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Module MA2E02 (Frolov), Multivariable Calculus Tutorial Sheet 8 Due: at the end of the tutorial session Tuesday/Thursday, 22/24 March 2016 Name and student number: 1. Consider the lamina that is the portion of the surface 6z − x2 − y 2 − 1 = 0 inside the surface x2 + y 2 − 9 = 0. The density of the lamina is δ(x, y, z) = δ0 (9 + x2 + y 2 ) where δ0 is a constant (a) What is the surface 6z − x2 − y 2 − 1 = 0? (b) What is the surface x2 + y 2 − 9 = 0? (c) Sketch the projection of the lamina onto the xy-plane. (d) Find the mass of the lamina. Show the details of your work. Solution: (a) The surface 6z − x2 − y 2 − 1 = 0 is a paraboloid. (b) The surface x2 + y 2 − 9 = 0 is a circular cylinder of radius 3. (c) Below is the lamina σ, and its projection onto the xy-plane 3 2 1 1.5 1.0 0.5 0.0 -3 -2 1 -1 2 3 -1 2 -2 -2 0 0 2 -2 -3 The projection of the surface onto the xy-plane is the disk R: x2 + y 2 ≤ 9. (d) The mass of the lamina is equal to the following surface integral s 2 2 ZZ ZZ ZZ ∂z ∂z 1 2 2 M= δ(x, y, z) dS = δ0 (9+x +y ) 1 + + dA = δ0 (9+x2 +y 2 )3/2 dA . ∂x ∂y 3 σ R R The simplest way to compute the double integral over R is to use the polar coordinates 3 ZZ Z 3 1 2π 2π 1 2 2 3/2 2 3/2 2 5/2 M = δ0 (9 + x + y ) dA = δ0 (9 + r ) rdr = δ0 (9 + r ) 3 3 3 5 R 0 0 √ √ 2π 162π = δ0 35 (4 2 − 1) = δ0 (4 2 − 1) ≈ 474.01 δ0 15 5 1 2. Consider the surface σ which is the portion of the surface z = z = 5/3, oriented by downward unit normals. (a) Sketch the projection of the surface onto the xy-plane. (b) Find the flux of the vector field F across σ. x2 +y 2 +1 6 below the plane F(x, y, z) = 3y j + 2k ; Show the details of your work. Solution: (a) The surface σ is the same as the lamina in Problem 1. Thus, the projection of the surface onto the xy-plane is the disk R: x2 + y 2 ≤ 9. (b) Taking into account that through the surface z = f (x, y) oriented by down, the vector field F(x, y, z) = M i + N j + P k has ZZ ZZ ∂f ∂f +N − P dA , M flux = F · n dS = ∂x ∂y σ R we get by using the polar coordinates (M = 0 , N = 3y , P = 2) ZZ Z 2π Z 3 2 flux = y − 2 dA = (r2 sin2 φ − 2) rdr dφ R 0 0 Z 2π 4 34 3 9 = ( sin2 φ − 9)dφ = π − 18π = π . 4 4 4 0 3. (a) Express rectangular coordinates in terms of spherical coordinates (b) Use triple integral and spherical coordinates to derive the formula for the volume of a ball of radius R. (c) Use the Divergence Theorem to find the flux of the vector field F(x, y, z) = (2x + 15xy 2 + 5z 3 ) i − (y 2 + 3y + 3x2 ) j + (5 + 4z + 2yz − 5y 3 ) k across the surface σ with outward orientation where σ is given by the equation x2 + y 2 + z 2 = 4. Show the details of your work. Solution : (a) We have (draw a picture) x = r cos θ sin φ , y = r sin θ sin φ , z = r cos φ , 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π (b) We use the spherical coordinates to get (you must give more details here) ZZZ Z 2π Z π Z R 4 2 V = dV = r dr sin φ dφ dθ = πR3 . 