Calculus 2 Tutor Worksheet 12 Surface Area of Revolution In

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Calculus 2 Tutor

Worksheet 12

Surface Area of Revolution in Parametric Equations Worksheet for Calculus 2 Tutor, Section 12:

Surface Area of Revolution in Parametric Equations

1. For the function given by the parametric equation x = t, y = 1:

(a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1;

(b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0;

(d) What is the surface of rotation in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape.

c 2018 MathTutorDVD.com 1 2. For the function given by the parametric equation x = t, y = 2t + 1:

(a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1;

(b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0;

(d) What is the surface of rotation in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape.

3. For the function given by the parametric equation x = 2t + 1, y = t:

(a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1;

c 2018 MathTutorDVD.com 2 (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0;

(d) What is this function in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape.

4. For the function given by the parametric equation x = t, y = −at + 1 for a > 0:

1 (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = a ;

c 2018 MathTutorDVD.com 3 1 (b) What is the above surface area (from t = 0 to t = a ) as a → ∞? Explain your results in geometric terms.

π the surface area of f rotated about the x-axis as θ goes from 0 to 2 . Assume a ≤ 1. Explain your results geometrically, especially considering the case a = 1. Hint: to simplify, use the expressions cos 2θ = 2 cos2 θ − 1, sin 2θ = 2 sin θ cos θ.

5. Find the surface area of a sphere by rotating the unit circle around the x-axis: √ (a) The unit circle (the upper semi-circle) can be parametrized as x = t, y = 1 − t2, −1 ≤ t ≤ 1. Rotate this parametrization about the x-axis to ﬁnd the surface area of the unit sphere.

c 2018 MathTutorDVD.com 4 (b) The upper semi-circle of the unit circle can also be parametrized as x = cos θ, y = sin θ, 0 ≤ θ ≤ π. Rotate this parametrization about the x-axis. to ﬁnd the surface area of the unit sphere.

6. Find the surface area of the following surfaces of revolution: √ (a) The function x = t, y = 2 t between t = 0 and 9.

√ √ 2 (b) The function x = 2 t, y = 3 t t between t = 0 and t = 4.

c 2018 MathTutorDVD.com 5 √ √ 2 (c) The function x = 3 t t, y = 2 t between t = 0 and t = 1.

t3 2 (d) The function x = 3 − t, y = t between t = 2 and t = 5.

√ 2 t3 (e) The function x = 2t , y = 3 − 2t between t = 3 and t = 4.

7. Find the surface area of the following surfaces of revolution: √ (a) The function x = sin−1 t, y = 1 − t2 between t = −1 and t = 1. What is this function in Cartesian terms?

c 2018 MathTutorDVD.com 6 √ (b) The function x = sinh−1 t, y = 1 + t2 between t = −1 and t = 1. What is this function in Cartesian terms?

(d) The function x = sech t, y = tanh t from t = 0 to t = 1.

8. Find the surface area of the following surfaces of revolution:

2 π (a) The function x = cos θ, y = sin θ between θ = 0 and θ = 2 .

1 2 π (b) The function x = 2 sec θ, y = sec θ between θ = 0 and θ = 4 . Hint: make a

c 2018 MathTutorDVD.com 7 u-substitution related to a derivative that you computed earlier in this problem.

c 2018 MathTutorDVD.com 8 Answer key.

1. The parametric equation x = t, y = 1:

1(a). Answer: 2π . To ﬁnd the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Here since x(t) = t, dx = 1 dt Since y(t) = 1, dy = 0 dt Then, we compute

Z t=1 √ Z 1 A = 2π · 1 · 12 + 02dt = 2πdt t=0 0 which is 2π .

c 2018 MathTutorDVD.com 9 1(b). Answer: 2πT . To ﬁnd the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Here, x0(t) = 1, y0(t) = 0

as in the last problem. However, the bounds α and β are different. Before α = 0, β = 1, but now α = 0, β = T . Then the surface area is

Z T √ Z T A = 2π1 12 + 02dt = 2πdt 0 0 which is 2πT .

c 2018 MathTutorDVD.com 10 1(c). Answer: y = 1 . If x = t and y = 1, this means that x can be anything but y is always 1. This is a horizontal line. Between t = 0 and 1, then x can vary from 0 to 1, which is a line segment of length 1. Between t = 0 and T , then x can vary from 0 to T , which is a line segment of length T .

