Contents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9

Transcription

1 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Contents Multiple Integrals 3 2 Vector Fields 9 3 Line and Surface Integrals 5 4 The Classical Integral Theorems 9

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3 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables MULTIPLE INTEGRALS Problem.. Use the substitution x r cos θ and y (/2)r sin θ to evaluate x 2 x 2 + 4y 2 da, where D is the region between the two ellipses x 2 + 4y 2 and x 2 + 4y 2 4. D Solution. When using a substitution (change of variables) to evaluate an area (or volume) integral, we need to first ensure that we write down the transformed integral correctly. Transforming the original integrand. We substitute and simplify to get x 2 x 2 + 4y 2 (r cos θ) 2 (r cos θ) 2 + 4( 2 r sin cos2 θ. θ)2 Transforming the region of integration. In terms of x and y, the region D is defined by In terms of r and θ, then, we have { D D {(x, y) x 2 + 4y 2 4}. (r, θ) (r cos θ) ( ) 2 2 r sin θ 4} {(r, θ) r 2 4} {(r, θ) r 2}. Transforming the area element. The area element transforms as ( ) da dx dy det r x θ x dr dθ r y θ y ( ) det cos θ (/2)r sin θ dr dθ sin θ (/2)r cos θ D r dr dθ. 2 Putting these pieces together, x 2 x 2 + 4y 2 da 2 {(r,θ) r 2} r2 θ2π π 4. r dr r 2π cos 2 θ r dr dθ 2 θ cos 2 θ dθ + cos(2θ) dθ 2 3

4 Practice Problems - Solutions MATH 32B-2 (8W) Problem.2. A solid cone is bounded by the surface φ α (in spherical coordinates) and the surface z a. Its mass density is τ cos φ (in spherical coordinates), where τ is a constant. By evaluating a volume integral, find the total mass of the cone. Solution. We will compute with spherical coordinates. The integrand of the volume integral is the mass density τ cos φ, while the volume element is ρ 2 sin φ dρ dφ dθ. For the bounds on the integral, θ ranges from to 2π while φ ranges from to α. The bounds on r depend on φ, and from some geometry, one can find that r ranges from to a sec φ. a r max α φ Thus the total mass of the cone is τ cos φ dv cone θ2π φα ρa sec φ θ θ2π τ dθ θ φ ρ φα φ τ cos φ ρ 2 sin φ dρ dφ dθ ( ) ρa sec φ sin φ cos φ ρ 2 dρ dφ ρ φα 2πτ sin φ cos φ 3 a3 sec 3 φ dφ 2πτ a 3 3 2πτ a 3 3 φ φα φ sin φ cos 2 φ dφ ). ( cos α If we use cylindrical polar coordinates to set up the integral instead, we get θ2π za rz tan α θ z r ρ z r dr dz dθ. r2 + z2 4

5 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Problem.3. A tetrahedron V has vertices (,, ), (,, ), (,, ), and (,, ). Find the center of volume, defined by r dv, r (x, y, z). volume of V V Solution. By symmetry, the components of the center of volume will all be equal, so it is enough to look at one component, say the x-component. From the definition given, the x-component of the center of volume is x COV x dx dy dz. volume of V The (solid) tetrahedron is the region so {(x, y, z) x, y, z and x + y + z } x COV {(x, y, z) x, y, z, and y + z x} V {(x, y, z) x, y x, and z x y}, The volume of the tetrahedron is volume of V volume of V volume of V 2 volume of V x y x z x y x y x y x x x x x x 2 volume of V 2 volume of V 24 volume of V. y z x dz dy dx x( x y) dy dx [x( x)y 2 ] y x xy2 dx x( x) 2 dx (x 2x 2 + x 3 ) dx ] [ area of base height 3 y ( ) 2 6, so the x COV /4. Thus the center of volume itself is (/4, /4, /4). The answer is the average of the vertices (in either a component-wise or a vector sense). If the tetrahedron has constant mass density, then the center of volume coincides with the center of mass. 5

