16.6 Parametric Surfaces and Their Areas

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1 SECTION 6.6 PARAMETRIC SURFACES AND THEIR AREAS i j k (b) From (a), v = w r = =( ) i +( ) j +( ) k = i + j i j k (c) curl v = v = = () () i + ( ) () j + () ( ) k =[ ( )] k = k =w 9. For any continuous function on R,defineavectorfield G( ) =h( ) i where ( ) = ( ). Then div G = (( )) + () + () = ( ) = ( ) by the Fundamental Theorem of Calculus. Thus every continuous function on R is the divergence of some vector field. 6.6 Parametric Surfaces and Their Areas. (7 4) lies on the parametric surface r( ) =h i if and only if there are values for and where + =7, +5 =,and+ + =4.Butsolvingthefirst two equations simultaneously gives =, =and these values do not satisfy the third equation, so does not lie on the surface. (5 5) lies on the surface if + =5, +5 =,and+ + =5for some values of and. Solving the first two equations simultaneously gives =4, = and these values satisfy the third equation, so lies on the surface.. r( ) =( + ) i +( ) j +(+4 +5) k = h i + h 4i + h 5i. From Example, we recognize this as a vector equation of a plane through the point ( ) and containing vectors a = h 4i and b = h 5i. Ifwe i j k wish to find a more conventional equation for the plane, a normal vector to the plane is a b = 4 =4i j k 5 andanequationoftheplaneis4( ) ( ) ( ) = or 4 = r( ) =, so the corresponding parametric equations for the surface are =, =, =.Forany point ( ) onthesurface,wehave =. With no restrictions on the parameters, the surface is =,which we recognize as a hyperbolic paraboloid. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

2 CHAPTER 6 VECTOR CALCULUS 7. r( ) = +,,. The surface has parametric equations =, =, = +,,. In Maple, the surface can be graphed by entering plotd([uˆ,vˆ,u+v],u=-..,v=-..);. In Mathematica we use the ParametricPlotD command. If we keep constant at, =, a constant, so the corresponding grid curves must be the curves parallel to the -plane. If is constant, we have =, a constant, so these grid curves are the curves parallel to the -plane. 9. r( ) = cos sin 5. The surface has parametric equations = cos, = sin, = 5,,. Note that if = is constant then = 5 is constant and = cos, = sin describe a circle in, of radius, so the corresponding grid curves are circles parallel to the -plane. If =, a constant, the parametric equations become = cos, = sin, = 5.Then =(tan ), so these are the grid curves we see that lie in vertical planes = through the -axis.. =sin, =cossin 4, =sinsin 4,,. Note that if = is constant, then =sin is constant, so the corresponding grid curves must be parallel to the -plane. These are the vertically oriented grid curves we see, each shaped like a figure-eight. When = is held constant, the parametric equations become =sin, =cos sin 4, =sin sin 4. Since is a constant multiple of,the corresponding grid curves are the curves contained in planes = that pass through the -axis.. r( ) = cos i + sin j + k. The parametric equations for the surface are = cos, = sin, =. We look at the grid curves first; if we fix,then and parametrizeastraightlineintheplane = which intersects the -axis. If is held constant, the projection onto the -plane is circular; with =, each grid curve is a helix. The surface is a spiraling ramp, graph IV. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

3 SECTION 6.6 PARAMETRIC SURFACES AND THEIR AREAS 5. r( ) =sin i +cos sin j +sin sin k. Parametric equations for the surface are =sin, =cos sin, =sin sin. If = is fixed, then =sin is constant, and =(sin )cos and =(sin )sin describe a circle of radius sin, so each corresponding grid curve is a circle contained in the vertical plane =sin parallel to the -plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to holding constant, in which case =(cos )sin, =(sin )sin = (tan ),soeachgridcurveliesina plane = that includes the -axis. 7. =cos cos, =sin cos, =sin.if = is held constant then =sin 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 the family = cos, = sin and are called astroids.) The vertical grid curves we see on the surface correspond to = held constant, as then we have =cos cos, =sin cos so the corresponding grid curve lies in the vertical plane =(tan ) through the -axis. 9. From Example, parametric equations for the plane through the point ( ) that contains the vectors a = h i and b = h i are =+() + () =, =+( ) + () =, =+() + ( ) =.. Solving the equation for gives =+ + 4 = (We choose the positive root since we want the part of the hyperboloid that corresponds to.) If we let and be the parameters, parametric equations are =, =, = Since the cone intersects the sphere in the circle + =, = and we want the portion of the sphere above this, we can parametrize the surface as =, =, = 4 where +. Alternate solution: Using spherical coordinates, =sincos, =sinsin, =coswhere and Parametric equations are =, =4cos, =4sin, 5,. 7. The surface appears to be a portion of a circular cylinder of radius with axis the -axis. An equation of the cylinder is + =9, and we can impose the restrictions 5, to obtain the portion shown. To graph the surface on a CAS, we can use parametric equations =, =cos, =sin with the parameter domain 5,. Alternatively, we can regard and as parameters. Then parametric equations are =, =, = 9,where 5 and. 9. Using Equations, we have the parametrization =, = cos, = sin,,. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