3 V 0 0 0 (c) The surface is a sphere of radius 2. We have ZZ ZZZ ZZZ flux = F · n dS = div F dV = (2 + 15y 2 − 2y − 3 + 4 + 2y) dV V Z ZσZ ZZZ ZVZ Z ZZZ 4 3 2 2 = (3 + 15y ) dV = 3 dV + 15y dV = 3 π 2 + 5 r2 dV 3 V V V V ZZZ = 32π + 5 r2 dV , V 2 where we used that the volume of a ball of radius R is 43 πR3 and that ZZZ ZZZ ZZZ 2 2 2 2 3y dV = (x + y + z ) dV = r2 dV V V V due to the symmetry between x, y, z coordinates for a sphere (rotational symmetry). To compute the last integral we use again spherical coordinates ZZZ Z 2π Z π Z 2 4 128 4 2 r dr sin φ dφ dθ = π25 = π. r dV = 5 5 0 0 0 V Thus the flux is flux = 160π . 4. Let C be the triangle in the plane z = 12 y with vertices (2, 0, 0), (0, 2, 1) and (0, 0, 0) with a counterclockwise orientation looking down the positive z-axis. (a) Sketch the triangle and its projection H onto the xy-plane. (b) Use Stokes’ Theorem to evaluate C F · dr. F(x, y, z) = (5x − 2z 2 ) i − (y − 4x) j + (2xz + 4y) k . (c) Represent C as a union of three line segments C1 , C2 , C3 connecting the points (0, 0, 0) and (2, 0, 0), (2, 0,H0) and (0, 2, 1), (0, 2, 1) and (0, 0, 0), respectively. Parameterize C1 , C2 and C3 , and evaluate C F · dr. Show the details of your work. Solution : (a) Below is the triangle, σ, and its projection, R, onto the xy-plane 2.0 1.5 1.0 2.0 0.5 1.0 1.5 0.0 0.0 1.0 0.5 0.5 0.5 1.0 1.5 2.0 0.0 0.5 1.0 (b) According to Stokes’ Theorem I ZZ ZZ ∂z ∂z F · dr = (curl F) · n dS = (curl F) · − i − j + k dA , ∂x ∂y C σ R 3 1.5 2.0 where n is the normal vector to the surface z = 12 y looking up the positive z-axis. To use Stokes’ Theorem we first compute ∂(2xz + 4y) ∂(4x − y) ∂(5x − 2z 2 ) ∂(2xz + 4y) curl F = − i+ − j ∂y ∂z ∂z ∂x ∂(4x − y) ∂(5x − 2z 2 ) + − k ∂x ∂y = 4 i − 6z j + 4 k , and − ∂z ∂z 1 i− j + k = − j + k. ∂x ∂y 2 Thus we have Z 2 I ZZ Z 2 Z 2−x 3 3 ( (2 − x)2 + 4(2 − x)) dx F · dr = (3z + 4) dA = ( y + 4) dy dx = 2 4 0 C 0 0 Z 2R 3 1 ( x2 + 4x) dx = 23 + 2 · 22 = 10 . = 4 4 0 (c) The line segments C1 is parametrized as x = t, y = 0, z = 0, 0 ≤ t ≤ 2 and one gets Z Z 2 5 F · dr = 5t dt = 22 = 10 . 2 C1 0 The line segments C2 is parametrized as x = 2 − t, y = t, z = t/2, 0 ≤ t ≤ 2 and one gets Z Z F · dr = (5x − 2z 2 ) dx − (y − 4x) dy + (2xz + 4y)dz C2 C Z 21 Z 2 t(2 − t) t2 + 2t) dt = (−2 + 3t) dt (−5(2 − t) + − t + 4(2 − t) + = 2 2 0 0 3 = −4 + 22 = 2 . 2 The line segments C3 is parametrized as x = 0, y = 2 − t, z = 1 − 2t , 0 ≤ t ≤ 2 and one gets Z Z F · dr = (5x − 2z 2 ) dx − (y − 4x) dy + (2xz + 4y)dz C3 C Z 21 Z 2 = ((2 − t) − 2(2 − t)) dt = (−t) dt 0 0 1 = − 22 = −2 . 2 Summing up the contributions one gets I F · dr = 10 C which agrees with the result obtained by using Stokes’ Theorem. 4