6 Practice Problems - Solutions MATH 32B-2 (8W) Problem.4. By using a suitable change of variables, calculate the volume within a solid ellipsoid x 2 a 2 + y2 b 2 + z2 c 2. Solution. The change of variables we use is a linear scaling of the usual spherical coordinates, x ar sin φ cos θ, y br sin φ sin θ, z cr cos φ. Then r ranges from to, φ ranges from to π, and θ ranges from to 2π, and r x φ x θ x dv dx dy dz det r y φ y θ y dr dφ dθ abc r 2 sin φ dr dφ dθ, r z φ z θ z so the volume of the (solid) ellipsoid is ellipsoid dv abc r r abc 3 r 2 dr θπ θ 4πabc 2 2π. 3 sin θ dθ φ2π φ dφ If we are allowed to use without proof the result that the unit sphere has volume 4π/3, then if we use the substitution (x, y, z) (t, u, v), x at, y bu, z cv, we obtain abc times the integral for the volume of the unit sphere. 6

7 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Problem.5. The region D is bounded by the segments {x, y }, {y, x }, {y, x 3/4}, and by an arc of the parabola y 2 4( x). Consider a mapping into the (x, y) plane from the (u, v) plane defined by the transformation x u 2 v 2 and y 2uv. Sketch D and also the two regions in the (u, v) plane which are mapped into it. Hence evaluate da (x 2 + y 2 ). /2 Solution. We start with a sketch of D (below left). D y v D x u From this, we see that D {(x, y) y and x y 2 /4} {(u, v) 2uv and u 2 v 2 u 2 v 2 }. The condition 2uv is satisfied when uv /2, which is the cyan region. The condition u 2 v 2 is satisfied when u v, which is the magenta region. Finally, for the condition u 2 v 2 u 2 v 2, we rearrange and factor to get (u 2 )(v 2 + ). Since v 2 + >, this holds if and only if u 2, so u, which is the yellow region. Putting these together, the region E which corresponds to D is the overlap of all three (above right). The integral over D becomes ( ) da D (x 2 + y 2 ) /2 E ((u 2 v 2 ) 2 + (2uv) 2 ) /2 det u x v x du dv u y v y ( ) E (u 4 + 2u 2 v 2 + v 4 ) /2 det 2u 2v du dv 2v 2u E u 2 + v 2 4 (u 2 + v 2 ) du dv 4 du dv 4 area of E. E 7

8 Practice Problems - Solutions MATH 32B-2 (8W) To compute the area of E, note that the two regions comprising E are symmetric via rotation by π about the origin, so we only need to compute the area of one of them, say the upper right one. To do this, divide it into two parts as shown below. v u v u uv /2 / 2 u The first part is a right triangle, so we need to find the base and height, which are the coordinates of the intersection of u v and uv /2. This is solved by u v / 2, so the area of the triangular part is uv/2 /4. The second part is the area under the graph of v /(2u) from u / 2 to u, which can be computed by an integral, so ( ) area of E du / 2 2u 2 + ln 2. 2 Hence D da 4 area of E ln 2. (x 2 + y 2 ) /2 8

9 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Problem 2.. Verify that the vector field VECTOR FIELDS u(x, y, z) e x (x cos y + cos y y sin y)i + e x ( x sin y sin y y cos y)j is irrotational and express it as the gradient of a scalar field φ. Check that u is also solenoidal and find a function ψ such that u (ψk). Solution. Writing u u x i + u y j + u z k, so that we have u x e x (x cos y + cos y y sin y), u y e x ( x sin y sin y y cos y), u z, u ( y u z z u y )i + ( z u x x u z )j + ( x u y y u x )k ([e x ( x sin y sin y y cos y) + e x ( sin y)] [e x ( x sin y sin y sin y y cos y)]) k, so u is irrotational. Therefore, we can write u φ for some φ satisfying x φ e x (x cos y + cos y y sin y), y φ e x ( x sin y sin y y cos y), z φ. Integrating the first equation with respect to x (integration by parts is useful for this), we have φ(x, y, z) (xe x e x ) cos y + e x (cos y y sin y) + ψ(y, z) e x (x cos y y sin y) + Φ(y, z) for some function Φ(y, z) of y, z only. To satisfy z φ, we require that Φ(y, z) has no dependence on z, i.e. Φ(y, z) Φ(y). Then to satisfy y φ, we have e x ( x sin y sin y y cos y) y [e x (x cos y y sin y)] + y Φ e x ( x sin y sin y y cos y) + Φ (y), so Φ (y), meaning Φ(y) C for some constant C. In particular, since we just need to find some φ for which u φ, we can take C, so φ(x, y, z) e x (x cos y y sin y). To see that u is solenoidal, we compute u x u x + y u y + z u z [e x (x cos y + cos y y sin y) + e x (cos y)] + [e x ( x cos y cos y cos y + y sin y]. If u (ψk), then expanding the curl gives e x (x cos y + cos y y sin y) y ψ, e x ( x sin y sin y y cos y) x ψ. Integrating the first equation with respect to y gives ψ(x, y, z) e x (x sin y + y cos y) + Ψ(x, z), and as in the first part of the question, the choice Ψ(x, z) gives a valid solution to the other equation, so we may take ψ(x, y, z) e x (x sin y + y cos y). 9