4 4 CHAPTER 6 VECTOR CALCULUS. (a) Replacing cos by sin and sin by cos gives parametric equations =(+sin)sin, =(+sin)cos, = +cos. From the graph, it appears that the direction of the spiral is reversed. We can verify this observation by noting that the projection of the spiral grid curves onto the -plane, given by =(+sin)sin, =(+sin)cos, =,drawsacircleintheclockwise direction for each value of. The original equations, on the other hand, give circular projections drawn in the counterclockwise direction. The equation for is identical in both surfaces, so as increases, these grid curves spiral up in opposite directions for the two surfaces. (b) Replacing cos by cos and sin by sin gives parametric equations =(+sin)cos, =(+sin)sin, = +cos. From the graph, it appears that the number of coils in the surface doubles within the same parametric domain. We can verify this observation by noting that the projection of the spiral grid curves onto the -plane, given by =(+sin)cos, =(+sin)sin, =(where is constant), complete circular revolutions for while the original surface requires for a complete revolution. Thus, the new surface winds around twice as fast as the original surface, and since the equation for is identical in both surfaces, we observe twice as many circular coils in the same -interval.. r( ) =( + ) i + j +( ) k. r = i +6j + k and r = i k,sor r = 6 i +j 6k. Since the point ( ) corresponds to =, =,a normal vector to the surface at ( ) is 6 i +j 6k, and an equation of the tangent plane is = 6 or + =. 5. r( ) = cos i + sin j + k r =. r =cosi +sinj and r = sin i + cos j + k, so a normal vector to the surface at the point is r r = i + is j i + j + k + 7. r( ) = i + sin j + cos k r( ) = ( ). r = i +sin j +cos k and r = cos j sin k, so a normal vector to the surface at the point ( ) is r ( ) r ( ) = ( i + k) ( j) = i +4k. Thus an equation of the tangent plane at ( ) is ( ) + ( ) + 4( ) = or + =. = i j + k. Thus an equation of the tangent plane at = or + =. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

5 SECTION 6.6 PARAMETRIC SURFACES AND THEIR AREAS 5 9. The surface is given by = ( ) =6 which intersects the -plane in the line + =6,so is the triangular region given by ( ). By Formula 9, the surface area of is ()= + + = +( ) +( ) = 4 = 4 () = 4 = 4 4. Here we can write = ( ) = and is the disk +, so by Formula 9 the area of the surface is ()= + + = 4 () = 4 = 4 = + + = 4 4. = ( ) = ( + ) and = {( ) }. Then =, = and ()= +( ) + = + + = = ( + +) = ( +) ( +) = = ( 5 +)5 ( 5 +)5 = 4 5 ( )= 4 5 (5 7 +) 45. = ( ) = with +,so =, = ()= = + + = = += = ( +) = 47. A parametric representation of the surface is =, =4 +, = with,. Hence r r =(i +4j) ( j + k) =4i j +k. Note: In general, if = ( ) then r r = i j + k and () = + + ()= = = ln = r = h i, r = hi,andr r =.Then ()= r r = = ( + ) = = ln + ln 7 ( + ). Then + = = 8 +4 = =4 = 5. From Equation 9 we have () = +() +( ). Butif and then ( ), ( ) +( ) +( ) +( ) +( ). By Property 5.., +( ) +( ) () () () (). c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

6 6 CHAPTER 6 VECTOR CALCULUS 5. = ( ) = with + 4. () = + + = +4( + ) ( + ) = +4 = +4 = (a) () = + + Using the Midpoint Rule with ( ) = () = = (b) Using a CAS we have () = to the first decimal place. 57. =+ + +4,so () = + = ( + + ) , =, =we have ( + + ) 4 =4[( ) + ( ) + ( ) + ( ) + (5 ) + (5 )] Using a CAS, we have = 45 8 or ln = This agrees with the estimate in part (a) ( + + ) 4 +4+(+8) = ln ln (a) = sin cos, = sin sin, = cos (b) + + =(sin cos ) +(sin sin ) +(cos) =sin +cos = and since the ranges of and are sufficient to generate the entire graph, the parametric equations represent an ellipsoid. (c) From the parametric equations (with =, =,and =), we calculate r =cos cos i +cos sin j sin k and r = sin sin i +sincos j. Sor r =6sin cos i +sin sin j +sincos k, and the surface area is given by () = r r = 6 sin 4 cos +9sin 4 sin +4cos sin 6. To find the region : = + implies + =4 or =.Thus =or =are the planes where the surfaces intersect. But + + =4 implies + +( ) =4,so =intersects the upper hemisphere. Thus ( ) =4 or =+ 4. Therefore is the region inside the circle + +( ) =4, c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

7 SECTION 6.6 PARAMETRIC SURFACES AND THEIR AREAS 7 that is, = ( ) +. () = +[( )(4 ) ] +[( )(4 ) ] = + 4 = (4 = ) = 4 = = ( +4) = =4 6. Let ( ) be the surface area of that portion of the surface which lies above the plane =.Then() =( ). Following Example, a parametric representation of is = sin cos, = sin sin, = cos and r r = sin. For, and for each fixed, + or sin cos + sin sin () implies sin sin cos or sin (sin cos ). But,socos sin or sin + sin or. Hence = ( ),.Then Thus () = ( ). ( () )= () sin = ( )sin = [( cos ) ( cos +sin)] = ( ) Alternate solution: Working on we could parametrize the portion of the sphere by =, =, =. Then r r = + ( )= + = and ( ()) + () = Thus () =4 = ( ). Notes: ( ) = cos = = cos = [ ( cos ) ] = ( sin ) = ( sin ) = () Perhaps working in spherical coordinates is the most obvious approach here. However, you must be careful in setting up. () In the alternate solution, you can avoid having to use sin by working in the first octant and then multiplying by 4. However, if you set up as above and arrived at ( )=, you now see your error. c Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.