Spring 2004 Math 253/ Vector Calculus 14.7 Surface Integrals Tue, 13/Apr c 2004, Art Belmonte

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1 pring Math / Vector Calculus.7 urface Integrals Tue, /Apr c, Art Belmonte ummar efinitions Recall that a parametric surface in -space is the graph of a vector function s : R R of two parameters. s(u,v) =(u,v),(u,v),(u,v),(u,v) R Often the parameters are chosen to be two of the rectangular coordinates,, ; the polar/clindrical coordinates r,θ;the spherical coordinates θ,φ; or the clindrical coordinates,θ. The surface integral of a scalar field f over a parametric surface s is given b fd= f(s(u,v)) s u s v du dv. The surface integral of a vector field w over a oriented parametric surface s (also called the flu of w across s) is given b w d = w(s(u,v)) (s u s v ) du dv. An oriented surface has two sides; e.g., top and bottom; left and right; front and back; inward and outward; etc. If our chosen surface has an orientation s u s v that is opposite to what is desired in a given problem, simpl negate the value of the surface integral computed. This has the effect of using (s u s v ), the other orientation. Applications of urface Integrals Let δ be the mass densit and σ the charge densit. The center of mass is also called the centroid when the densit is constant. Application urface Integrals scalar differential d = s u s v du dv measure A() = d surface area total mass m = δ d electric charge Q = σ d moments M = δ d center of mass, ȳ, = m,, δ d moments of inertia I = ( + ) δ d radii of gration = I /m, ȳ = I /m, = I /m Table Notes. Other first-order moments are smmetricall defined. M = δd, M = δd. Other second-order moments are smmetricall defined. ( I = + ) ( δd, I = + ) δd Hand / MATLAB Eamples 9/ Evaluate the surface integral d, where the surface is the triangular patch with vertices (,, ), (,, ), and(,,) tewart 9/: Triangular patch tewart 9/: hadow region. The surface is pictured at left above. Its projection onto the -plane appears to the right of it. The routine planept on our TI-89 or in MATLAB gives + + = or= as an equation of the plane through the three noncollinear vertices. A rectangular parameteriation of the surface is s(, ) =,,,,. Compute s s =,,,, =,,. Then s s =. With f (,, ) =,wehave fd = f(s(,)) s s dd = ( ) dd =.7. 9/ revisited Evaluate the surface integral d, where the surface is the triangular patch with vertices (,, ), (,, ), and(,,).

2 Clearl there is a lot of work involved in computing a surface integral b hand. (It is a nice review of all the material in the course, however!) Recall from ection.6 that the sis routine (available on the TI-89 and in MATLAB) computes the eact value of the surface integral of a scalar field automaticall. Here s the needful. (If the surface integral of a scalar field cannot be computed eactl, just use sisn in MATLAB.) = r, r cos θ,r sin θ. Thus s r s θ = r r +. efine f (,, ) = +.Then f(s(r,θ)) = r. Hence fd = f(s(r,θ)) s r s θ dr dθ π = r r r + dr dθ ( = 6 π 9 ) tewart 9/ sms f = *; s = --; M = -; ; = sis(f,s,m); prett() double().7 echo off; diar off / / 9/6 revisited Evaluate the surface integral + d, where the surface is the part of the circular paraboloid = that lies in front of the plane =, the -plane. This time we ll just use sis. 9/6 Evaluate the surface integral + d, where the surface is the part of the circular paraboloid = that lies in front of the plane =, the -plane. tewart 9/6 sms r t f = ˆ + ˆ; s = -rˆ r*cos(t) r*sin(t); M = r ; t *pi; = sis(f,s,m); prett() tewart 9/6 tewart 9/6: Projection onto plane double() 8.6 echo off; diar off / /6 pi (9 7 + ) A plot of the surface appears to the left above. On the right is its projection ontothe -plane. A parameteriation polar/clindrical of the surface is s(r,θ)= r,rcos θ,r sin θ, r, θ π. Now s r s θ = r, cos θ,sin θ, r sin θ,r cos θ 9/ Evaluate the surface integral + + d,wherethe surface consists of the part of the circular clinder + = 9 between the planes = and= together with its top and bottom circular disks; in other words, a closed tin can! OK, campers: we have three surface integrals to compute. Let s set up parameteriations for the side, bottom, and top of the tin can, then dispatch the whole shooting match with repeated invocations of sis added together!