10 Practice Problems - Solutions MATH 32B-2 (8W) Problem 2.2. Let ψ(r) be a scalar field and v(r) be a vector field, where r (x, y, z). Show that (ψv) ( ψ) v + ψ( v) and (ψv) ( ψ) v + ψ( v). Evaluate the divergence and curl of the vector fields where ρ r and a, b are constant vectors. ρr, a(r b), a r, r/ρ 3, Solution. For the first identity, let v (v x, v y, v z ). Then (ψv) x (ψv x ) + y (ψv y ) + z (ψv z ) ( x ψ)v x + ψ( x v x ) + ( y ψ)v y + ψ( y v y ) + ( z ψ)v z + ψ( z v z ) [( x ψ)v x + ( y ψ)v y + ( z ψ)v z ] + ψ( x v x + y v y + z v z ) ( ψ) v + ψ( v). For the second identity, the x-component of (ψv) is [ (ψv)] x y (ψv z ) z (ψv y ) ( y ψ)v z + ψ( y v z ) ( z ψ)v y ψ( z v y ) [( y ψ)v z ( z ψ)v y ] + ψ( y v z z v y ) [( ψ) v] x + [ψ( v)] x [( ψ) v + ψ( v)] x, and similar computation shows that the other two components agree, proving the identity. To compute the divergence and curl of ρr, we use the identities we just proved along with the intermediate computations (exercises) (ρ) r, r 3, r. ρ This gives (ρr) ( (ρ)) r + ρ( r) r ρ r + 3ρ ρ2 ρ + 3ρ 4ρ, (ρr) ( (ρ)) r + ρ( r) r ρ r + +. For a(r b), we take r b to be the scalar field and a to be the vector field. Then (exercises) (r b) b, a, a, so ((r b)a) (r b) a + (r b)( a) b a, ((r b)a) (r b) a + (r b)( a) b a.

11 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables For a r, we compute directly. If a (a, b, c), then so a r (bz cy, cx az, ay bx), (a r) + +, (a r) (a ( a), b ( b), c ( c)) 2a. For r/ρ 3, we take /ρ 3 for the scalar field and r for the vector field. Then ( ) ρ 3 3 (r) 3r ρ4 ρ 5, so we have ( ) r ρ ( 3 ) r ρ 3 ( ρ 3 ( ρ 3 ) r + ( r) 3r ρ3 ρ 5 r + 3 ρ 3, ) r + ( r) 3r ρ3 ρ 5 r +.

12 Practice Problems - Solutions MATH 32B-2 (8W) Problem 2.3 ( ). Maxwell s equations for electric and magnetic fields E(r, t) and B(r, t) are E ρ ɛ, B, E B t, B µ j + ɛ µ E t, where ρ(r, t) and j(r, t) are the charge density and current density, and ɛ and µ are constants. Show that these imply the conservation equation j ρ/ t. Show also that if j, then U ) (ɛ E 2 + µ B 2 and P E B 2 µ satisfy P U/ t. Solution. The only way we can obtain j from the given equations is to take the divergence of the curl equation for B (Ampère s law with Maxwell s correction), which gives us ( ) E ( B) µ j + ɛ µ t ( ) E µ j + ɛ µ. t For reasonable smoothness assumptions on E, we can interchange the order of the divergence and the time derivative and then use the divergence equation for E (Gauss s law) to get ( ) j ɛ t ( E) ɛ ρ ρ t t. To be completed. ɛ 2

13 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Problem 2.4 ( ). Show that (u v) u( v) + (v ) v( u) (u )v for vector fields u and v. Show also that (u )u (u 2 /2) u ( u), where u u. The notation a, where a (a x, a y, a z ), refers to the differential operator a a x x + a y y + a z z, which can be applied to a scalar field or a vector field (by applying it to each component separately). Solution. To be written. 3

