0.1 Practical Guide - Surface Integrals. C (0,0,c) A (0,b,0) A (a,0,0)

Transcription

1 . Practical Guide - urface Integrals urface integral,means to integrate over a surface. We begin with the stud of surfaces. The easiest wa is to give as man familiar eamles as ossible ) a lane surface an lane is comletel de ned b - 3 oints (belonging to that lane), ( when the 3 oints reresent the intersections of the lane with O; O; O, then we have an eas drawing) C (,,c) A (,b,) A (a,,) - the "normal" to the lane and one oint (belonging to the lane), let the normal unit vector be! n = (a; b; c) and ( ; ; ) a oint that belongs to the lane.

2 A(o,o,o) n M (,,) O the normal vector! n is orthogonal on ever vector in the lane, we use the scalar roduct! n?! AM, <! n ;! AM >=! n! AM =, (a; b; c) ( ; ; ) = a + b + c + ( a b { c ) = }, a + b + c + d = d - vectors in the lane and one oint (belonging to the lane), let! u ;! v be (linearl indeendent) vectors in the lane and ( ; ; ) a oint that belongs to the lane. We use the vector roduct (cross roduct) to get a vector normal to the lane, since! u! v?! u,! u! v?! v consequentl al the revious comutation with! n =! u! v ) a clinder, we consider a simle case, the clinder is orthogonal on the O lane, has radius = r and its ais is O

3 the rojection of the clinder onto the O lane is clearl - a circle ( for a circular clinder), which equation is + = r - an ellise ( for an ellitical clinder) which equation is a + b = b r a this elains wh in man tet books ou will nd the "equation" of such a clinder as + = r or a + b = which is a bit confusing, since these are equations for a circle or an ellise. Actuall the "full" equation of such a clinder is + = r and or a + b = and 3

4 4) a cone, we consider a simle case, the ais of the cone is O and it is circular or ellitical, the equation is = a ( + ) or = a + b 3) a shere or an ellisoid centered at the origin (; ; ) r c r b r a the equations are + + = r or a + b + c = 4) a araboloid ( circular, ellitical,...) in a simle case, with ais O 4

5 the equations are = a ( + ) or = a + b o we have to memorise all these equations? No! We have two eas roblems. i) what kind of surface does an equation reresent? ii) what is the equation for a surface de ned b its geometric roerties? Both roblems are easil solved b cutting (intersecting) the surface with horiontal lanes and with the "vertical" lane O Eamles.. What kind of surface reresents the equation = 4( + )? Cut the surface with horiontal lanes, that is =constant= c, since in this case we get 4( + ) = c, + = c which are clearl circles, "located" at = c, the greater is, the greater the radius of these (horiontal) circles 5

6 now cut the surface with the O lane, whose equation is =, ut = in = 4( + ), we get = 4 which is clearl a arabola in the O lane = 4² nall the surface is "generated" b all these (horiontal) circles along a arabola, and we get a circular araboloid 6

7 . what kind of surface reresents the equation = 4( + )? cut the surface with horiontal lanes, that is =constant= c, we get 4( + ) = c, + = c which are clearl (horiontal) circles now cut the surface with the O lane, whose equation is =, ut = in = 4( + ), we get = 4, 4 =, ( )( + ) =, = or + = which are clearl the equations of two straight lines in the O lane 7

8 = = nall the surface is "generated" b all these (horiontal) circles along (between) two straght lines, and we get a cone Now for the converse roblem, how to determine the equation of a surface de ned b its geometric roerties we roceed somehow "backwards" as before. Eamles. 8

9 3. a shere centered at (a; b; c) with radius r. All the oints (; ; ) on the shere have a constant distance = r to the center (a; b; c), which leads to distance = ( a) + ( b) + ( c) = r, ( a) + ( b) + ( c) = r 4. A circular cone with ais O. We need just the radius of one horiontal circle, sa r = and the corresonding = 3, we get the "cut" (intersection) with the O lane, as two straight lines = 3, = 3, = (3/) r = = (3/) = 3 which means that for = the equation of the cone should be ( 3 )( + 3 ) =, 9 4 =, 4 = 9 since b cutting with horiontal lanes ( =constant ) we get circles, the "full" equation of the cone should be 9( + ) = 4 Now we get to "arametried" surfaces. In simle words, it means we de ne a surface using two coordinates (variables), that is we "ma" the surface onto a lane surface (the "sace" of the coordinates) To get a better idea, think of the surface of the Earth and the "geograhic" coordinates : longitude and latitude. The "ma" the (curved) surface of the Earth onto a "lane", (actuall the ma from a geograhic atlas) de ned b the two coordinates. The "standard" de nition is the following. or e nition. A function (corresondence, ma) h :! 3 = (u; v) = (u; v) = (u; v) is called a arametriation of the surface, or is a arametried surface. We assume the function h to be injective (one to one). Also to be of class C, that is, it has artial derivatives which are continuous. as h(u; v) = ((u; v); (u; v); (u; v)) 9