1(d). Answer: a cylinder . Rotating the line y = 1 (between any two bounds of x) about the x-axis yields a cylinder. The surface area of a cylinder is

2πr2 + 2πrh

However, the 2πr2 accounts for the circles on the top and bottom. The surface area of the proper surface of rotation (the “face” of the cylinder) is

2πrh

The line y = 1 rotated about the x-axis (that is, y = 0) has radius 1, so

r = 1

The height is the distance between x = 0 and x = T , which is

h = T

Therefore, the surface area is 2πT

which matches the result we derived for the surface area of this surface of rotation.

c 2018 MathTutorDVD.com 11 2. The parametric equation x = t, y = 2t + 1: √ 2(a). Answer: 4π 5 . To ﬁnd the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Since x(t) = t, then dx = 1 dt Since y(t) = 2t + 1, then dy = 2 dt The bounds of t are between 0 and 1, so α = 0 and β = 1. Then,

Z 1 √ Z 1 A = 2π (2t + 1) p(12 + 22)dt = 2π 5 2t + 1dt 0 0 According to the technique of Calculus I, this integral is Z 2t + 1dt = t2 + t + C

so we have √ √ 2 1 A = 2π 5 t + t 0 = 2π 5 (2 − 0) √ which is 4π 5 .

c 2018 MathTutorDVD.com 12 √ 2(b). Answer: 2π 5 T 2 + T . This is the same integral as the last problem but with a difference in the boundaries of integration. The integral is

Z T √ √ Z T √ 2 2 2 T A = 2π(2t + 1) 1 + 2 dt = 2π 5 2t + 1dt = 2π 5 t + t 0 0 0 √ which is 2π 5 T 2 + T .

2(c). Answer: y = 2x + 1 . Substituting x = t into y = 2t + 1 gives the answer y = 2x + 1 . Between t = 0 and t = 1 (that is, between x = 0 and x = t), this is the line between the points

(0, 1) → (1, 3)

Between t = 0 and t = T (that is, between x = 0 and x = T ), this is the line between the points (0, 1) → (T, 2T + 1)

c 2018 MathTutorDVD.com 13 2(d). Answer: a cone segment or truncated cone. Rotating the line segment from (0, 1) to (1, 3) about the x-axis yields a truncated cone (cone segment). Rotating

1 the longer line segment from − 2 , 0 to (1, 3) would be an entire cone, but since the line segment is shorter, the cone is truncated. The surface area of the face of a truncated cone (not counting the base circles) is

p 2 2 π(r2 + r1) h + (r2 − r1)

where r2 is the larger radius, r1 is the smaller radius, and h is the height. The

smaller radius r1 is just

r1 = 1

since it is the circle made by rotating the point (0, 1) around the x-axis. The larger

radius r2 is

r2 = 2T + 1

since it is the circle made by rotating the point (T, 2T + 1) around the x-axis. The height is h = T

since the height of the cone is the distance along the x-axis between (0, 1) and (T, 2T + 1). Therefore, the surface area found according to geometry is q √ √ √ π(2T +1+1) T 2 + (2T + 1 − 1)2 = π(2T +2) T 2 + 4T 2 = 2π(T +1) 5T 2 = 2π 5 T 2 + T

which matches the result that we found using the calculus of parameric equations.