14 Practice Problems - Solutions MATH 32B-2 (8W) Problem 2.5 ( ). (a) The vector field B(r) is everywhere parallel to the normals to a family of surfaces f(r) constant. (That is, there is some function f(r) for which at any position (a, b, c), the vector B(a, b, c) is perpendicular to the level surface of f that passes through (a, b, c).) Show that B ( B). (b) The tangent vector at each point on a curve in space is parallel to a non-vanishing vector field H(r). (Non-vanishing means that H is not equal to at any point.) Show that the curvature of the curve is given by H 3 H (H )H. Solution. To be written. 4

15 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables LINE AND SURFACE INTEGRALS Problem 3.. Evaluate explicitly each of the line integrals (x dx + y dy + z dz), (y dx + x dy + dz), along (a) the straight line path from the origin to (,, ), and (y dx x dy + e x+y dz), (b) the parabolic path given parametrically by (x, y, z) (t, t, t 2 ) from t to t. For which of these integrals do the two paths give the same results, and why? Solution. (a) This path may be parametrized as (x, y, z) (t, t, t) with t. Then (x dx + y dy + z dz) (y dx + x dy + dz) (y dx x dy + e x+y dz) (t dt + t dt + t dt) (t dt + t dt + dt) (t dt t dt + e 2t dt) 3t dt 3 2, (2t + ) dt 2, e 2t dt e 2. (b) In this case, dx dy dt and dz d(t 2 ) 2t dt, so we have (x dx + y dy + z dz) (y dx + x dy + dz) (y dx x dy + e x+y dz) (t dt + t dt + t 2 2t dt) (t dt + t dt + 2t dt) (t dt t dt + e 2t 2t dt) (2t 3 + 2t) dt 3 2, 4t dt 2, 2te 2t dt 2 ( + e2 ). The first two integrals give the same results for the two paths, as the endpoints are the same and they are exact differentials, with ( ) x dx + y dy + z dz d 2 (x2 + y 2 + z 2 ), y dx + x dy + dz d(xy + z). 5

16 Practice Problems - Solutions MATH 32B-2 (8W) Problem 3.2. Consider forces F (3x 2 yz 2, 2x 3 yz, x 3 z 2 ) and G (3x 2 y 2 z, 2x 3 yz, x 3 y 2 ). Compute the work done, given by the line integrals F dr and G dr, along the following paths, each of which consist of straight line segments joining the specified points: (a) (,, ) (,, ); (b) (,, ) (,, ) (,, ) (,, ); (c) (,, ) (,, ) (,, ) (,, ). Solution. (a) Parametrizing the path as (x, y, z) (t, t, t), so then dr (,, ) dt, F dr G dr (3t 5 + 2t 5 + t 5 ) dt (3t 5 + 2t 5 + t 5 ) dt 6t 5 dt, 6t 5 dt. (b) Along the first two segments, all components of F and G are zero, so the integrals will be zero. Parametrizing the third segment as (t,, ), so then dr (,, ) dt, F dr G dr (3t 2 + 2t 3 + t 3 ) dt (3t 2 + 2t 3 + t 3 ) dt 3t 2 dt, 3t 2 dt. (c) Along the first two segments, all components of F are zero, so the integral of F will be zero. Similarly, the first two components of G are zero, and dr has a third component of zero, so the integral of G will be zero. Parametrizing the third segment as (,, t), so the dr (,, ) dt, F dr G dr (3t 2 + 2t + t 2 ) dt (3t + 2t + ) dt t 2 dt /3, dt. 6

17 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Problem 3.3. Check, by calculating its curl, that the force field ( ) F(x, y, z) 3x 2 tan z y 2 e xy2 sin y, (cos y 2xy sin y)e xy2, x 3 sec 2 z is conservative. Find the most general scalar potential for F and hence, or otherwise, find the work done by the force as it acts on a particle moving from (,, ) to (, π/2, π/4). Solution. Write F (F x, F y, F z ), so that Then F F x 3x 2 tan z y 2 e xy2 sin y, F y (cos y 2xy sin y)e xy2, F z x 3 sec 2 z. ( Fz y F y z, F x z F z x, F y x F ) x y (, 3x 2 sec 2 z 3x 2 sec 2 z,, [2xy 3 sin y y 2 cos y 2y sin y]e xy2 [ 2y sin y + 2xy 2 sin y y 2 cos y]e xy2 ) so F is conservative as it is defined on all of R 3. Let ψ be a potential for F, so that ψ x 3x2 tan z y 2 e xy 2 sin y, Integrating the first equation, ψ y 2 (cos y 2xy sin y)e xy, ψ z x3 sec 2 z. Differentiating with respect to y and z, ψ x 3 tan z + e xy2 sin y + A(y, z). ψ y 2 (cos y 2xy sin y)e xy + A y, ψ z x3 sec 2 z + A z. Comparing to our earlier equations, we see that A(y, z) is in fact a genuine constant, so the most general potential is ψ x 3 tan z + e xy2 sin y + C. Taking the potential where C, the work done by F is F dr ψ(, π/2, π/4) ψ(,, ) ( + e π2 /4 ) + e π2 /4. 7

