Lecture 14: 7/24/2003, SURFACE AREA Maths21a, O. KnillSURFACE AREAR RR|~ru(u, v) ×~rv(u, v)| dudvis the area of the surface.INTEGRAL OF A SCALAR FUNCTION ON A SURFACE. IfS is a surface, thenR RSf(x, y) dS should be an average of f onthe surface. If f (x, y) = 1, thenR RSdS should be the area ofthe surface. If S is the image of ~r under the map (u, v) 7→ ~r(u, v),then dS = |~ru×~rv| dudv.DEFINITION. Given a surface S = ~r(R), where R is a domain in the plane and where ~r(u, v) =(x(u, v), y(u, v), z(u, v)). The surface integral of f(u, v) on S is defined asZ ZSf dS =Z ZRf(u, v)|~ru×~rv| dudv .INTERPRETATION. If f(x, y) measures a quantity thenR RSf dS is the average of the function f on S.EXPLANATION OF |~ru×~rv|. The vector ~ruis a tangent vector to the curve u 7→ ~r(u, v), when v is fixed andthe vector ~rvis a tangent vector to the curve v 7→ ~r(u, v), when u is fixed. The two vectors span a parallelogramwith area |~ru×~rv|. A little rectangle spanned by [u, u + du] and [v, v + dv] is mapped by ~r to a parallelogramspanned by [~r, ~r + ~ru] and [~r, ~r + ~rv].A simple case: consider ~r(u, v) = (2u, 3v, 0). This surface is part of the xy-plane. The parameter region R justgets stretched by a factor 2 in the x coordinate and by a factor 3 in the y coordinate. ~ru×~rv= (0, 0, 6) and wesee for example that the area of ~r(R) is 6 times the area of R.POLAR COORDINATES. If we take ~r(u, v) = (u cos(v), u sin(v), 0), then the rectangle [0, R] × [0, 2π] ismapped into a flat surface which is a disc in the xy-plane. In this case ~ru× ~rv= (cos(v), sin(v), 0) ×(−u sin(v), u cos(v), 0) = (0, 0, u) and |~ru× ~rv| = u = r. We can explain the integration factor r in polarcoordinates as a special case of a surface integral.THE AREA OF THE SPHERE.The map ~r : (u, v) 7→ (L cos(u) sin(v), L sin(u) sin(v), L cos(v)) maps the rectangle R : [0, 2π] × [0, π] ontothe sphere of radius L. We compute ~ru× ~rv= L sin(v)~r(u, v). So, |~ru× ~rv| = L2|sin(v)| andR RR1 dS =R2π0Rπ0L2sin(v) dvdu = 4πL2.SURFACE AREA OF GRAPHS. For surfaces (u, v) 7→(u, v, f(u, v)), we have ~ru= (1, 0, fu(u, v)) and ~rv=(0, 1, fv(u, v)). The cross product ~ru× ~rv= (−fu, −fv, 1)has the lengthp1 + f2u+ f2v. The area of the surface above aregion R isR RRp1 + f2u+ f2vdA.EXAMPLE. The surface area of the paraboloid z = f(x, y) = x2+y2is (use polar coordinates)R2π0√1 + 4r2r drdθ = 2π(2/3)(1 +4r2)3/2/8|10= π(53/2− 1)/6.AREA OF SURFACES OF REVOLUTION. If we rotatethe graph of a function f(x) on an interval [a, b] aroundthe x-axis, we get a surface parameterized by (u, v) 7→~r(u, v) = (v, f(v) cos(u), f(v) sin(u)) on R = [0, 2π] × [a, b]and is called a surface of revolution. We have ~ru=(0, −f(v) sin(u), f (v) cos(u)), ~rv= (1, f0(v) cos(u), f0(v) sin(u))and ~ru× ~rv= (−f(v)f0(v), f(v) cos(u), f (v) sin(u))= f(v)(−f0(v), cos(u), sin(u)). The surface area isR R|~ru×~rv| dudv = 2πRba|f(v)|p1 + f0(v)2dv.EXAMPLE. If f(x) = x on [0, 1], we get the surface area of a cone:R2π0R10x√1 + 1 dvdu = 2π√2/2 = π√2.P.S. In computer graphics, surfaces of revolutions are constructed from a few prescribed points (xi, f(xi)). Themachine constructs a function (spline) and rotatesGABRIEL’S TRUMPET. Take f(x) = 1/x on the interval [1, ∞).Volume: The volume is (use cylindrical coordinates in the x di-rection):R∞1πf(x)2dx = πR∞11/x2dx = π.Area: The area isR2π0R∞11/xp1 + 1/x4dx ≥ 2πR∞11/x dx =2π log(x)|∞0= ∞.The Gabriel trumpet is a surface of finite volume but with infinite surface area! You can fill the trumpet witha finite amount of paint, but this paint does not suffice to cover the surface of the trumpet!Question. How long does a Gabriel trumpet have to be so that its surface is 500cm2(area of sheet of paper)?Because 1 ≤p1 + 1/x4≤√2, the area for a trumpet of length L is between 2πRL11/x dx = 2π log(L) and√22π log(L). In our case, L is between e500/(√22π)∼ 2 ∗ 1024cm and e500/(2π)∼ 4 ∗ 1034cm. Note that theuniverse is about 1026cm long (assuming that the universe expanded with speed of light since 15 Billion year).It could not accommodate a Gabriel trumpet with the surface area of a sheet of paper.M¨OBIUS STRIP. The surface ~r(u, v) = (2+v cos(u/2) cos(u), (2+v cos(u/2)) sin(u), v sin(u/2)) parametrized by R = [0, 2π]×[−1, 1]is called a M¨obius strip.The calculation of |~ru×~rv| = 4+3v2/4+4v cos(u/2)+v2cos(u)/2 isstraightforward but a bit tedious. The integral over [0, 2π]×[−1, 1]is 17π.QUESTION. If we build the Moebius strip from paper. What isthe relation between the area of the surface and the weight of thesurface?REMARKS.1) An OpenGL implementation of an Escher theme can be ad-mired with”xlock -inwindow -mode moebius”on an X-terminal. 2) A patent was once assigned to the idea touse a Moebius strip as a conveyor belt. It would last twice aslong as an ordinary
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