# Review guide for the final exam in Math 233

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1 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered beginning pproximtely with chpter 15.4 of the book. We first recll polr coordintes formuls given in chpter 1.3. The coordintes of point (x, y) R 3 cn be described by the equtions: x r cos(θ) y r sin(θ), (1) where r x + y is the distnce from the origin nd ( x, y ) is (cos(θ), sin(θ)) r r on the unit circle. Note tht r nd θ cn be tken to lie in the intervl [, π). To find r nd θ when x nd y re known, we use the equtions: r x + y tn(θ) y x. () Exmple 1 onvert the point (, π ) from polr to rtesin coordintes. 3 Since r nd θ π, Equtions 1 give 3 x r cos(θ) cos π y r sin(θ) sin π Therefore, the point is (1, 3) in rtesin coordintes. Exmple Represent the point with rtesin coordintes (1, 1) in terms of polr coordintes. If we choose r to be positive, then Equtions give r x + y 1 + ( 1) tn(θ) y x 1. Since the point (1, 1) lies in the fourth qudrnt, we cn choose θ π 4 or θ 7π 4. Thus, one possible nswer is (, π 4 ); nother is (r, θ) (, 7π 4 ). The next theorem describes how to clculte the integrl of function f(x, y) over polr rectngle. Note tht da r dr dθ.

2 1 BASI MATERIAL. Theorem 3 (hnge to Polr oordintes in Double Integrl) If f is continuous on polr rectngle R given by r b, α θ β, where β α π, then R f(x, y)da β b α f(r cos(θ), r sin(θ))r dr dθ. Exmple 4 Evlute R (3x + 4y )da, where R is the region in the upper hlf-plne bounded by the circles x y 1 nd x + y 4. The region R cn be described s R {(x, y) y, 1 x + y 4}. It is hlf-ring nd in polr coordintes it is given by 1 r, θ π. Therefore, by Theorem 3, R π (3x + 4y )da π 1 π [r 3 cos(θ) + r 4 sin (θ)] r r1 dθ π 1 (3r cos(θ) + 4r sin (θ))r dr dθ (3r cos(θ) + 4r 3 sin (θ)) dr dθ π [7 cos(θ) + 15 (1 cos(θ))] dθ 7 sin(θ) + 15θ 15 4 sin(θ) (7 cos(θ) + 15 sin (θ)) dθ ] π 15π. Exmple 5 Find the volume of the solid bounded by the plne z nd the prboloid z 1 x y. If we put z in the eqution of the prboloid, we get x +y 1. This mens tht the plne intersects the prboloid in the circle x +y 1, so the solid lies under the prboloid nd bove the circulr disk D given by x + y 1. In polr coordintes D is given by r 1, θ π. Since 1 x y 1 r, the volume is V D π 1 (1 x y )da π 1 [ r (r r 3 ) dr dθ π (1 r )r dr dθ r4 4 ] 1 π. The next theorem extends our previous ppliction of Fubini s theorem for type II regions.

3 1 BASI MATERIAL. 3 Theorem 6 If f continuous on polr region of the form then D D {(r, θ) α θ β, h 1 (θ) r h (θ)} f(x, y)da β h (θ) α h 1 (θ) f(r cos(θ), r sin(θ))r dr dθ The next definition describes the notion of vector field. We hve lredy seen n exmple of vector field ssocited to function f(x, y) defined on domin D R, nmely the grdient vector field f(x, y) f x (x, y), f y (x, y). In nture nd in physics, we hve the fmilir exmples of the velocity vector field in wether nd force vector fields tht rise in grvittionl fields, electric nd mgnetic fields. Definition 7 Let D be set in R ( plne region). A vector field on R is function F tht ssigns to ech point (x, y) in D two-dimensionl vector F(x, y). Definition 8 Let E be subset of R 3. A vector field on R 3 is function F tht ssigns to ech point (x, y, z) in E three-dimensionl vector F(x, y, z). Note tht vector field F on R 3 cn be expressed by its component functions. So if F (P, Q, R), then: F(x, y, z) P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k. We now describe our first kind of line integrl. These type of integrls rise form integrting function long curve in the plne or in R 3. The type of line integrl described in the next definition is clled line integrl with respect to rc length. Definition 9 Let be smooth curve in R. Given n, consider n equl subdivisions of lengths s i ; let (x i, yi ) denote the midpoints of the i-th subdivision. If f is rel vlued function defined on, then the line integrl of f long is n f(x, y) ds lim f(x i, yi ) s i, n if this limit exists. The following formul cn be used to evlute this type of line integrl. Theorem 1 Suppose f(x, y) is continuous function on differentible curve (t), : [, b] R. Then b (dx ) ( ) dy f(x, y) ds f(x(t), y(t)) + In the bove formul, is the speed of (t) t time t. (dx ) + j1 ( ) dy,