c 2018 MathTutorDVD.com 14 3. For the function given by the parametric equation x = 2t + 1, y = t: √ 3(a). Answer: π 5 . To ﬁnd the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Since x(t) = 2t + 1 then dx = 2 dt Since y(t) = t then dy = 1 dt Then, the surface area is

Z 1 √ √ Z 1 A = 2πt 22 + 12dt = 2π 5 tdt 0 0 This is √ t2 1 √ 2π 5 = π 5 2 0 √ The surface area of this surface of revolution is π 5 . The arc length of this function is the same as that of the function in the last problem (since switching x and y does not change the symmetric calculation for arc length), but the surface area is different.

c 2018 MathTutorDVD.com 15 √ 3(b). Answer: π 5T 2 . The integral is the same as in the last part, but the boundaries are different. Changing the boundaries, we have

Z T √ √ Z T √ t2 T A = 2πt 22 + 12dt = 2π 5 tdt = 2π 5 0 0 2 0 √ which is π 5T 2 .

1 1 3(c). Answer: y = x − . Since y = t, we can plug y = t into x = 2t + 1 to get 2 2

x = 2y + 1

1 1 In terms of y, this means that y = x − . 2 2

c 2018 MathTutorDVD.com 16 3(d). Answer: a cone. At t = 0, then y = 0, so the line is at the point

(1, 0)

At t = T , then y = T , so the line is at the point

(2T + 1,T )

This is a line segment. Since the place where the segment touches the x-axis is included, the revolution of the segment about the x-axis will include the point of the cone. Therefore the surface of revolution is a cone, not a cone segment or truncated cone. The surface area of the face of a cone (not counting the base circle) is √ A = πr h2 + r2

Here, the height is the distance along the x-axis between (1, 0) and (2T + 1,T ), which is h = 2T + 1 − 1 = 2T

The radius is the radius of the base circle, which is the path traced out by rotating (2T + 1,T ) around the x-axis. Then, the radius is

r = T

According to the geometric formula, the area of the face is √ A = πT pT 2 + (2T )2 = π 5T 2

which matches the result we found using calculus to study the parametric func- tions.

c 2018 MathTutorDVD.com 17 4. For the function given by the parametric equation x = t, y = −at + 1 for a > 0:

r 1 4(a). Answer: π 1 + . The surface area of f rotated about the x-axis is given by a2 s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Since x = t, dx = 1 dt Since y = −at + 1, dy = −a dt Then the area is

1 1 Z a √ Z a 2π(−at + 1)p12 + (−a)2dt = 2π 1 + a2 −at + 1dt 0 0 The antiderivative of this function is

Z at2 −at + 1dt = t − 2

so the integral is

1 √ 2 a √ 2 at 2 1 a A = 2π 1 + a t − = 2π 1 + a − 2 2 0 a 2a This simpliﬁes to √ 1 r 1 A = 2π 1 + a2 = π 1 + 2a a2 r 1 The area of this surface is π 1 + . a2

c 2018 MathTutorDVD.com 18 4(b). Answer: π . Taking the limit of the above quantity,

r 1 lim π 1 + = π a→∞ a2

We can calculate the equation of the line segment as a → ∞. At t = 0, the point is always (0, 1), since y = −at + 1 is always equal to 1 at t = 0, no matter how large a is. Since a is always a ﬁnite quantity even in a large limit, −a · 0 is zero. At

1 t = a , the point is 1 , 0 a As a → ∞, this point goes to (0, 0) (0, 0). The rotation of the limiting line segment (the segment from (0, 1) to (0, 0)) about the x-axis is just a circle of radius 1. The surface area of this circle is πr2 = π

Geometrically, the rotation of this line segment is a cone; as a → ∞ the height of the cone goes to zero, so the cone becomes a circle. The circle’s area can be calculated using geometry.

