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urface and Volume Integrals 9. Introduction A vector or scalar field (including one formed from a vector derivative (div, grad or curl)) can be integrated over a surface or volume. This section shows how

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urface and Volume Integrals 9. Introduction A vector or scalar field (including one formed from a vector derivative (div, grad or curl)) can be integrated over a surface or volume. This section shows how to carry out such operations. Prerequisites Before starting this ection you should... Learning Outcomes After completing this ection you should be able to... 1 be familiar with vector derivatives be familiar with surface and volume integrals be able to carry out operations involving integrations of vector fields. 1. urface integrals involving vectors The unit normal For the surface of any three-dimensional shape, it is possible to find a vector lying perpendicular to the surface and with magnitude 1. The unit vector points outwards from the surface and is usually denoted by ˆn. Example If is the surface of the sphere x + y + z a find the unit normal ˆn olution The unit normal at the point (x, y, z) points away from the centre of the sphere i.e. it lies in the direction of xi + yj + zk. Tomake this a unit vector it must be divided by its magnitude x + y + z i.e. the unit vector ˆn x x i + y +y +z x j + z +y +z x k x +y +z a i + y a j + z a k where a x + y + z z n y x Example For the cube x 1, y 1, z 1, find the unit normal ˆn olution On the face given by x,the unit normal points in the negative x-direction. Hence the unit normal is i. imilarly :- On the face x 1the unit normal is i On the face y the unit normal is j On the face y 1the unit normal is j On the face z the unit normal is k On the face z 1the unit normal is k HELM (VERION 1: April 14, 4): Workbook Level 1 d and the unit normal The vector d is a vector, an element of the surface with magnitude dudv and direction perpendicular to the surface. If the plane in question is the Oxy plane, then d ˆn du dv k dx dy. If the plane in question is not one of the three coordinate planes (Oxy, Oxz, Oyz), appropriate adjustments must be made to express d in terms of dx and dy (or dz and either dx or dy). Example The rectangle OABC lies in the plane z y. The vertices are O (,, ), A (1,, ),B (1, 1, 1) and C (, 1, 1). Find a unit vector ˆn normal to the plane and an appropriate vector d expressed in terms of dx and dy. z C y B O E A D x olution Note two vectors in the rectangle are OA i and OC j + k. Avector perpendicular to the plane is i (j + k) j + k. However, this vector is of magnitude sothe unit normal vector is ˆn 1 ( j + k) 1 j + 1 k. The vector d is therefore ( 1 j + 1 k) du dv where du and dv are increments in the plane of the rectangle OABC. Now, one increment, say du, maypoint in the x direction while dv will point in a direction up the plane, parallel to OC. Thusdu dx and (by Pythagoras) dv (dy) +(dz).however, as z y, dz dy and hence dv dy. Thus, d ( 1 j + 1 k) dx dy ( j + j) dx dy. Note :- the factor of could also have been found by comparing the area of rectangle OABC, i.e. 1, with the area of its projection in the Oxy plane i.e. OADE or area 1. Integrating a scalar field A function can be integrated over a surface in a manner similar to that shown in sections 9.1 and 9.. Often, such integrals can be carried out with respect to an element containing the unit normal. 3 HELM (VERION 1: April 14, 4): Workbook Level 1 Example Evaluate the integral A 1 1+x d where is the unit normal over the area A and A is the square x 1, y 1, z. olution In this integral, becomes k dx dy i.e. the unit normal times the surface element. Thus the integral is k dx dy 1+x π 4 k k tan 1 x ] 1 dy k( π 4 ) ] 1 dy π 4 k dy Example Find ud where u r x + y + z and is the surface of the unit cube x 1, y 1, z 1. olution The unit cube has six faces and the normal vector ˆn points in a different direction on each face. The surface integral must be evaluated for each face separately and the results summed. On the face x,the normal ˆn i and the surface integral is ( + y + z )( i) dz dy i i y z z3 ] 1 y + 1 ] 3 On the face x 1,the normal ˆn i and the surface integral is (1 + y + z )(i) dz dy i i dy z + y z z3 ] 1 y + 4 ] 3 1 dy i 3 y3 + 1 ] 1 3 y 3 i dy 1 dy i 3 y3 + 4 ] 1 3 y 5 3 i The net contribution from the faces x and x 1is 3 i i i. Due to the symmetry of the scalar field u and the unit cube, the net contribution from the faces y and y 1isj while the net contribution from the faces z and z 1isk. The sum i.e. the surface integral ud i + j + k HELM (VERION 1: April 14, 4): Workbook Level 1 4 Key Point A scalar function integrated with respect to a unit normal gives a vector quantity When the surface does not lie in one of the planes (Oxy plane, Oxz plane, Oyz plane), extra care must be taken when finding d. Example Find fd where f is the function x and is the surface of the triangle bounded by (,, ), (, 1, 1) and (1,, 1). z Area 3 y Area 1 x olution The unit vector n is perpendicular to two vectors in the plane e.g. (j+k) and (i+k). The vector (j+k) (i+k) i+j+k and has magnitude 3. Hence the normal vector n 1 3 i+ 1 3 j 1 3 k. As the area of the triangle is 3 and the area of its projection in the Oxy plane is 1, the vector d 3/ n dy dx (i + j + k) dy dx. 1/ Thus x fd x dydx(i + j + k) xy] 1 x dx (i + j + k) (x x ) dx (i + j + k) x 3 x3 ] 1 (i + j + k) 1 (i + j + k) 3 The scalar function being integrated may be the divergence of a suitable vector function. 