4 1 BASI MATERIAL. 4 Exmple 11 Evlute x ds, where consists of the rc 1 of the prbol y x from (, ) to (1, 1). become Therefore, We cn choose x s the time prmeter nd the equtions for 1 x x y x x 1 1 x ds 1 (dx ) ( ) dy x + dx dx dx x 1 + 4x dx (1 + 4x ) 3 ] Actully for wht we will studying next, nother type of line integrl will be importnt. These line integrls re clled line integrls of f long with respect to x nd y. They re defined respectively for x nd y by the following limits: n f(x, y) dx lim f(x i, yi ) x i n f(x, y) dy lim n i1 n f(x i, yi ) y i. The following formuls show how to clculte these new type line integrls. Note tht these integrls depend on the orienttion of the curve, i.e., the initil nd terminl points. Theorem 1 f(x, y) dx f(x, y) dy b b i1 f(x(t), y(t))x (t) f(x(t), y(t))y (t). Exmple 13 Evlute y dx+x dy, where 1 is the line segment from ( 5, 3) to (, ) A prmetric representtion for the line segment is x 5t 5, y 5t 3, t 1 Then dx 5, dy 5, nd Theorem 1 gives 1 y dx + x dy (5t 3) (5) + (5t 5)(5 ) (5t 5t + 4) [ 5t t + 4t ]

5 1 BASI MATERIAL. 5 Exmple 14 Evlute y dx+x dy, where is the rc of the prbol x 4 y from ( 5, 3) to (, ). Since the prbol is given s function of y, let s tke y s the prmeter nd write s x 4 y y y, 3 y. Then dx y dy nd by Theorem 1 we hve y dx + x dy y ( y) dy + (4 y ) dy 3 3 ( y 3 y + 4) dy [ y4 y y ] One cn lso define in similr mnner the line integrl with respect to rc length of function f long curve in R 3. Theorem 15 f(x, y, z) ds b (dx ) f(x(t), y(t), z(t)) + ( ) dy + P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz, where f(x, y, z) P (x, y, z), Q(x, y, z), R(x, y, z). ( ) dz The next exmple demonstrtes how to clculte line integrl of function with respect to x, y nd z. Exmple 16 Evlute y dx + z dy + x dz,, where consists of the line c segment 1 from (,, ) to (3, 4, 5) followed by the verticl line segment from (3, 4, 5) to (3, 4, ). We write 1 s or, in prmetric form, s r(t) (1 t),, + t 3, 4, 5 + t, 4t, 5t x + t y 4t z 5t t 1. Thus y dx + z dy + x dz 1 1 (4t) + (5t)4 + ( + t)5