18 Practice Problems - Solutions MATH 32B-2 (8W) Problem 3.4 ( ). A curve C is given parametrically in Cartesian coordinates by r(t) (cos(sin nt) cos t, cos(sin nt) sin t, sin(sin nt)), t 2π, where n is some fixed integer. Using spherical polar coordinates, or otherwise, sketch or describe the curve. Show that ( H dr 2π, where H(x, y, z) y ) x 2 + y 2, x x 2 + y 2, C and C is traversed in the direction of increasing t. Can H be written as the gradient of a scalar function? Comment on your results. Solution. In spherical polar coordinates (ρ, φ, θ), so that the curve is given by The sketch of C is omitted. We can compute and dr dt so after some calculations, Hence x ρ sin φ cos θ, y ρ sin φ sin θ, z ρ cos φ, ρ, φ π/2 sin nt, θ t. ( H(r(t)) sin t ) cos(sin nt), cos t cos(sin nt), ( n sin(sin nt) cos(nt) cos t cos(sin nt) sin t, n sin(sin nt) cos(nt) sin t + cos(sin nt) cos t, n cos(sin nt) cos nt), C H dr dt. H dr Away from the z-axis, where H is not defined, 2π ( ( y H arctan x)) dt 2π. ( arccot ( )) x. y These functions are only simultaneously undefined on the z-axis, which is not in the domain of H, so at any point, H is the gradient of a function defined near that point. However, the requirement for the circulation of H to be around any closed curve is that H is the gradient of a single function defined on the entire domain, so there is nothing contradictory going on in this example. 8

19 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables THE CLASSICAL INTEGRAL THEOREMS Problem 4.. The closed curve C in the plane consists of the arc of the parabola y 2 4ax (where a > ) between the points (a, ±2a) and the straight line joining (a, 2a). The region enclosed by C is A. Show, by calculating the integrals explicitly, that (x 2 y dx + xy 2 dy) (y 2 x 2 ) da 4 5 a4, where C is traversed anticlockwise. C Solution. For the line integral, we have two parts C and C 2, where C is the parabolic arc from (a, 2a) to (a, 2a) and C 2 is the straight line segment from (a, 2a) to (a, 2a). Parametrizing C as (y 2 /4a, y) with y running from 2a to 2a, we have dx (y/2a) dy, so C (x 2 y dx + xy 2 dy) 2a 2a 2 A [ ( ) y 2 2 y 4a [ (2a) a 3 + (2a)5 5 4a y 2a + y2 4a y2 ] ] a4. 2a [ ] y 6 dy 2a 32a 3 + y4 dy 4a Parametrizing C 2 as (a, y) with y running from 2a to 2a, we have dx, so 2a (x 2 y dx + xy 2 dy) ay 2 dy a (2a)3 a ( 2a)3 6 C a4. Hence C 2a (x 2 y dx + xy 2 dy) ( ) a a4. For the area integral, A is bounded on the left by the parabolic arc and on the right by the line segment, so A {(x, y) 2a y 2a and y 2 /4a x a}. This gives us A (y 2 x 2 ) da 2a y 2a 2a 2a 2a 2a xa xy 2 /4a ( [y 2 a y2 4a [ (y 2 x 2 ) dx dy ) ( a 3 3 y a 3 y4 4a + ay2 a3 3 [ (2a) a 3 (2a)5 5 4a + a (2a)3 3 [ ] a a4. ( ) y 2 3 )] dy 4a ] dy ] a3 2a 3 9

