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; (c) What is the Cartesian equation for this function? (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; (c) What is the Cartesian equation for this function? (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; (c) What is the Cartesian equation for this function? (d) What is this function in geometric terms? Compare the results of the above ques- tions 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 ! 1? Explain your results in geometric terms. (c) Challenge. Parametrize the same function as x = cos2 θ; y = −a cos2 θ + 1. Find 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: p (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. (c) Find the surface area of the sphere of radius r by rotating x = r cos θ; y = r sin θ about the x-axis. 6. Find the surface area of the following surfaces of revolution: p (a) The function x = t; y = 2 t between t = 0 and 9. p p 2 (b) The function x = 2 t; y = 3 t t between t = 0 and t = 4. c 2018 MathTutorDVD.com 5 p p 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. p 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: p (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 p (b) The function x = sinh−1 t; y = 1 + t2 between t = −1 and t = 1. What is this function in Cartesian terms? (c) The function x = ln t; y = t between t = 1 and 2. (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 p 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 p 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 rota- tion. c 2018 MathTutorDVD.com 11 2. The parametric equation x = t; y = 2t + 1: p 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 p 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 p p 2 1 A = 2π 5 t + t 0 = 2π 5 (2 − 0) p which is 4π 5 . c 2018 MathTutorDVD.com 12 p 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 p p Z T p 2 2 2 T A = 2π(2t + 1) 1 + 2 dt = 2π 5 2t + 1dt = 2π 5 t + t 0 0 0 p 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.