10 Also "nonsingular", that is, the Jacobi matri has rank at ever oint (u; It looks a bit comlicated, but actuall @v A = Comments. We clearl need "one to one", ever oint on the surface (; ; ) should not "overla". emember a arametried ath should have its own coordinates, the γ (t) γ(t) : [a; b]! 3, (t) = ((t); (t); (t)) is the "osition" vector, and b taking the derivatives (t) = ( (t); (t); (t)) is the "velocit" vector, which is tangent to the ath. Now for a arametried surface, we have a "similar" situation. Let! r = (; ; ) be the "osition" vector.! r = (; ; ) =! i +! j +! k = ((u; v), (u; v), (u; v))

11 ru rv rv ru The lane which is tangent to the surface is de ned b two vectors, the "two" artial derivatives, with resect to u and v! ru @u ),! rv @v ) To actuall de ne a lane these two vectors should be "linearl indeendent", which means their matri has @v A = And this is how we get the de nition as stated before. Clearl the vector roduct! r u! r v is orthogonal to the tangent lane, thus a normal direction. We should also add that is a "domain", that is ever oint (; ) is "surounded" b an oen disk (; ) f(u ) + (v ) < rg Each surface has in nitel man arametriations. In the following we basicall consider just two kinds of arametriations: - rojecting on the lanes O, or O, or O, which works for almost all surfaces in "school" roblems - using sherical coordinates for a shere or an ellisoid To better understand, consider the surfaces we alread studied in the begining. e nition. Now we clearl state: the domains and are bounded. The function f :!, is a continuous function on a neiborhood of The "standard" notation for surface integral is f(; ; )ds or f(; ; )d or Z f(; ; )d for the constant function the surface integral reresents the area of the surface f(; ; ) = for all (; ; ), the "area" area() = ds

12 and nall the surface integral is reduced to a double integral f(; ; )ds = f((u; v); (u; v); (u; v)) k! r u! r v k dudv The value of the integral does not deend on the arametriation. Comments. To comute a surface integral we basicall need to - nd a suitable arametriation, - determine the domain, - comute k! r u! r v k Case I. The surface is "rojectable" onto one of the lanes O, O or O, that is the rojection is a "one to one" function. Projecting the surface onto the O lane. (Whenever this is ossible) Let be the rojection of on the lane O. An eas wa to get the rojection is to set = in the equation de ning the surface (whenever this equation is given) Then = f(; ; ) 3, (; ), = (; )g The surface is arametried as 3 (; )! (; ; (; )) (we actuall identif with the O lane b (; ) $ (; ; ) ) (,,(,)) = (,) (,,) In this case! r = (; ; ) = (; ; (; )) and we take derivatives with resect to and! r (; ; (; )) = ),! r (; ; (; )) = ) and the vector roduct is! r! r =!!! i = )! i )! j + ()! k ;

13 and the norm is k! r! r k = s = s and nall the surface integral comutes as s f(; ; )ds = f(; ; (; )) k! r! r k dd = f(; ; (; )) + we de nitel ma use the nal formula directl an time we need. emark. A simle question rises naturall. If the surface is a lane surface in the O lane, is there an di erence between the surface integral and the double integral? The answer is No! No di erence at all. In this case the rojection of onto the lane O is itself, the surface is arametried as (; )! (; ; ), therefore = (; ) = is a constant function and the derivatives are = consequentl the surface intergal comutes as f(; ; )ds = f(; ; )dd and the surface intergal turns out to be eactl a double integral. The same haens if is a lane surface arallel to the O lane, de ned b = c =constant. Again the artial derivaties are = The cases when the surface is rojectable onto the O or O lanes are similar. Eamles. ) Comute the area of the hemishere = f(; ; ) 3, + + =, g This surface is rojectable onto the O lane and the rojection is the disc = f + g ( for = ) You should notice that this hemishere is not rojectable onto the lanes O or From the equation of the surface + + = we get = (; ) =, (; ) 3