3 tewart 9/: Tin can = /*pi prett() double().888 echo off; diar off / pi tewart 9/ sms r t f = ˆ + ˆ + ˆ; s = *cos(t) *sin(t), ; side s = r*cos(t) r*sin(t), ; bottom s = r*cos(t) r*sin(t), ; top M = t *pi; ; side ranges M = r ; t *pi; bottom ranges M = r ; t *pi; top ranges = sis(f,s,m) + sis(f,s,m) + sis(f,s,m); prett() Add em up! double() 77.8 echo off; diar off pi 9/6 Evaluate the surface integral w d of the vector field w =,, over the part of the circular paraboloid = + 9 that lies below the rectangle, and has downward orientation. That is, compute the flu of w across this surface. 6 tewart 9/6 9/ 8 Evaluate the surface integral + + d, wherethe surface is the helicoid (spiral ramp). s(u,v) =ucosv, u sin v, v, u, v π. tewart 9/ The surface appears above. You are handed a rectangular parameteriation of the surface: s(, ) =,, + 9,,. Again we use sis.. tewart 9/ sms u v f = sqrt( + ˆ + ˆ); s = u*cos(v) u*sin(v), v; M = u ; v pi; = sis(f,s,m) Compute s s =,,,, =,,, an upward orientation. (Look at its k component!) ince the author requested a downward orientation, we ll negate the double integral we compute below. w d = w(s(,)) (s ) s d d =,,,, d d 9/6 revisited = +6 dd =. Evaluate the surface integral w d of the vector field w =,, over the part of the circular paraboloid

4 = + 9 that lies below the rectangle, and has downward orientation. That is, compute the flu of w across this surface. This time we invoke siv, which computes the eact value of the surface integral of a vector field over a surface. We still need to check the orientation, but MATLAB sure beats doing it b hand! NOTE WELL: Internall, siv computes the orientation as s u s v, where u is the inner variable in the range matri M and v is the outer variable. Accordingl, ou should use this same order when computing orientation to decide whether or not to negate the result returned b siv. tewart 9/6 sms w = ˆ* -**ˆ *ˆ; s = ˆ+ˆ-9; M = ; ; For our parameteriation, the normal points UPWAR. The author wants OWNWAR. s = diff(s,); s = diff(s,); cross(s,s) -*, -*, Accordingl, negate the result that siv returns! = -siv(w,s,m) = - 96/8 prett() double() - echo off; diar off Evaluate the surface integral w d of the vector field w =,, over the part of the cone = + between the planes = and=with upward orientation. That is, compute the flu of w across this surface. tewart 9/8 - Here we just use siv. Thefluis 7 6 π 8.. tewart 9/8 sms r t w = - - ˆ; s = r*cos(t) r*sin(t) r; M = r ; t *pi; For our parameteriation, the normal points UPWAR, as the author has requested. sr = diff(s,r); st = diff(s,t); simple(cross(sr,st)) -r*cos(t), -r*sin(t), r Accordingl, use the result that siv returns! = siv(w,s,m) = 96/ 7/6*pi prett() double() 8.7 echo off; diar off 7/6 pi Find the mass (in kg) of a thin funnel in the shape of a cone = +,, if its variable densit is δ =. tewart 96/ The mass is δ d = 8 π 79.8 kg, the surface integral of a scalar field, dispatched with sis.. tewart 96/ sms r t f = - ; s = r*cos(t) r*sin(t) r; M = r ; t *pi;

5 96/6 = sis(f,s,m) = 8*ˆ(/)*pi prett() double() 79.8 echo off; diar off / 8 pi A fluid has densit δ = and velocit v =,,. Find the rate of flow outward through the sphere + + =. tewart 96/6 96/6 revisited A fluid has densit δ = and velocit v =,,. Find the rate of flow outward through the sphere + + =. That hand work we just did was a world o hurt! Instead, let s use siv to compute the surface integral of the vector field w = δv over the sphere. Herewith the needful in MATLAB. Much better! tewart 96/6 sms p t v = - *; w = *v; s = *sin(p)*cos(t) *sin(p)*sin(t) *cos(p); M = p pi; t *pi; sp = diff(s,p); st = diff(s,t); simple(cross(sp,st)) -*(-+cos(p)ˆ)*cos(t), -*(-+cos(p)ˆ)*sin(t), *cos(p)*sin(p) = siv(w,s,m) = *pi prett() Parameterie the surface in a spherical fashion. s(φ, θ) = sin φ cos θ,sinφsin θ,cosφ, φ π, θ π Compute s φ s θ : cos φ cos θ,cosφsin θ, sinφ sinφsin θ,sinφcos θ, or sin φ cos θ, sin φ sin θ, sin φ cos φ,whichis an outward orientation. (Look at its k component while in the first octant.) This is what the author requested. Now compute w(s(φ, θ)) (s φ s θ ) : 7 sin φ sin θ,7 sin φ cos θ, cosφ sin φ cos θ, sin φ sin θ, sin φ cos φ or 7, sin φ cos φ. Now put the pieces together to compute the integal. w d = w(s(φ, θ)) (s ) φ s θ dφ dθ π π = 7, sin φ cos φ dφ dθ =,π.7 6. double().78e+6 echo off; diar off pi