Calculus Answers. , where π u π

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1 Calculus Answers 13. r(u, v) u cosvi + u sinvj + vk. The parametric equations for the surface are x u cosv y u sin v z v We look at the grid curve first; if we fix v, then x and y parametrize a straight line in the plane z v which intersects the z axis.if u is held constant,the projection onto the xy plane is circular;with z v, each grid curve is a helix. The surface is a spiraling ramp,graph IV. 14. r(u, v) u cosvi + u sin vj + sin uk. The corresponding parametric equations for the surface are x u cosv y u sin v z sin u, where π u π If u u is held constant, then x u cosv,y u sin v so each grid curve is a circle of radius u in the horizontal plane z sin u. If v v is constant, then x u cosv, y u sin v y (tanv )x, so the grid curve lie in the vertical planes y kx through the z axis. In fact, since x and y are constant multiples of u and z sin u, each of these is a sine wave. The surface is graph I. 15. r(u, v) sin vi + cosusin vj + sin u sinvk. Parametric equtaions for the surface are x sin v y cos u sinv z sin u sinv. If v v is fixed, then x sin v is constant, and y (sin v ) cosu and z (sin v ) sinu describe a circle of radius sinv, so each corresponding grid curve is a circle contained in the vertical plane x sin v parallel to yz plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to holding u constant, in which case y (cosu ) sin v, z sin u sin v z (tanu )y, so each grid curve lies in a plane z ky that includes the x axis.

2 16. x (1 u)(3 + cosv) cos 4πu y (1 u)(3 + cosv) sin 4πu z 3u + (1 u) sin v. These equations corresponding to graph V: When u, then x 3 + cosv y z sin v, which are equations of a circle with radius 1 in the xz plane centered at (3,, ). When u 1, then x cosv y z sin v, which are equations of a circle with radius 1 in the xz plane centered at ( 3,, 3 ). When u 1, then x y and z 3, giving the topmost point shown in the graph.this suggests that the grid curves with u constant are the vertically oriented circles visible on the surface. The spiralling grid curves correspond to keeping v constant. 17. x cos 3 u cos 3 v y sin 3 u cos 3 v z sin 3 v. If v v is held constant then z sin 3 v is constant, so the corresponding grid curve lies in a horizontal plane. Several of the graphs exhibit horizontal grid curves, but the curves for this surface are neither circles nor straight lines, so graph III is the only possibility. (In fact, the horizontal grid curves here are members of family x a cos 3 u, y a sin 3 u and are called astroids.) The vertical grid curves we see on the surface correspond to u u held constant, as then we have { x cos 3 u cos 3 v y sin 3 u cos 3 v so the corresponding grid curve lies in the vertical plane y (tan 3 u )x through the z axis. 18. x (1 u ) cosv y (1 u ) sinv z u.

3 Then x + y (1 u ) cos v + (1 u ) sin v (1 u ), so if u is held constant, each grid curve is a circle of radius (1 u ) in the horizontal plane z u. The graph then must be graph VI. If v is held constant, so v v, we have x (1 u ) cosv and y (1 u ) sinv. Then y (tan v )x, so the grid curve we see running vertically along the surface in the planes y kx correspond to keeping v constant. 3. First we graph the surfaces as viewed from the front, then from two additional viewpoints. The surface appears as a twisted sheet, and is unusual because it has only one side. (The Möbius strip is discussed in more detail in Section 17.7 [ET 16.7].) 54. (a) r u r v r u r v a cosvi + b sin vj + uk au sin vi + bu cosvj + k bu cosvi au sin vj + abuk. A(S) r u r v dudv 4b u 4 cos v + 4a u 4 sin v + a b u dudv (b) x a u cos v, y b u sin v, z u x a + y b u z which is an elliptic paraboloid. To find D, notice that u z 4 x a + y b 4

4 (c) Therefore, using Formula 9, we have A(S) a a b Õ4 ( x a ) b Õ4 ( x a ) 1 + ( x a ) + ( y b ) dydx (d) We substitute a, b 3 in the integral in part(a) to get A(S) u 9u cos v + 4u sin v + 9dudv. We use a CAS to estimate the integral accurate to four decimal places. To speed up the calculation, we can set Digit: 7; (in Maple) or use the approximation command N (in Mathematica). We find that A(S) We first find the area of the face of the surface that intersects the positive y-axis. A parametric representation of the surface is with x + z 1. Then x x y z z, r(x, z) x, z, z r x 1,,,r z,, 1 and z r x r z, 1, r x r z 1 + z 1. A(S) x +z 1 r x r z da z 1 1 z 1 1 z 1 dxdz 1 dxdz

5 by the symmetry of the surface. The integral is improper[when z 1], so A(S) lim t 1 4 t 4 t 1 t lim lim t 1 4 t 1 z dz 1 dxdz dz lim t 1 4t 4 Since the complete surface consists of four congruent faces, the total surface is (4(4)) 16. Alternate solution: The face of the surface that intersects the positive y-axis can also be parametrized as r(x, θ) x, cosθ, sin θ for π θ π and x + z 1 x + sin θ 1 1 sin θ x cosθ x cosθ. 1 sin θ Then r x 1,,, r θ, sin θ, cosθ and r x r z, cosθ, sin θ r x r z 1, so A(S) π π cos θ cos θ 1dxdθ cosθdθ sinθ] π π Again, the area of the complete surface is 4(4) Let A(S 1 ) be the surface area of that portion of the surface which lies above the plane z. Then A(S) A(S 1 ). Following Example 1, a parametric representation of S 1 is x a sin φ cosθ y a sin φ sin θ z a cosφ, and r φ r θ a sin φ. For D, φ π and for each fixed φ, (x a ) + y ( a ) or [a sin φ cosθ a ] + a sin φ sin θ ( a ) a sin φ a sin φ cosθ or sin φ(sin φ cosθ).

6 But φ π, so cosθ sin φ or sin(π + θ) sin φ or φ π θ π φ. Hence D {(φ, θ) φ π, φ π θ π φ}. Then A(S1 ) φ φ π a sin φdθdφ a [( π cosφ) ( φ cosφ + sin φ)] π a (π ). Thus A(S) a (π ). Alternate solution: Working on S 1 we could parametrize the portion of the sphere by x x y y z a x y. Then r x r y 1 + x a x y + y a x y a a x y and A(S 1 ) a a x y da (x ( a ) ) +y ( a ) π acos θ π π π π a a r drdθ a(a r ) 1/ ] racos θ r dθ a [1 (1 cos θ) 1/ ]dθ a (1 sinθ )dθ a Thus A(S) 4a ( π 1) a (π ). (1 sin θ)dθ a ( π 1). Notes: (1) Perhaps working in spherical coordinates is the most obvious approch here. However, you must be careful in setting up D. () In the alternate solution, you can avoid having to use sin θ by working in the first octant and then multiplying by 4. However, if you set up S 1 as above and arrived at A(S 1 ) a π, you now see your error.