# Arc Length and Surface of Revolution Worksheet

Arc Length and Surface of Revolution Worksheet
• Page 1
1.
Find the arc length of the curve $y$ = 5 - $\frac{3}{2}$$x$ from $x$ = 3 to $x$ = 5.
 a. 4$\sqrt{13}$ units b. $\frac{2}{\sqrt{13}}$ units c. $\frac{\sqrt{13}}{2}$ units d. $\sqrt{13}$ units e. $\frac{\sqrt{13}}{4}$ units

#### Solution:

y = 5 - 3 / 2x
[Write the equation.]

dydx = - 3 / 2
[Find dydx.]

Arc length of the curve = 351+(dydx)2 dx
[Formula for arc length.]

= 351+(-32)2 dx
[Substitute dydx = - 3 / 2.]

= 35132 dx

= 132 [ x ]35
[Evaluate the integral.]

= 132 (5 - 3) = 13

So, the arc length of the curve = 13 units

2.
Find the arc length of the curve $y$ = $\frac{3}{2}$${\left(x\right)}^{\frac{2}{3}}$ from $x$ = 1 to $x$ = 8.
 a. 7$\sqrt{5}$ units b. 3$\sqrt{2}$ units c. 5$\sqrt{5}-2\sqrt{2}$ units d. $\sqrt{5}-\sqrt{2}$ units e. 5$\sqrt{5}+2\sqrt{2}$ units

#### Solution:

y = 3 / 2x23
[Write the equation.]

dydx = 3 / 2 × 2 / 3 x-13 = x-13
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= 181+x-23 dx
[Substitute dydx = x-13 .]

= 181+1x23 dx

= 18x23+1x13 dx

= [(1+x23)32 ]18
[Use substitution method to evaluate Integral.]

= 532-232 = 55-22

So, the arc length of the curve = 55-22 units

3.
Find the arc length of the curve $y$ = $\frac{1}{2}$$x$2 + 5 from $x$ = 0 to $x$ = 2.
 a. ($\frac{1}{2}$ln (2 + $\sqrt{5}$)) units b. ($\sqrt{5}$ + $\frac{1}{2}$ln (2 + $\sqrt{5}$)) units c. (ln (2 + $\sqrt{5}$)) units d. ($\sqrt{5}$ + 2 ln (2 + $\sqrt{5}$)) units e. ($\sqrt{5}$ + ln (2 + $\sqrt{5}$)) units

#### Solution:

y = 1 / 2x2 + 5
[Write the equation.]

dydx = x
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= 021+x2 dx
[Substitute dydx = x.]

= [ x21+x2+12sin h-1 (x) ]02

= [1+4+12ln (2 + 22+1) - 0]
[Use sin h-1(x) = ln (x + x2+1) .]

= [5 + 1 / 2ln (2 + 5) ]

So, the arc length of the curve = (5 + 1 / 2ln (2 + 5)) units.

4.
Find the arc length of the curve $y$ = $\frac{{x}^{3}}{12}+\frac{1}{x}$ from $x$ = 2 to $x$ = 3.
 a. $\frac{37}{12}$units b. $\frac{25}{12}$units c. $\frac{3}{4}$units d. $\frac{7}{4}$units e. $\frac{29}{12}$units

#### Solution:

y = x312+1x
[Write the equation.]

dydx = x24-1x2 = x4-44x2
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

231+(x4-44x2)2 dx
[Substitute dydx = x4-44x2.]

2316x4+x8+16-8x416x4 dx

= 23(x4+4)216x4 dx

= 23x4+44x2 dx

= 23x24 dx + 23x-2 dx

= [x312]23 - [1x ]23

= 2712-812-13+12

= 1912-13+12 = 19-4+612

= 21 / 12 = 7 / 4
[Simplify.]

So, the arc length of the curve = 7 / 4units

5.
Find the arc length of the curve $y$ = ln sin $x$ from $\frac{\pi }{6}$ to $\frac{\pi }{3}$.
 a. 0.767 units b. 1.865 units c. 0.549 units d. 0 units e. 1.316 units

#### Solution:

y = ln sin x
[Write the equation.]

dydx = 1sinx cos x = cot x
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= π/6π/31+cot2x dx
[Substitute dydx = cot x.]

= π/6π/3 cosec x dx

= [ ln | tan (x2) | ]π/6π/3

= ln | tan (π6) | - ln | tan (π12) |

= ln (13) - ln (2 - 3)

= - 0.549 + 1.316

0.767

So, the length of the curve 0.767 units

6.
Find the arc length of the curve $y$ = 1 + ln cos $x$ from $x$ = 0 to $x$ = $\frac{\pi }{4}$.
 a. 0 units b. 0.414 units c. 1.414 units d. 0.3464 units e. 0.8814 units

#### Solution:

y = 1 + ln cos x
[Write the equation.]

dydx = 1cosx (- sin x) = - tan x
[Find dydx]

Arc length of the curve = 0π/41+tan2x dx
[Use formula for arc length.]

