16.6 Parametric Surfaces and their Areas Can define surfaces similarly to spacecurves: need two parameters u, v instead of just t. Definition. Let x, y, z be functions of two variables u, v, all with the same domain D. The parametric surface defined by the co-ordinate functions x, y, z is the collection S of position vectors r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, for all (u, v) 2 D v r D u (u, v) We think of a parameterized surface as a function r : D ! R3. Example The surface S parameterized by r(u, v) = u cos vi + u sin vj + u2 cos 2vk is drawn for 0 ≤ u ≤ 1 and 0 ≤ v < 2p. Since x2 − y2 = u2 cos2 v − u2 sin2 v = u2 cos 2v = z we recognize S as the saddle surface z = x2 − y2 Definition. The grid lines of a parametric surface S are the two families of curves for which either u or v is constant. Important: surface distinct from parameterization! All surfaces can be parameterized in infinite many ways. Below are two different parameterizations of the saddle surface with the grid lines marked. Note how the grid lines look like a map for the surface. If you rotate the images to look at the (x, y)- plane, the grid lines should look familiar. 0 u cos v 1 0 u 1 r1(u, v) = @ u sin v A r2(u, v) = @ v A u2 cos 2v u2 − v2 1 Standard Parameterized Surfaces Planes The plane through a point with position vector r0 and containing the fixed vectors u and v has parametric equation r(u, v) = r0 + uu + vv, u, v 2 R − · u×v The grid lines are parallel to u, v. Equivalent to (r r0) n = 0 where n = ju×vj is the unit normal vector. The blue lines in the picture are the grid lines with u = 0, u = 1 and u = 2 respectively. The orange lines are v = 0, v = 1 and v = 2. Any other choice of r0, u, v leads to a new parameterization. Cylinders A cylinder is a parametric surface r(u, v) = p(u) + vv, u0 ≤ u ≤ u1, v 2 R, where p(u) is a spacecurve and v is a constant vec- tor. The grid lines come in two families: copies of the original spacecurve translated in the v-direction, and lines parallel to v. A right-circular cylinder is a cylinder where p(u) is a circle and v is perpendicular to the plane of the circle. Graphs of Functions The graph of any function z = f (x, y) is automatically parameterized: take (u, v) = (x, y) to obtain r(u, v) = ui + vj + f (u, v)k We can also consider graphs parameterized using polar-co-ordinates (u, v) = (r, q) r(u, v) = u cos vi + u sin vj + f (u, v)k 2 The saddle surface z = x2 − y2 was parameterized earlier using both these methods. It is often easiest to think of a surface as a graph: for example. Parameterize the surface S, being the part of the paraboloid z = 10 − x2 − y2 lying inside the cylinder (x − 1)2 + y2 = 4. Viewing S as a graph, we first project onto the xy-plane to obtain the domain of the function: the disk D of radius 2 cen- tered at (1, 0). Now parameterize D using modified polar co-ordinates: (x, y) = (1 + u cos v, u sin v) Finally, compute z using the formula for the paraboloid to obtain the parameterization 0 1 + u cos v 1 r(u, v) = @ u sin v A 10 − (1 + u cos v)2 − u2 sin2 v 0 1 + u cos v 1 = @ u sin v A 9 − u2 − 2u cos v where 0 ≤ u ≤ 2, 0 ≤ v < 2p. Spheres The sphere of radius R has three common parameterizations: p • As two graphs in Cartesian co-ordinates: z = ± f (x, y) = ± R2 − x2 − y2. p • As two graphs in polar co-ordinates: z = ± f (r, q) = ± R2 − r2. 0sin u cos v1 • Using spherical co-ordinates: (u, v) = (f, q) =) r(u, v) = R @sin u sin vA cos u The grid lines for each are drawn below. 