c 2018 MathTutorDVD.com 19 √ 4(c). Answer: π 1 + a2 (2 − a) . The function can be parametrized by

x = cos2 θ, y = −a cos2 θ + 1

This satisﬁes the same linear equality,

y = −ax + 1

However, this parametrization in terms of cos2 θ can only parametrize a limited segment of the line, since cos2 θ is always between 0 and 1. As θ goes from 0 to

π 2 , the surface area of the rotated line segment is

π s 2 2 Z 2 dx dy A = 2π −a cos2 θ + 1 + dθ 0 dθ dθ Since x = cos2 θ we can use the chain rule to calculate

dx d = 2 cos θ cos θ = −2 cos θ sin θ dθ dθ

Since y = −a cos2 θ + 1 we can calculate

dy = 2a cos θ sin θ dt

Then, the integral is

π Z 2 p A = 2π −a cos2 θ + 1 4 cos2 θ sin2 θ + 4a2 cos2 θ sin2 θdθ 0

π √ Z 2 = 4π 1 + a2 −a cos2 θ + 1 cos θ sin θdθ 0 Since cos 2θ = 2 cos2 θ − 1

we can write 1 + cos 2θ cos2 θ = 2 Also, we can rewrite sin 2θ cos θ sin θ = 2

c 2018 MathTutorDVD.com 20 Then, the integral is

π √ Z 2 1 + cos 2θ sin 2θ 4π 1 + a2 −a + 1 dθ 0 2 2

π √ Z 2 a 1 a = 4π 1 + a2 − cos 2θ sin 2θ + − sin 2θdθ 0 4 2 4 We can integrate each of these terms separately. Combining

1 cos 2θ sin 2θ = sin 4θ 2 we can write

Z π Z π π 2 a 2 a h a i 2 − cos 2θ sin 2θdθ = − sin 4θdθ = cos 4θ 0 4 0 8 32 0

π a a Next, since cos 4 · 2 = cos 0 this is equal to 32 − 32 = 0. We can solve

π π Z 2 1 a 1 a 2 − sin 2θdθ = − · (− cos 2θ) 0 2 4 4 8 0 Since cos π = −1 and cos 0 = 1 this is

1 a 1 a − (1 − +1) = − 4 8 2 4

Therefore the area is

√ 1 a √ A = 4π 1 + a2 − = π 1 + a2 (2 − a) 2 4 √ So the answer is π 1 + a2 (2 − a) . The line segment from (1, −a + 1) to (0, 1) rotates to produce a truncated cone. In the special case a = 1, the cone is not truncated because it intersects the x-axis, so the point is included. There, the area is √ √ π 2 (2 − 1) = π 2

In this special case, we also have that

y = − cos2 θ + 1 = 1 − cos2 θ = sin2 θ

c 2018 MathTutorDVD.com 21 5. The surface area of a sphere using a parametrization:

5(a). Answer: 4π . To compute the surface area of the unit sphere, we compute s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Since x(t) = t, then dx = 1 dt √ Since y(t) = 1 − t2, then by the chain rule

dy 1 d −t = √ 1 − t2 = √ dt 2 1 − t2 dt 1 − t2

Then,

Z 1 √ r 2 Z 1 √ r 2 2 2 2 t 2 1 − t + t A = 2π 1 − t 1 + 2 dt = 2π 1 − t 2 dt −1 1 − t −1 1 − t Since the numerators in the square root cancel, the integral is

Z 1 √ r Z 1 2 1 A = 2π 1 − t 2 dt = 2πdt = 4π −1 1 − t −1 The parametric computation of the surface area of the unit sphere is 4π, which matches the computation of the surface area of the unit sphere as a surface of revolution of a Cartesian function.

c 2018 MathTutorDVD.com 22 5(b). Answer: 4π . To compute the surface area of the unit sphere using this trigonometric parametrization, we compute s Z π d 2 d 2 Z π q A = 2π sin θ cos θ + sin θ dθ = 2π sin θ (− sin θ)2 + cos2 θdθ 0 dθ dθ 0 Z π = 2π sin θdθ 0 The antiderivative of sin θ is − cos θ so the surface area of this shape is given by