6 1 BASI MATERIAL. 6 or 1 (1 + 9t) 1t + 9 t Likewise, cn be written in the form Then dx dy, so ] r(t) (1 t) 3, 4, 5 + t 3, 4, 3, 4, 5 5t x 3 y 4 z 5 5t dz 5, t 1. y dx + z dy + x dz 1 Adding the vlues of these integrls, we obtin ( 5) y dx + z dy + x dz We now get to our finl type of line integrl which cn be considered to be line integrl of vector field. This type of integrl is used to clculte the work W done by force field F in moving prticle long smooth curve. Theorem 17 If is given by the vector eqution r(t) x(t)i + y(t)j + z(t)k on the intervl [, b], then the work W cn be clculted by where is the dot product. W b F(r(t)) r (t), In generl, we mke the following definition which is relted to the formul in the bove theorem. Definition 18 Let F be continuous vector field defined on smooth curve given by vector function r(t), t b. Then the line integrl of F long is b F dr F(r(t)) r (t) F T ds; here, T is the unit tngent vector field to the prmeterized curve. Exmple 19 Find the work done by the force field F(x, y) x i xyj in moving prticle long the qurter-circle r(t) cos t i + sin t j, t π.

7 1 BASI MATERIAL. 7 Since x cos t nd y sin t, we hve nd F(r(t)) cos ti cos t sin tj r (t) sin ti + cos tj. Therefore, the work done is π F dr F(r(t)) r (t) ] π cos3 t 3 π 3. ( cos t sin t) Exmple Evlute F dr, where F(x, y, z) xyi + yzj + zxk nd is the twisted cubic given by Thus, We hve x t y t z t 3 t 1. 1 r(t) ti + t j + t 3 k r (t) i + tj + 3t k F(r(t)) t 3 i + t 5 j + t 4 k. F dr 1 F(r(t)) r (t) (t 3 + 5t 6 ) t t7 7 ] Theorem 1 If in R 3 is prmeterized by r(t) nd F P i+qj+rk, then F dr P dx + Qdy + Rdz. We now pply the mteril covered so fr on line integrls to obtin severl versions of the fundmentl theorem of clculus in the multivrible setting. Recll tht the fundmentl theorem clculus cn be written s b when F (x) is continuous on [, b]. F (x)dx F (b) F (), Theorem Let be smooth curve given by the vector function r(t), t b. Let f be differentible function of two or three vribles whose grdient vector f is continuous on. Then f dr f(r(b)) f(r()).

8 1 BASI MATERIAL. 8 For further discussion, we mke the following definitions. Definition 3 A curve r : [, b] R 3 (or R 3 ) closed if r() r(b). Definition 4 A domin D R 3 (or R ) is open if for ny point p in D, smll bll (or disk) centered t p in R 3 (in R ) is contined in D. Definition 5 A domin D R 3 (or R ) is connected if ny two points in D cn be joined by pth contined inside D. Definition 6 A curve r: [, b] R 3 (or R ) is simple curve if it doesn t intersect itself nywhere between its end points (r(t 1 ) r(t ) when < t 1 < t < b). Definition 7 An open, connected region D R is simply-connected region if ny simple closed curve in D encloses only points tht re in D. Definition 8 A vector field F is clled conservtive vector field if it is the grdient of some sclr function f(x, y); the function f(x, y) is clled potentil function for F. For exmple, for f(x, y) xy+y, f y, x+y nd so, F(x, y) yi + (x + y)j is conservtive vector field. Definition 9 If F is continuous vector field with domin D, we sy tht the line integrl F dr is independent of pth if 1 F dr F dr for ny two pths 1 nd in D with the sme initil nd the sme terminl points. We now sttes severl theorems tht you should know for the finl exm. Theorem 3 F dr is independent of pth in D if nd only if F dr for every closed pth in D. Theorem 31 Suppose F is vector field tht is continuous on n open connected region D. If F dr is independent of pth in D, then F is conservtive vector field on D; tht is, there exists function f such tht f F. Theorem 3 If F(x, y) P (x, y)i + Q(x, y)j is conservtive vector field, where P nd Q hve continuous first-order prtil derivtives on domin D, then throughout D we hve P y Q x. Theorem 33 Let F P i + Qj be vector field on n open simply-connected region D. Suppose tht P nd Q hve continuous first-order derivtives nd P y Q x throughout D. Then F is conservtive.

