MATH2000 Flux integrals and Gauss divergence theorem (solutions)

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1 DEPARTMENT O MATHEMATIC MATH lux integrals and Gauss divergence theorem (solutions ( The hemisphere can be represented as We have by direct calculation in terms of spherical coordinates. = {(r, θ, φ r, θ π, φ π/}. div d = 3 div = 3(x + y + z = 3r π/ π ( π/ ( π = 3 sin φ dφ = 3 π 5 = 6π 5. r r sin φ dr dθ dφ ( dθ r dr Now to evaluate n d. In this case the surface comprises of two parts: the base of the hemisphere which lies in the x-y plane, denoted, and the part of the sphere itself, denoted. o that n d = n d + n d where n and n are outwardly pointing unit normal vectors to the surfaces and respectively. We expect the integral n d to be zero since the k component of is when is restricted to the x-y plane, so there is no flux across that surface. To verify this by direct calculation, we must first parametrise the surface. ince it is just a circular disc in the x-y plane, we have We then take the tangent vectors r(r, θ = r cos θi + r sin θj, r, θ π. r θ = r sin θi + r cos θj r r = cos θi + sin θj

2 We can calculate r r r θ = rk. However, this is directed into the solid. We require this vector to be directed outwards from the solid, so instead we ll take r θ r r = rk. In terms of our parametrisation, (r, θ = r 3 cos 3 θi + r 3 sin 3 θj, so that the dot product (r, θ (r θ r r = which tells us that the flux across will be zero as originally thought. To calculate the flux across, we parametrise (compare with the spherical coordinate transformation as r(θ, φ = cos θ sin φi + sin θ sin φj + cos φk, θ π, φ π/. The tangent vectors are r θ = sin θ sin φi + cos θ sin φj r φ = cos θ cos φi + sin θ cos φj sin φk. To find a vector normal to the surface, take r φ r θ = cos θ sin φi + sin θ sin φj + sin φ cos φk. We should check the direction to make sure it is directed outwards from the surface. Take for example the parameter values φ = π/ and θ =. This gives r φ r θ = i which is directed out, so the direction is ok. In terms of the parameters, we can write so that the dot product (θ, φ = cos 3 θ sin 3 φi + sin 3 θ sin 3 φj + cos 3 φk. (θ, φ (r φ r θ = cos θ sin 5 φ + sin θ sin 5 φ + sin φ cos φ. The flux integral we need to evaluate is then π π/ Using cos θ = ( + cos θ we have ((cos θ + sin θ sin 5 φ + sin φ cos φ dφ dθ. cos θ = ( + cos θ( + cos θ = ( + cos θ + cos θ = ( + cos θ + ( + cos θ = cos θ + cos θ. 8

3 It follows that cos θ + sin θ = cos θ + ( cos θ( cos θ = cos θ + cos θ = ( + cos θ cos θ + cos θ = 3 + cos θ. inally note that we can write sin 5 φ = sin φ( cos φ + cos φ. Putting this together, the flux across is = π π/ + π π/ ( π + ( π ( 3 + cos θ sin φ( cos φ + cos φ dφ dθ sin φ cos φ dφ dθ ( π/ cos θ dθ sin φ( cos φ + cos φdφ ( 3 + ( π/ dθ sin φ cos φ dφ Using the substitution u = cos φ in both φ integrals gives ( [3 = θ + ] π sin θ ( u + u du + 6 = 3π [ u 3 u3 + ] [ ] 5 u5 + π 5 u5 = 3π ( π 5 5 = 6π 5. ( π u du Therefore n d = n d + n d = + 6π 5 = 6π 5. o we have shown that for this example both sides of the equation in Gauss theorem are equal. 3

4 ( Note that in this case we cannot use Gauss divergence theorem since the vector field = i is undefined at any point in the y-z plane (ie. when x =, part of which lies in x the region enclosed by the surface. We must evaluate n d directly. ince the surface is the unit sphere, the position vector r = xi + yj + zk will also be an outwardly pointing unit normal (since x + y + z = on the surface. Taking n = r, we have that n =. Therefore the flux evaluates to n d = d (3 A diagram of the solid is as follows: = surface area of the unit sphere = π. Z X.5.5 Y The outward flux can be calculated as n where is the closed surface of the box, is the vector field, and n is an outwardly pointing unit normal vector. The surface consists of six open surfaces: the six faces of the box. We can evaluate the flux integral directly by calculating the outward flux through each face: n = n d+ n d+ 3 n d+ n d+ 5 n d+ 6 n d.