π A = 2π [− cos θ]0 = 2π (− cos π + cos 0) = 4π

The surface area of the unit sphere is 4π . This matches the result from the Cartesian parametrization.

c 2018 MathTutorDVD.com 23 5(c). Answer: 4πr2 . We can compute the surface area of any sphere in the same way. We need to rotate the upper semi-circle of this circle about the x-axis. The bounds of the upper semi-circle are from θ = 0 to θ = π. We compute

Z π q Z π √ Z π A = 2π (r sin θ) (−r sin θ)2 + (r cos θ)2dθ = 2πr sin θ r2 = 2πr2 sin θdθ 0 0 0 Since the antiderivative is − cos θ, this is

2 π 2 2πr [− cos θ]0 = 4πr

The surface area of a sphere of radius r is 4πr2 .

c 2018 MathTutorDVD.com 24 6. Finding the surface area of revolution for parametrics in algebraic form: √ 80π 10 8π 6(a). Answer: − . We use the fomula 3 3 s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Taking the relevant derivatives, we compute s s Z 9 √ d 2 d √ 2 Z 9 √ 1 2 A = 2π 2 t t + 2 t dt = 4π t 12 + √ dt 0 dt dt 0 t Combining the two square roots, we can reduce this to Z 9 √ 4π t + 1dt 0 This can be evaluated using the techniques of Calculus I as

9 2 3 A = 4π (t + 1) 2 3 0

At t = 9, this is √ 2 3 20 10 (10) 2 = 3 3 √ 80π 10 8π At t = 0, this is just 2 . Then, multiplying by 4π, the answer is − . 3 3 3

c 2018 MathTutorDVD.com 25 4π √ 6(b). Answer: 17 17 − 1 . To ﬁnd the surface area of revolution for the para- 9 √ √ 2 metric function x = 2 t, y = 3 t t, we use the formula: s Z t=β dx2 dy 2 A = 2πy(t) + dt t=α dt dt We compute dx d √ 2 1 = 2 t = √ = √ dt dt 2 t t by the power rule. We compute

dy d 2 3 1 √ = t 2 = t 2 = t dt dt 3

also by the power rule. Then,

s 2 r Z 4 2 √ 1 √ 2 4π Z 4 √ 1 4π Z 4 √ A = 2π t t √ + t dt = t t + tdt = t 1 + t2dt 0 3 t 3 0 t 3 0 This integral allows the u-substitution of

u = 1 + t2

Then, du = 2t dt and 1 dt = du 2t The integral then becomes 4π Z t=4 √ 1 2π Z t=4 √ A = t u du = udu 3 t=0 2t 3 t=0 which we can evaluate as

t=4 4 2π 2 3 4π h 2 3 i A = u 2 = (t + 1) 2 3 3 t=0 9 0 At t = 4, √ 2 3 3 (t + 1) 2 = 17 2 = 17 17 4π √ At t = 0, it is 1. Then, the answer is 17 17 − 1 . 9

c 2018 MathTutorDVD.com 26 √ √ 6(c). Answer: 2π 2 + 2π ln 1 + 2 . To ﬁnd the surface area of revolution for the √ √ 2 parametric function x = 3 t t, y = 2 t, we use the formula: s Z t=β dx2 dy 2 A = 2πy(t) + dt t=α dt dt We compute

dx d 2 3 1 √ = t 2 = t 2 = t dt dt 3 by the power rule. We compute

dy d √ 2 1 = 2 t = √ = √ dt dt 2 t t

also by the power rule. (These are the same derivatives as in the previous exer- cise, but reversed.) Then,

s 2 r Z 1 √ √ 2 1 Z 1 √ 1 A = 2π 2 t t + √ dt = 4π t t + dt 0 t 0 t We combine the square roots to reduce to

