(θ) from θ = α to θ = β about the line θ = π2 is
s = 2π αβ f(θ) · cosθ · (f(θ))2+(f(θ))2

f ′(θ) = cosθ is continuous in the interval [0, π2].

Surface area of revolution = 2π 0π/2 sinθ cosθ sin2θ+cos2θ dθ

= π 0π/2 sin 2θ dθ

= π [-cos2θ2]0π/2

= π2 (2)

= π

So, the area of surface formed by revolving the curve r = sinθ is π.

Correct answer : (4)
8.
Find the area of surface formed by revolving the curve $r$ = $e$$\theta$, 0 ≤ $\theta$$\frac{\pi }{4}$ about the line $\theta$ = $\frac{\pi }{2}$.
 a. $\frac{2\pi }{5}\left({3e}^{\frac{\pi }{2}}+2\sqrt{2}\right)$ b. $\frac{\pi }{5}\left({3e}^{\frac{\pi }{2}}+2\sqrt{2}\right)$ c. $\frac{2\pi }{5}\left({3e}^{\frac{\pi }{2}}$ + 2) d. $\frac{\pi }{5}\left({3e}^{\frac{\pi }{2}}-2\sqrt{2}\right)$ e. $\frac{2\pi }{5}\left({3e}^{\frac{\pi }{2}}-2\sqrt{2}\right)$

#### Solution:

r = f(θ) = eθ
[Write the polar equation.]

If the derivative of the function f(θ) is continuous on the interval α ≤ θ ≤ β, then the area of the surface formed by revolving the graph of f(θ) from θ = α to θ = β about the line θ = π2 is
s = 2π αβ f(θ) · cosθ · (f(θ))2+(f(θ))2

f ′(θ) = eθ is continuous in the interval [0, π4].

Surface area of revolution = 2π 0π/4 eθ cosθ e2θ+e2θ dθ

= 22π 0π/4 e2θ cosθ dθ

= 22π5 [2 e2θ cosθ + e2θ sinθ]0π/4

= 22π5(2eπ2+eπ212 - 2)

= 2π5(3eπ2-22)

So, the area of surface formed by revolving the curve r = eθ is 2π5(3eπ2-22).

Correct answer : (5)
9.
Find the area of surface formed by revolving the curve $r$ = 1, 0 ≤ $\theta$$\frac{\pi }{2}$ about polar axis.
 a. $\sqrt{2}\pi$ b. 4$\pi$ c. 8$\pi$ d. 4$\sqrt{2}\pi$ e. 2$\pi$

#### Solution:

r = f(θ) = 1
[Write the polar equation.]

If the derivative of the function f(θ) is continuous on the interval α ≤ θ ≤ β, then the area of the surface formed by revolving the graph of f(θ) from θ = α to θ = β about the polar axis is
s = 2π αβ f(θ) · sinθ · (f(θ))2+(f(θ))2

f ′(θ) = 0 is continuous on the interval [0, π2].

Surface area of revolution = 2π 0π/2 sinθ dθ

= 2π [- cosθ]0π/2

= 2π (1)

= 2π

So, the area of surface formed by revolving the curve r = θ is 2π.

Correct answer : (5)
10.
Find the area of surface formed by revolving the curve $r$ = sin$\theta$ + cos$\theta$, 0 ≤ $\theta$$\frac{\pi }{2}$ about the polar axis.
 a. $\frac{{\pi }^{2}}{\sqrt{2}}$ b. $\frac{\pi }{2\sqrt{2}}$ c. $\frac{\pi }{\sqrt{2}}$ d. $\frac{\pi }{4\sqrt{2}}$ e. $\frac{{\pi }^{2}}{2\sqrt{2}}$

#### Solution:

r = sinθ + cosθ
[Write the polar equation.]

If the derivative of the function f(θ) is continuous on the interval α ≤ θ ≤ β, then the area of the surface formed by revolving the graph of f(θ) from θ = α to θ = β about the polar axis is
s = 2π αβ f(θ) · sinθ · (f(θ))2+(f(θ))2

f ′(θ) = cosθ - sinθ is continuous in the interval [0, π4].

Surface area of revolution = 2π 0π/4 sinθ(sinθ + cosθ) (sinθ+cosθ)2+(cos θ-sin θ)2 dθ

= 2π 0π/4 (2 sin2θ + 2 sinθ cosθ) dθ

= 2π 0π/4 [(1 - cos 2θ) + sin 2θ] dθ

= 2π [θ - sin2θ2-cos2θ2]0π/4

= π222

So, the area of the surface formed by revolving the curve r = sinθ + cosθ is π222.

Correct answer : (5)

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