5 We represent each open surface follows: is the base of the box which lies in the plane z = 3 (and is therefore parallel to the x-y plane. An outwardly pointing unit normal is n = k. Restricted to, the vector field is given by Therefore over, = xi + yj + 9k, for x 3, y. n = (xi + yj + 9k ( k = 9. n d = ( 9 d = 9 d. ince d is just the area of, which is a rectangle of area =, so d = n d = 9 = 8. is the lid of the box which lies in the plane z = 5. An outwardly pointing unit normal is n = k. Restricted to, the vector field is given by Therefore over, = xi + yj + 5k, for x 3, y. n = (xi + yj + 5k k = 5. n d = (5 d = 5 d. ince d is just the area of, which is a rectangle of area =, so d = n d = 5 = 3. 3 is the back of the box which lies in the plane x =. An outwardly pointing unit normal is n = i. Restricted to 3, the vector field is given by = i + yj + 3zk, for y, 3 z 5. 5

6 Therefore over 3, n = (i + yj + 3zk ( i = n d = ( d = d. ince d is just the area of 3, which is a rectangle of area =, so 3 3 d = 3 n d =. is the front of the box which lies in the plane x = 3. An outwardly pointing unit normal is n = i. Restricted to, the vector field is given by Therefore over, = 3i + yj + 3zk, for y, 3 z 5. n = (3i + yj + 3zk (i = 3. n d = (3 d = 3 d. ince d is just the area of, which is a rectangle of area =, so d = n d = 3 = 6. 5 is the left side of the box which lies in the plane y = (the x-z plane. An outwardly pointing unit normal is n = j. Restricted to 5, the vector field is given by = xi + 3zk, for x 3, 3 z 5. Therefore over 5, n = (xi + 3zk ( j =. n d = ( d =

7 6 is the right side of the box which lies in the plane y =. An outwardly pointing unit normal is n = j. Restricted to 6, the vector field is given by Therefore over 6, = xi + j + 3zk, for x 3, 3 z 5. n = (xi + j + 3zk (j = n d = ( d = d. ince d is just the area of 6, which is a rectangle of area =, so 6 6 d = 6 n d = = 8. Putting all of this information together gives n = n d+ n d+ n d+ 3 n d+ 5 n d+ 6 n d = ( = 6. Using the divergence theorem, we can also calculate the outward flux as div d, where is the region enclosed by (ie. the box. We can calculate The outward flux is then div = x (x + y (y + (3z = = 6. z div d = 6 d = 6 (vol. of box = 6 ( = 6. We have therefore verified the divergence theorem. In this case, it is a lot less work to calculate the volume integral compared to the flux integral. 7

8 ( Use the divergence theorem. The region (in this case a sphere of radius 5 can be represented as = {(r, θ, φ r 5, φ π, θ π} in term of spherical polar coordinates. We also have div = x (3x + y (y + (5z = =. z Hence by the divergence theorem the flux out of the surface is π div d = = ( π π 5 ( π dθ = π 5 3 = π. r sin φ dr dφ dθ ( 5 sin φ dφ r dr Alternatively, we could make the observation that div d = d = volume of sphere of radius 5 ( = 3 π53 = π. (5 We need to find n d. By Gauss divergence theorem this is equal to div d. div = x (x + y (3y + z (6z = =. In cylindrical polar coordinates, the cone is z = (r cos θ + (r sin θ = r z = r in this case since z. The region in R 3 is = {(r, θ, z z, θ π, r z}. 8

9 o flux = = = 5 = 5 π z π [ 3 z3 = 8π 3. z dz ] r dr dθ dz z dθ dz π π (6 = (x 3 + xy + xz i + (x y + y 3 + yz j + (x z + y z + z 3 k so The sphere is described by div = x (x3 + xy + xz + y (x y + y 3 + yz + z (x z + y z + z 3 dθ = (3x + y + z + (x + 3y + z +(x + y + 3z = 5r. = {(r, θ, φ r a, θ π, φ π}. o by Gauss divergence theorem, the flux across the surface of the sphere = a π π a = 5 r dr [ ] a = 5 5 r5 (5r r sin φ dφ dθ dr π = 5 5 a5 π = πa 5. dθ π [θ] π [ cos φ]π sin φ dφ 9

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