= 0π/4sec x dx
[Substitute dydx = - tan x.]

= [ ln | sec x + tan x | ]0π/4

= ln | 2 + 1 |

0.8814

So, the arc length of the curve 0.8814 units

7.
Find the arc length of the curve $y$ = 2${x}^{\frac{3}{2}}$ + 3 from $x$ = 0 to $x$ = 9.
 a. [${\left(82\right)}^{\frac{3}{2}}$ - 1] units b. $\frac{2}{27}$ [${\left(82\right)}^{\frac{3}{2}}$ + 1] units c. $\frac{2}{27}$ ${\left(82\right)}^{\frac{3}{2}}$ - 1 units d. $\frac{2}{27}$ [${\left(82\right)}^{\frac{3}{2}}$ - 1] units e. $\frac{2}{27}$ ${\left(82\right)}^{\frac{3}{2}}$ + 1 units

#### Solution:

y = 2x32 + 3
[Write the equation.]

dydx = 3x12
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= 091+9x dx
[Substitute dydx = 3x12.]

= 2 / 27 (1+9x)32 |09
[Evaluate the integral.]

= 2 / 27 (82)32-227

= 2 / 27 [(82)32 - 1]

So, the arc length of the curve = 2 / 27 [(82)32 - 1] units

8.
Find the arc length of the curve $y$ = $\frac{{x}^{4}}{8}+\frac{1}{{4x}^{2}}$ from $x$ = 1 to $x$ = 2.
 a. $\frac{35}{16}$units b. $\frac{37}{16}$units c. $\frac{39}{16}$units d. $\frac{15}{8}$units e. $\frac{33}{16}$units

#### Solution:

y = x48+14x2
[Write the equation.]

dydx = x32-12x3 = x6-12x3
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= 121+(x6-12x3)2 dx
[Substitute dydx = x6-12x3.]

= 124x6+x12-2x6+14x6 dx

= 12(x6+1)24x6 dx

= 12x6+12x3 dx

= 1 / 212x3 dx + 1 / 212x-3 dx

= [x48 ]12 - [14x2 ]12

= 168-18-116+14

= 32-2-1+416 = 33 / 16

So, the arc length of the curve = 33 / 16units

9.
Find the arc length of the curve $y$ = $\frac{{x}^{5}}{10}+\frac{1}{{6x}^{3}}$ from $x$ = 1 to $x$ = 2.
 a. $\frac{779}{240}$units b. $\frac{709}{240}$units c. $\frac{7}{48}$units d. $\frac{31}{10}$units e. $\frac{1}{48}$units

#### Solution:

y = x510+16x3
[Write the equation.]

dydx = x42-12x4 = x8-12x4
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= 121+(x8-12x4)2 dx
[Substitute dydx = x8-12x4.]

= 124x8+x16+1-2x84x8 dx

= 12(x8+1)24x8 dx

= 12x8+12x4 dx

= 1 / 212x4 dx + 1 / 212x-4 dx

= [x510 ]12 - [16x3 ]12

= 3210-110-148+16

= 3110+748 = 779 / 240

So, the arc length of the curve = 779 / 240units

10.
Find the arc length of the curve $y$ = $\frac{{e}^{x}+{e}^{-x}}{2}$ from $x$ = 0 to $x$ = 2.
 a. $\frac{{e}^{4}+1}{{e}^{2}}$ units b. $\frac{{e}^{4}-1}{{2e}^{2}}$ units c. $\frac{{e}^{4}+1}{{2e}^{2}}$ units d. $\frac{{e}^{4}-1}{{e}^{2}}$ units e. $\frac{{e}^{4}-1}{2}$ units

#### Solution:

y = ex+e-x2
[Write the equation.]

dydx = ex-e-x2
[Find dydx.]

Arc length of the curve = ab1+(dydx)2 dx
[Formula for arc length.]

= 021+(ex-e-x2)2 dx
[Substitute dydx = ex-e-x2.]

= 024+e2x+e-2x-24 dx

= 02(ex+e-x)24 dx

= 02ex+e-x2 dx

= 1 / 2 [ex-e-x ]02

= 1 / 2 [e2-e-2-e0+e-0]

= 1 / 2 [e4-1e2] = e4-12e2

So, the arc length of the curve = e4-12e2 units