Z 1 √ A = 4π t2 + 1dt 0 Since the antideriative of this is

1 √ 1 t t2 + 1 + sinh−1 t + C 2 2

we can evaluate:

1 √ 1 1 1√ 1 1 A = 4π t t2 + 1 + sinh−1 t = 4π 2 + sinh−1 1 − 0 − sinh−1 0 2 2 0 2 2 2 Since √ sinh−1 1 = ln 1 + 2

and sinh−1 0 = 0 √ √ the answer is 2π 2 + 2π ln 1 + 2 .

c 2018 MathTutorDVD.com 27 6576π 6(d). Answer: . Again, we use the formula 5 s Z β Z 5 d t3 2 d A = 2πy(t)px0(t)2 + y0(t)2dt = 2πt2 − t + t2 dt α 2 dt 3 dt Calculating the derivatives, this is

Z 5 q Z 5 √ Z 5 √ 2π t2 (t2 − 1)2 + (2t)2dt = 2π t2 t4 − 2t2 + 1 + 4t2dt = 2π t2 t4 + 2t2 + 1dt 2 2 2 We factor the square root as

t4 + 2t2 + 1 = t2 + 12

so this is Z 5 t5 t3 5 2π t2(t2 + 1)dt = 2π + 2 5 3 2 At t = 5, this is 55 53 125 + = 625 + 5 3 3

32 8 At t = 2, this is 5 + 3 . Then, we evaluate 125 32 8 117 32 32 3288 625 + − + = 625 + − = 664 − = 3 5 3 3 5 5 5

6576π and the ﬁnal answer (multiplying by 2π) is . 5

c 2018 MathTutorDVD.com 28 2065π 6(e). Answer: . We ﬁrst evaluate 9 dx √ = 2 2t dt and dy = t2 − 2 dt Then, we compute r Z 4 t3 √ 2 Z 4 t3 √ A = 2π − 2t 2 2t + (t2 − 2)2dt = 2π − 2t 8t2 + t4 − 4t2 + 4dt 3 3 3 3 We compute the square root as √ √ q 8t2 + t4 − 4t2 + 4 = t4 + 4t2 + 4 = (t2 + 2)2 = t2 + 2

Then the integral is Z 4 t3 Z 4 t5 4t3 2π − 2t t2 + 2 dt = 2π − − 4tdt 3 3 3 3 3 which we evaluate using the techniques of Calculus I as t6 t4 4 2π − − 2t2 18 3 3 t6 The evaluation of the ﬁrst term, 18 , is 4096 729 3367 − = 18 18 18

t4 The evlauation of the second term, − 3 , is 256 81 175 1050 − + = − = − 3 3 3 18 The third term, −2t2, yields −32 + 18 = −14

Then, the total value of the deﬁnite integral is 3367 1050 2317 2317 252 2065 − − 14 = − 14 = − = 18 18 18 18 18 18 2065π and after multiplying by 2π we get as a ﬁnal answer . 9

c 2018 MathTutorDVD.com 29 7. Finding the surface area of revolution for hyperbolic and logarithmic parametrics: √ √ π π 7(a). Answer: 2π 2 + 2π ln 1 + 2 and the curve is y = cos x, − ≤ x ≤ . We 2 2 calculate the area of this surface of revolution using s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Since x(t) = sin−1 t, then dx 1 = √ dt 1 − t2 √ Since y(t) = 1 − t2, we can use the chain rule to compute

dy 1 d −t = √ · 1 − t2 = √ dt 2 1 − t2 dt 1 − t2 Then, we can compute s Z 1 √ 1 2 −t 2 Z 1 √ r 1 t2 Z 1 √ r1 + t2 A = 2π 1 − t2 √ + √ dt = 2π 1 − t2 + dt = 2π 1 − t2 dt 2 2 2 2 2 −1 1 − t 1 − t −1 1 − t 1 − t −1 1 − t The denominator in the square root cancels with the other square root to yield Z 1 √ 2π 1 + t2dt −1 which is 1 √ 1 1 2π t t2 + 1 + sinh−1 t 2 2 −1 Since √ sinh−1 1 = ln 1 + 2

