11/14/2002, SURFACE AREA, POLAR COORDINATES Math 21a, O. KnillHomework: Section 12.4: Numbers 14, 22, 30 section 12.6 Numbers 8, 22POLAR COORDINATES. For many regions, it is better to use Polar coordinates for integration:R RRf(x, y) dxdy =R RRf(r, θ)r drdθEXAMPLE. The area of the disc {x2+y2≤ 1} can be computed by treating the region as a type I region and do-ing the integral with x = sin(u), dx = cos(u)du:R1−1R√1−x2−√1−x21 dydx =R1−12√1 − x2dx =Rπ/2−π/22 cos2(u) du =π. It is easier to do that integral in polar coordinates:Z2π0Z10r drdθ = 2πr2/2|10= π .WHERE DOES THE FACTOR ”r” COME FROM?A small rectangle with dimensionsdθdr in the (r, θ) plane is mapped byT : (r, θ)7→(r cos(θ), r sin(θ)) to a sector segmentin the (x, y) plane. It has approxi-mately the area rdθdr. It is small forsmall r.SURFACE AREAR RR|Xu(u, v) × Xv(u, v)| dudvis the area of the surface.INTEGRAL OF A SCALAR FUNCTION ON A SURFACE. If Sis a surface, thenR RSf(x, y) dS should be an average of f on thesurface. If f (x, y) = 1, thenR RSdS should be the area of thesurface. If S is the image of X under the map (u, v) 7→ X(u, v),then dS = |Xu× Xv|dudv.DEFINITION. Given a surface S = X(R), where R is a domain in the plane and where X(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)|Xu× Xv| dudv .INTERPRETATION. If f(x, y) measures a quantity thenR RSf dS is the average of the function f on S.EXPLANATION OF |Xu× Xv|. The vector Xuis a tangent vector to the curve u 7→ X(u, v), when v is fixedand the vector Xvis a tangent vector to the curve v 7→ X(u, v), when u is fixed. The two vectors span aparallelogram with area |Xu×Xv|. A little rectangle spanned by [u, u + du] and [v, v + dv] is mapped by X toa parallelogram spanned by [X, X + Xu] and [X, X + Xv].A simple case: consider X(u, v) = (2u, 3v, 0). This surface is part of the x-y plane. The parameter region Rjust gets stretched by a factor 2 in the x coordinate and by a factor 3 in the y coordinate. Xu× Xv= (0, 0, 6)and we see for example that the area of X(R) is 6 times the area of R.THE AREA OF THE SPHERE. The map X = (u, v) 7→ (cos(u) sin(v), sin(u) sin(v), cos(v)) maps the rectangleR : [0, 2π] × [0, π] onto the sphere. Xu× Xv= sin(v)X(u, v). So, |Xu× Xv| = |sin(v)| andR RR1dS =R2π0Rπ0sin(v) dvdu = 4π.AREA OF GRAPHS. For surfaces (u, v) 7→ (u, v, f (u, v), we have Xu= (1, 0, fu(u, v)) and Xv= (0, 1, fv(u, v)).The cross product Xu× Xv= (−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-axes, we get a surface parameterized by (u, v) 7→X(u, v) = (v, f (v) cos(u), f(v) sin(u)) on R = [0, π] × [a, b]and is called a surface of revolution. We have Xu=(0, −f(v) sin(u), f(v) cos(u)), Xv= (1, f0(v) cos(u), f0(v) sin(u))and Xu× Xv= (−f(v)f0(v), f(v) cos(u), f(v) sin(u)) =f(v)(−f0(v), cos(u), sin(u)) which has the length |Xu× Xv| =|f(v)|p1 + f0(v)2dudv.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.GABRIEL’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! It could hardly accomodate a Gabriel trumpet in our universe if it should havethe surface area of a sheet of paper!M¨OBIUS STRIP. The surface X(u, v) = (2 +v cos(u/2) cos(u), (2 + v cos(u/2)) sin(u), v sin(u/2)) parametrizedby R = [0, 2π] × [−1, 1] is called a M¨obius strip.The calculation of |Xu×Xv| = 4+3v2/4+4v cos(u/2)+v2cos(u)/2is straightforward 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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