20 Practice Problems - Solutions MATH 32B-2 (8W) Problem 4.2. Verify Stokes theorem for the hemispherical surface {ρ, z } and the field F(r) (y, x, z), r (x, y, z). Solution. We compute F (,, 2) and ds r ds whenever r is on the unit sphere (hence on the hemisphere S given), so θ2π φπ/2 ( F) ds 2z ds 2 cos φ sin φ dφ dθ 2π. S S θ The boundary of S is the unit circle in the xy-plane, traversed counterclockwise, so it can be parametrized as r(t) (cos t, sin t, ). Then F(r(t)) (sin t, cos t, ) and dr ( sin t, cos t, ) dt, so we get 2π F dr (sin 2 t + cos 2 t) dt 2π. S φ Problem 4.3. Let F(r) (x 3 + 2y + z 2, y 3, x 2 + y 2 + 3z 2 ) and let S be the open surface z x 2 + y 2, z. Use the divergence theorem (and cylindrical polar coordinates) to evaluate result by calculating the integral directly. S F ds. Verify your Solution. We compute F 3x 2 + 3y 2 + 6z 3r 2 + 6z. Let T be the surface {(x, y, ) r 2 }, so then S and T together bound the volume V { z r 2 }. By the divergence theorem, θ2π r z r 2 F ds + F ds ( F) dv (3r 2 + 6z) r dz dr dθ S T 2π V θ r z [3r 2 ( r 2 ) + 3( r 2 ) 2 ] r dr 2π On T, we have ds (,, ) r dr dθ and F (x 3 + 2y, y 3, x 2 + y 2 ), so θ2π r F ds (x 2 + y 2 ) r dr dθ r 3 dr dθ π/2. Hence S T F ds 3π/2 + π/2 2π. T θ r ( 3r 3 + 3r) dr 3π/2. To compute the integral over S directly, we parametrize S as (x, y, z x 2 y 2 ), from which we get ds (2x, 2y, ) dx dy. Then (making use of some symmetry) F ds (2x 4 + 4xy + 2xz 2 + 2y 4 + x 2 + y 2 + 3z 2 ) dx dy S S θ2π r θ θ2π θ r ( cos 4 θ 3 [2r 4 cos 4 θ + 2r 4 sin 4 θ + r 2 + 3( r 2 ) 2 ] r dr dθ + sin4 θ ) dθ 2π. 2 2

21 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Problem 4.4. Consider the line integral C x 2 y dx + xy 2 dy for C a closed curve traversed anticlockwise in the plane. (a) Evaluate this integral when C is a circle with radius R centered at the origin. Use Green s theorem to relate the results for R b and R a to an area integral over the region a 2 x 2 + y 2 b 2, and calculate the area integral directly. (b) Now suppose that C is the boundary of a square centered at the origin with sides of length l. Show that the line integral is independent of the orientation of the square in the plane. Solution. C (a) Parametrize C as (R cos t, R sin t) with t 2π. Then x 2 y dx + xy 2 dy 2π [ R 3 cos 2 t sin t ( R sin t) + R 3 cos t sin 2 t (R cos t)] dt πr4 2. Green s theorem gives us, if C R denotes the circle of radius R and A {a 2 x 2 + y 2 b 2 }, (y 2 + x 2 ) da A ( x 2 y dx + xy 2 dy) C b ( x 2 y dx + xy 2 dy). C a Computing the area integral with polar coordinates, A (y 2 + x 2 ) da θ2π rb θ ra This is consistent with the result for the line integrals. r 2 r dr dθ π(b2 a 2 ). 2 (b) Green s theorem gives us the same integral, but now with A a square centered at the origin. The integrand is r 2, which is unchanged by rotating the region. The area element da is also unchanged by rotation (since rotations preserve areas), so the integral is unchanged by rotating the square. 2

22 Practice Problems - Solutions MATH 32B-2 (8W) Problem 4.5 ( ). By applying the divergence theorem to the vector field a A, where a is an arbitrary constant vector and A(r) is a vector field, show that A dv A ds, V where the surface S encloses the volume V. Verify this result when S is the sphere r R and A(x, y, z) (z,, ). Solution. To be written. Problem 4.6 ( ). By applying Stokes theorem to the vector field a F, where a is an arbitrary constant vector and F(r) is a vector field, show that dr F (ds ) F, C S where the curve C bounds the open surface S. Verify this result when C is the unit square in the xy-plane, with opposite vertices at (,, ) and (,, ), and F(r) r. Solution. To be written. Problem 4.7 ( ). From an integral theorem, derive Green s second identity, (ψ 2 ϕ ϕ 2 ψ) dv [ψ ϕ ϕ ψ] ds, where the surface S encloses the volume V. Here 2 f ( f) is the Laplacian. Solution. To be written. V S S 22