and sinh−1 −1 = − sinh−1 1

this evaluates to √ 2π 2 + sinh−1 1 √ √ which is 2π 2 + 2π ln 1 + 2 . To ﬁnd the Cartesian expression of this function, since x = sin−1 t, we can write

t = sin x

c 2018 MathTutorDVD.com 30 Then, √ √ y = 1 − t2 = p1 − (sin x)2 = cos2 x = cos x

The function is the cosine curve! As t goes from −1 to 1, x = sin−1 t goes from π π − π to π . The curve is y = cos x, − ≤ x ≤ . 2 2 2 2

c 2018 MathTutorDVD.com 31 √ √ 7(b). Answer: 2π 2 + 2π ln 1 + 2 and the curve is √ √ y = cosh x, − ln 1 + 2 ≤ x ≤ ln 1 + 2 . We ﬁnd

dx d 1 = sinh−1 t = √ dt dt 1 + t2 By the chain rule, dy t = √ dt 1 + t2 Then we compute s Z 1 √ 1 2 t 2 Z 1 √ r 1 t2 A = 2π 1 + t2 √ + √ dt = 2π 1 + t2 + dt 2 2 2 2 −1 1 + t 1 + t −1 1 + t 1 + t The square root is 1, so the answer is Z 1 √ 2π 1 + t2dt −1 √ √ which we found in the last problem to be 2π 2 + 2π ln 1 + 2 . To ﬁnd this function in Cartesian terms, since x = sinh−1 t, we write

t = sinh x

Then, p y = 1 + sinh2 x

Since cosh2 x − sinh2 x = 1

then p y = cosh2 x = cosh x

This is the hyperbolic cosine curve. The bounds are from when x = sinh−1 −1 √ √ to sinh−1 1; that is, y = cosh x, − ln 1 + 2 ≤ x ≤ ln 1 + 2 . It’s interesting that rotating the cosine curve between sin−1 (−1) and sin−1 1 yields the same surface area as rotating the hyperbolic cosine curve between sinh−1 (−1) and sinh−1 1.

c 2018 MathTutorDVD.com 32 √ √ √ √ 7(c). Answer: 2π 5 + π ln 2 + 5 − π 2 − π ln 1 + 2 . We compute

dx d 1 = ln t = dt dt t

Since dy = 1 dt we can compute the area of the surface of revolution as s Z 2 2 Z 2 r 1 2 1 A = 2πt + 1 dt = 2πt 1 + 2 dt 1 t 1 t √ We can factor the t into the square root as t2 to get

Z 2 √ 2π t2 + 1dt 1 Since the antiderivative is

Z √ 1 √ 1 t2 + 1dt = t t2 + 1 + sinh−1 t 2 2

this is

1 √ 1 2 √ 1 1√ 1 2π t t2 + 1 + sinh−1 t = 2π 5 + sinh−1 2 − 2 − sinh−1 1 2 2 1 2 2 2 Since √ sinh−1 t = ln t + t2 + 1 √ √ √ √ this is 2π 5 + π ln 2 + 5 − π 2 − π ln 1 + 2 .

c 2018 MathTutorDVD.com 33 e2 − 2e + 1 7(d). Answer: 2π . We recall e2 + 1 dx sech t = − sech t tanh t dt and d tanh t = sech2 t dt Then,

1 1 Z q 2 Z p A = 2π tanh t (− sech t tanh t)2 + sech2 t dt = 2π tanh t sech2 t tanh2 t + sech4 tdt 0 0 √ Factoring out sech2 t = sech t we see that this is equal to Z 1 p 2π tanh t sech t tanh2 t + sech2 tdt 0 Now, tanh2 t + sech2 t = 1