9 1 BASI MATERIAL. 9 Exmple 34 Determine whether or not the vector field F(x, y) (x y)i + (x )j is conservtive. Let P (x, y) x y nd Q(x, y) x. Then P y 1 Q x 1. Since P Q y x, F is not conservtive by Theorem 3. Exmple 35 Determine whether or not the vector field F(x, y) (3+xy)i+ (x 3y )j is conservtive. Let P (x, y) 3 + xy nd Q(x, y) x 3y. Then P y x Q x. Also, the domin of F is the entire plne (D R ), which is open nd simplyconnected. Therefore, we cn pply Theorem 33 nd conclude tht F is conservtive. Attention! You will likely hve problem on the finl exm which is similr to the one described in the next exmple. Exmple 36 () If F(x, y) (3 + xy)i + (x 3y )j, find function f such tht F f. (b) Evlute the line integrl F dr, where is the curve given by r(t) e t sin t i + e t cos t j, t π. () From Exmple 35 we know tht F is conservtive nd so there exists function f with f F, tht is, Integrting (3) with respect to x, we obtin f x (x, y) 3 + xy (3) f y (x, y) x 3y (4) f(x, y) 3x + x y + g(y). (5) Notice tht the constnt of integrtion is constnt with respect to x, tht is, it is function of y, which we hve clled g(y). Next we differentite both sides of (5) with respect to y: f y (x, y) x + g (y). (6)

10 1 BASI MATERIAL. 1 ompring (4) nd (6), we see tht g (y) 3y. Integrting with respect to y, we hve g(y) y 3 + K where K is constnt. Putting this in (5), we hve s the desired potentil function. f(x, y) 3x + x y y 3 + K (b) To pply Theorem ll we hve to know re the initil nd terminl points of, nmely, r() (, 1) nd r(π) (, e π ). In the expression for f(x, y) in prt (), ny vlue of the constnt K will do, so let s choose K. Then we hve F dr f dr f(, e π ) f(, 1) e 3π ( 1) e 3π + 1. This method is much shorter thn the strightforwrd method for evluting line integrls described in Theorem 1. Definition 37 A simple closed prmeterized curve in R lwys bounds bounded simply-connected domin D. We sy tht is positively oriented if for the prmetriztion r(t) of, the region D is lwys on the left s r(t) trverses. Note tht this prmetriztion is the counterclockwise one on the boundry of unit disk D {(x, y) x + y 1}. The next theorem is version of the fundmentl theorem of clculus, since it llows one to crry out two-dimensionl integrl on domin D by clculting relted integrl one-dimensionl on the boundry of D. There will be t lest one finl exm problem relted to the following theorem. Theorem 38 (Green s Theorem) Let be positively oriented, piecewisesmooth, simple closed curve in the plne nd let D be the region bounded by. If P nd Q hve continuous prtil derivtives on n open region tht contins D, then ( Q P dx + Qdy x P ) da. y D An immedite consequence of Green s Theorem re the re formuls described in the next theorem.

11 1 BASI MATERIAL. 11 Theorem 39 Let D be simply-connected domin in the plne with simple closed oriented boundry curve. Let A be the re of D. Then: A x dy y dx 1 x dy y dx. (7) Exmple 4 Find the re enclosed by the ellipse x + y 1. b The ellipse hs prmetric equtions x cos t nd y b sin t, where t π. Using the third formul in Eqution 7, we hve A 1 x dy y dx 1 b π π ( cos t)(b cos t) (b sin t)( sin t) πb. Exmple 41 Use Green s Theorem to evlute (3y esin x ) dx + (7x + y4 + 1) dy, where is the circle x + y 9. The region D bounded by is the disk x +y 9, so let s chnge to polr coordintes fter pplying Green s Theorem: (3y e sin x ) dx + (7x + y 4 + 1) dy [ x (7x + y 4 + 1) ] y (3y esin x ) da 4 D π 3 π dθ (7 3) r dr dθ 3 r dr 36π.

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