3 Surfaces of revolution Rotate a curve y = f (x) for a ≤ x ≤ b around the x-axis. Letting u measure the distance along the x-axis and v the polar angle in the yz-plane, we obtain r(u, v) = ui + f (u)(cos vj + sin vk) The plot is the result when f (x) = 2 + cos x The grid lines come in two families: copies of the original curve rotated around the x-axis, and cir- cles centered on the x-axis. Curves can be rotated around any axis for alterna- tive versions. Tangent Planes Definition. Let r(u, v) be a parameterization of S. A point r(uˆ, vˆ) is regular if (ru × rv)(uˆ, vˆ) 6= 0. The tangent plane at a regular point r(uˆ, vˆ) is the plane through said point spanned by the tangent vectors ru(uˆ, vˆ) and rv(uˆ, vˆ). The unit normal vector is n(uˆ, vˆ) = ru×rv (uˆ, vˆ). jru×rvj At r(1, p ) = ( p1 , p1 , 1), the paraboloid 4 2 2 r(u, v) = u cos vi + u sin vj + (2 − u2)k has tangent vectors 0 1 0 p1 1 cos v 2 1 ru = sin v = B p C @ A @ 2 A −2u −2 0 1 0− p1 1 − sin v 2 1 rv = u cos v = B p C @ A @ 2 A 0 0 The unit normal vector can be easily found: p p 0 21 0 21 p 1 p ru × rv = @ 2A =) n = p @ 2A 1 5 1 The tangent plane therefore has equation p p 1 p p 2(x − p1 ) + 2(y − p ) + (z − 1) = 0 () 2x + 2y + z = 3 2 2 4 Surface area 3 Let S be parameterized by r : D ! R and consider a point P0 = r(a, b) on S. Increase each of the co-ordinates u, v by small quantities Du, Dv to give three new points P1 = r(a + Du, b), P2 = r(a + Du, b + Dv), P3 = r(a, b + Dv). These four points are the corners of a small piece of surface DS = fr(u, v) : a ≤ u ≤ a + Du, b ≤ v ≤ b + Dvg By the definition of partial derivative, we see that r(a + Du, b) − r(a, b) dr ≈ (a, b) =) r(a + Du, b) − r(a, b) ≈ r (a, b)Du Du du u Similarly r(a, b + Dv) − r(a, b) ≈ rv(a, b)Dv. It follows that the small piece of surface DS is approximately the parallelogram in the tangent plane at P0 with sides ruDu and rvDv. Since cross-products compute the area of a parallelogram, we see that the area of DS is A(DS) ≈ jru × rvj DuDv We can approximate the surface area of the entire surface S by summing the areas of small approxi- mating parallelograms. Explicitly, if the domain D is a rectangle, subdivided into mn subrectangles, each containing a sample point (ui, vj),, then the area of S is approximately the Riemann sum m n Area(S) ≈ ∑ ∑ ru(ui, vj) × rv(ui, vj) DuDv i=1 j=1 Torus approximated by 192 and 1728 parallelograms resp. Taking limits as m, n ! ¥, we obtain. Theorem. If a surface S is defined by the vector-valued function r(u, v) where u, v are co-ordinates on some domain D ⊆ R2, then the surface area of S is given by ZZ A(S) = jru × rvj du dv D 5 Definition. dS = jru × rvj du dv is the area element of a surface. Lemma. dS is independent of co-ordinates. Indeed, if u = u(a, b) and v = v(a, b) for some alternative co-ordinates (a, b), then 9 j × j = ¶(a,b) × ru rv ¶(u,v) ra rb = =) jr × r j du dv = r × r da db ¶(u,v) u v a b du dv = ¶(a,b) da db ; Surface Area of a Graph If S is the graph of z = f (x, y) for (x, y) 2 D, then we may parameterize with u = x and v = y: r(u, v) = ui + vj + f (u, v)k Therefore 0 1 0 1 0 1 1 0 − fx jru × rvj = rx × ry = @ 0 A × @ 1 A = @− fyA fx fy 1 q 2 2 = fx + fy + 1 Hence the surface area is ZZ q 2 2 A(S) = fx + fy + 1 dx dy D Example The plane z = y − 2x + 2 for 0 ≤ x, y ≤ 1 has surface area Z 1 Z 1 q p A(S) = (−2)2 + 12 + 1 dx dy = 6 0 0 Surfaces of Revolution If S is parameterized by u = x and v = q, then r(u, v) = ui + f (v)(cos vj + sin vk) Supposing that a ≤ x ≤ b and 0 ≤ q < 2p, we have 1 0 f 0(x) 0( ) cos q jru × rvj = jrx × rqj f x × f (x) − sin q = f (x) − cos q f 0(x) sin q cos q − sin q q = f (x) f 0(x)2 + 1 (supposing f (x) ≥ 0) The surface area is therefore Z 2p Z b q Z b q A(S) = f (x) f 0(x)2 + 1 dx dq = 2p f (x) f 0(x)2 + 1 dx 0 a a 6 p Example If f (x) = x for 0 ≤ x ≤ 1 then its surface area is s Z 1 1 2 A = 2p x1/2 p + 1 dx 0 2 x Z 1 p p = p (4x + 1)1/2 dx = (5 5 − 1) 0 6 More General Surfaces Expect hard calculations where you have parameterize the surface and cal- culate ru × rv yourself! For example.