Multiplying through by cosh2 t proves this formula since

sinh2 t + 1 = cosh2 t

Then the integral is Z 1 2π tanh t sech tdt 0 As we just found out, the antiderivative of tanh t sech t is just − sech t! So the integral is

1 2π [− sech t]0

We evaluate 1 2 sech 1 = = 1 cosh 1 e + e and 1 sech 0 = = 1 cosh 0 Then the answer is 2 e2 + 1 − 2e 2π − 1 + 1 = 2π 2 e + e e + 1

c 2018 MathTutorDVD.com 34 e2 − 2e + 1 which is 2π . e2 + 1

c 2018 MathTutorDVD.com 35 8. Finding the surface area of revolution for trigonometric parametrics: √ 5π 5 − π 8(a). Answer: . We compute 6 s Z β dx2 dy 2 A = 2πy(θ) + dθ α dθ dθ Since x = cos2 θ, then by the chain rule

dx = −2 sin θ cos θ dt

Likewise, dy = cos θ dt Then we can compute

π π Z 2 q Z 2 p A = 2π sin θ (−2 sin θ cos θ)2 + (cos θ)2dθ = 2π sin θ 4 sin2 θ cos2 θ + cos2 θdθ 0 0 Factoring out √ cos2 θ = cos θ

we get π Z 2 √ 2π sin θ cos θ 4 cos2 θ + 1dθ 0 Then we can make the u-substitution of

u = 4 cos2 θ + 1

This is helpful because du = −8 cos θ sin θ dθ so −1 dθ = du 8 cos θ sin θ Then we compute

π π Z 2 √ −1 −π Z 2 √ A = 2π sin θ cos θ u du = udu θ=0 8 cos θ sin θ 4 θ=0

c 2018 MathTutorDVD.com 36 According to the techniques of Calculus I, this is

π π 2 −π 2 3 −π h 3 i 2 2 2 u 2 = 4 cos θ + 1 4 3 θ=0 6 0 √ π At 2 , cosine is zero, so this is 1. At 0, cosine is 1, so this is 5 5. Then, the answer is −π √ 1 − 5 5 6 √ 5π 5 − π which is . 6

c 2018 MathTutorDVD.com 37 √ 4 √ 8(b). Answer: 2π 3 − π 2 . We compute 3

π s 2 2 Z 4 d 1 d A = 2π sec θ sec2 θ + sec θ dθ 0 dθ 2 dθ The derivative of sec θ is sec θ tan θ. Then, by the chain rule, the derivative of

1 2 2 sec θ is sec2 θ tan θ

The area is π Z 4 p A = 2π sec θ sec4 θ tan2 θ + sec2 θ tan2 θdθ 0 Factoring out √ sec2 θ tan2 θ = sec θ tan θ

we get π Z 4 √ A = 2π sec2 θ tan θ sec2 θ + 1dθ 0 It seems like a really difﬁcult integral, but then we remember that we already

2 1 2 found that sec θ tan θ is the derivative of 2 sec θ! This means that with the clever u-substitution of u = sec2 θ + 1

the term sec2 θ tan θ will cancel. We ﬁnd

du = 2 sec2 θ tan θ dθ

so 1 dθ = du 2 sec2 θ tan θ Then the area is

π π Z 4 Z 4 2 √ 1 √ A = 2π sec θ tan θ u 2 du = π udu θ=0 2 sec θ tan θ θ=0 This is π π 2π h 3 i 4 2π h 3 i 4 2 2 u 2 = sec θ + 1 3 θ=0 3 0

c 2018 MathTutorDVD.com 38 √ π At 4 , secant is 2. At 0, secant is 1. Then, this is

2π 3 3 2π √ √ (2 + 1) 2 − (1 + 1) 2 = 3 3 − 2 2 3 3 √ 4 √ which is 2π 3 − π 2 . 3

c 2018 MathTutorDVD.com 39