14 = sqrt(² ² ²) (,,) "sqrt" stands for "square root", =, because some grahic facilities are not installed et. Clearl we have = (; ) = The hemishere rojects onto the O lane into the disk = f(; ; ) 3 ; + g t 4

15 and the area is s area() = ds = + dd s = dd = s + + dd = dd = use now olar coordinates to comute the double integral = r cos t, = r sin t in order to cover the disc we need t [; ] and r [; ], so nall we get = Z Z r rdt A dr = which is indeed the area of an emishere with radius =. r rdr = r r= = = ) Comute the area of the surface = f(; ; ) 3 ; = + 4g, (a iece from a araboloid) This a araboloid. B rojecting it on the O lane we get a disk = f(; ; ) 3 ; + 4g, centered at (; ; ) with radius ( for = ) The artial derivatives are r= = (; ) = = The area of is s area() = ds = + = dd + + dd = q + () + () dd = net we use olar coordinates to comute the double integral = r cos t, = r sin t, and we need t [; ] and r [; ] to cover the disk, so nall we get Z = + r rdta dr = 4 Z + r rdr = 4 + r r= r= = 4( 5 ) 3) Comute the area of the surface which is the ortion from the hemishere f + + = 4, g "cut" b the clinder f(; ; ) 3, + =, g. 5

16 This surface is rojectable on the lane O and the rojection is the disc = f(; ), + g centered at (; ) and radius =, since +, + +, ( ) + and we ma de ne the surface as follows = f(; ; ) 3, (; ), = (; ) = 4 g. Consequentl the rojection of the surface is a "smaller" disk, centered at (; ), tangent to O and to the circle + = 4 6

17 Consequentl we =, = 4 s area() = ds = + dd s = dd = v u t + 4! + 4 dd = 4! = then use olar coordinates r cos t, r = r sin t to comute the double integral the equation of the disc in olar coordinates is r r cos t, r cos t, so cos t ) t [ and r cos t area() = ::: = Z cos t rdr A dt = 4 r Z ( ) 4 r r= cos t r= dt = ; ] = Z 4 4 cos t dt = 4 Z sin t dt = 4 Z ( jsin tj) dt = = 4 4 Z sin tdt 4 Z sin tdt = 4 4 cos tj 4 ( cos t)j = = 4 8 Case II. Using sherical coordinates The arametriation of a iece of a shere centered at (; ; ) with radius is = (; ') = cos ' sin, = (; ') = sin ' sin, = (; ') = cos 7

18 θ ϕ the artial derivatives r ; ; = ( cos ' cos ; sin ' cos ; sin ' ; ; = ( sin ' sin ; cos ' sin ; r r! i j k ' = cos ' cos sin ' cos sin sin ' sin cos ' sin = = sin ' cos sin cos ' sin! i cos ' cos sin sin ' sin! cos ' cos sin ' cos j + sin ' sin cos ' sin! k = = cos ' sin ; sin ' sin ; cos ' cos + sin ' cos A = sin ( cos ' sin ; sin ' sin ; cos ) { } cos k! r r! ' k = sin r ( cos ' sin ) + ( sin ' sin ) + (cos ) = sin (cos ' + sin ' { } ) sin + cos = sin and the surface integral becames f(; ; )ds = f( cos ' sin ; sin ' sin ; cos ) sin dd' All we need to remember, are the sherical coordinates and the fact that k! r! r ' k = sin Eamles. 4) Comute the area of the shere = f(; ; ) 3, + + = g, using sherical coordinates. To cover the whole (surface) shere we clearl need, [; ], ' [; ] 8

19 = Z area() = ds = [;][;] k! r! r ' k dd' = Z sin d' A d = sin d = ( cos j ) = ( cos + cos ) = 4 5) Comute the area of the surface from the shere f(; ; ) 3, + + = 8 g that is located "inside" the cone = +. θ ϕ We use sherical coordinates. Clearl the angle ' goes all around the O ais, that is ' [; ]. The radius is = We need to nd out the etent of the angle. Find the intersection between the shere and the cone + + = 8 = + ), = 8 ) = Therefore we have OA =, AB =, sin = = ) = 4 9

20 B A θ O C Finall the area of the surface "" is area() = ds = = =4 Z sin d' A d = [; 4 ][;] k! r! r ' k dd' = =4 Z sin d = [; 4 ][;] sin dd' = cos j ==4 = = ( )