But when φ uv v 2 3 u 6 v 2 u 2 v we see that t u φ

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But when Φ ( u,v ) = ( v + 2 , 3 u + 6 v, 2 u 2 v ) , we see that T u = Φ ∂u = (0 , 3 , 2) , while T v = Φ ∂v = (1 , 6 , 2) , In this case T u × T v = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 0 3 2 1 6 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 6 i 2 j 3 k ,
baek (mjb3994) – HW13 – goddard – (54330) 5 and so, bardbl T u × T v bardbl = radicalBig (6) 2 + ( 2) 2 + ( 3) 2 . Consequently, dS = 49 dudv = 7 dudv . 008 10.0 points The surface S shown in is the portion of the graph of z = f ( x,y ) = x 2 y 2 lying inside the cylinder x 2 + y 2 = 2 Determine the surface area of S . 1. Surface Area = 16 3 π sq. units 2. Surface Area = 5 π sq. units 3. Surface Area = 17 3 π sq. units 4. Surface Area = 14 3 π sq. units 5. Surface Area = 13 3 π sq. units correct Explanation: As S is enclosed by the cylinder x 2 + y 2 = 2 , it is the graph of the function f ( x,y ) = x 2 y 2 , over the disk D = { ( x,y ) : x 2 + y 2 2 } in the xy -plane. Its surface area element is dS = ( f 2 x + f 2 y + 1) 1 / 2 dxdy where f x = 2 x, f y = 2 y. Thus dS = (4 x 2 + 4 y 2 + 1) 1 / 2 dxdy, and so its surface area is given by the integral I = integraldisplay integraldisplay D (4 x 2 + 4 y 2 + 1) 1 / 2 dxdy. Because of rotational symmetry, the inte- gral is most easily evaluated using polar coor- dinates. For then D = { ( r,θ ) : 0 r 2 , 0 θ 2 π } so that I = integraldisplay 2 0 integraldisplay 2 π 0 (4 r 2 + 1) 1 / 2 rdθdr = 2 π integraldisplay 2 0 (4 r 2 + 1) 1 / 2 rdr. To evaluate this last integral we use the sub- stitution u 2 = 1 + 4 r 2 . For then I = π 2 integraldisplay 3 1 u 2 du = π 6 bracketleftBig u 3 bracketrightBig 3 1 = 13 3 π . 009 10.0 points
baek (mjb3994) – HW13 – goddard – (54330) 6 If S is the surface { Φ ( u,v ) : 0 u 1 , 0 v 3 } , determine the surface area of S when Φ ∂u = ( 6 , 1 , 2) , Φ ∂v = (3 , 0 , 2) , for all u,v . 1. surface area = 25 sq units 2. surface area = 21 sq units correct 3. surface area = 22 sq units 4. surface area = 23 sq units 5. surface area = 24 sq units Explanation: The surface area of any surface { Φ ( u,v ) : 0 u 1 , 0 v 3 } is given by the integral I = integraldisplay 1 0 integraldisplay 3 0 dS = integraldisplay 1 0 parenleftBig integraldisplay 3 0 bardbl T u × T v bardbl dv parenrightBig du But when T u = Φ ∂u = ( 6 , 1 , 2) , and T v = Φ ∂v = (3 , 0 , 2) , we see that T u × T v = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 6 1 2 3 0 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 2 i + 6 j + 3 k . Thus bardbl T u × T v bardbl = radicalBig (2) 2 + (6) 2 + (3) 2 . Consequently, I = 49 integraldisplay 1 0 integraldisplay 3 0 dvdu = 21 . 010 10.0 points The surface S shown in is the portion of the sphere x 2 + y 2 + z 2 = 16 where x 2 + y 2 12 . Determine the surface area of S . 1. Surface Area = 24 sq. units 2. Surface Area = 16 sq. units 3. Surface Area = 32 sq. units 4. Surface Area = 16 π sq. units 5. Surface Area = 32 π sq. units correct 6. Surface Area = 24 π sq. units Explanation: The sphere x 2 + y 2 + z 2 = 16 is parametrized in spherical polar coordinates by Φ ( θ,φ ) = 4(cos θ sin φ, cos θ sin φ, cos φ )
baek (mjb3994) – HW13 – goddard – (54330) 7 with 0 θ 2 π, 0 φ π. We have to determine what conditions need to be imposed on θ,φ so that only the points where x 2 + y 2 12 are included. Since S doesn’t change with rotation by θ from 0 to 2 π around the z -axis, there will be no restriction on θ . On the other hand, spherical caps at the north and south poles are missing, so restrictions on φ are needed. Now a vertical cross-section of S through the origin looks like π 3 π 3 z 4 2 3 and S is generated by rotating the arc of the circle between the two solid arrows around the z -axis. Thus S is parametrized by Φ ( θ,φ ) = 4(cos θ sin φ, cos θ sin φ, cos φ ) , with 0 θ 2 π, π 3 φ 2 π 3 , and the surface area element is dS = 16 sin φdθdφ.

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