5 HELM (VERION 1: April 14, 4): Workbook Level 1 Example Find ( F)d where F xi + yzj + xyk and is the surface of the triangle with vertices at (,, ), (1,, ) and (1, 1, ). olution Note that F +z asz everywhere along. As the triangle lies in the Oxy plane, the normal vector n k and d k dy dx. Thus, x ( F)d dy dx y] x dx x dx x ] Evaluate the integral 4xd where represents the trapezium with vertices at (, ), (3, ), (, 1) and (, 1).. Evaluate the integral xyd where is the triangle with vertices at (,, 4), (,, ) and (1,, ). 3. Find the integral xyz d where is the surface of the unit cube x 1, y 1, z Evaluate the integral (x i + yzj + x yk) ] d where is the rectangle with vertices at (1,, ), (1, 1, ), (1, 1, 1) and (1,, 1). 1.) (a) Find the vector d (b) write the surface integral as a double integral (c) evaluate this double integral (a) k (b) 3 y 4x dxdyk (c) 38 3 k HELM (VERION 1: April 14, 4): Workbook Level 1 6 .) 8 i + 4j + k ) 1 4 (x + y + z) 4.) 5 i 7 HELM (VERION 1: April 14, 4): Workbook Level 1 Integrating a vector field In a similar manner, a vector field may be integrated over a surface. Two common integrals are F (r) d and F (r) d which integrate to a scalar and a vector respectively. Again, when the unit normal d is expressed appropriately, the expression will reduce to a double integral. Example Evaluate the integral A (x yi + zj +(x + y)k) d where is the unit normal over the area A and A is the square x 1, y 1, z. olution On A, the unit normal is dx dy k so the integral becomes (x yi + zj +(x + y)k) (k dx dy) A (x + y) dx dy x + xy ] 1 dy y + 1 ] 1 y 3 (1 + y) dy Example Evaluate r d where the surface A represents the surface of the unit cube A x 1, y 1, z 1 and r represents the vector xi + yj + zk. HELM (VERION 1: April 14, 4): Workbook Level 1 8 olution The unit normal d will be a constant vector on each face but will be different for each face. On the face x (left), d dy dz i and the integral on this face is (i + yj + zk) ( dy dz i) dy dz imilarly on the face y (front), d dx dz j and the integral on this face is (xi +j + zk) ( dx dz j) dx dz Furthermore on the face z (bottom), d dx dy k and the integral on this face is (xi + yj +k) ( dx dy k) dx dy On these three faces, the contribution to the integral is zero. However, on the face x 1 (right), d +dy dz i and the integral on this face is (1i + yj + zk) (+ dy dz i) (using the techniques of double integration from Workbook 7). 1 dy dz 1 imilarly, on the face y 1(back), d +dx dz j and the integral on this face is (xi +1j + zk) (+ dx dz j) 1 dx dz 1 and finally,on the face z 1(top), d +dx dy k and the integral on this face is (xi + yj +1k) (+ dx dy k) 1 dx dy 1 Adding together the contributions from the various faces gives r d A Example If F x i + y j + z k,evaluate F d where is the part of the plane z bounded by x ±1, y ±1. 9 HELM (VERION 1: April 14, 4): Workbook Level 1 olution Here d dx dyk and hence i j k F d x y z y dx dyi x dx dyj and dx dy F d y 1 x 1 y dx dy i y 1 x 1 x dx dy j The integral y 1 x 1 y dx dy y 1 y 1 y x ] 1 x 1 dy y dy 3 y3 ] imilarly y 1 x 1 x dx dy 4 3. Thus F d 4 3 i 4 3 j Key Point 1. An integral of the form F (r) d evaluates to a scalar.. An integral of the form F (r) d evaluates to a vector. The vector function involved may be the gradient of a scalar or the curl of a vector. Example Integrate ( φ).d where φ x +yz and is the area between y and y x for x 1 and z. y x HELM (VERION 1: April 14, 4): Workbook Level 1 1 olution Here φ xi +zj +yk and d k dy dx. Thus( φ).d ydydxand ( φ).d x ] x5 y ] x y dydx dx 1 5 x 4 dx Forintegrals of the form F d, non-cartesian geometry e.g. cylindrical or spherical polar coordinates may be used. Once again, it is necessary to include any scale factors along with the unit normal. Example Using cylindrical polar coordinates, (see ection 8.3), find the integral F (r) d for F ρz ˆρ + z sin φẑ and being the complete surface (including ends) of the cylinder ρ a, z 1. z z 1 y ρ 1 x olution The integral F (r) d must be evaluated separately for the curved surface and the ends. For the curved surface, d ˆρ adφdz(with the a coming from ρ the scale factor for φ and the fact that ρ a on the curved surface). Thus, F d a zdφdzand a π F (r) d a zdφdz φ a 1 z]a πa zdzπa πa 4 11 HELM (VERION 1: April 14, 4): Workbook Level 1 olution On the bottom, z sof and the contribution to the integral is zero. On the top, z 1and d ẑ rdrdφand F d ρz sin φdφdρ ρh sin φdφdρand a π F (r) d ρh sin φdφdρ hπ o F (r) d πa4 + 1 hπa πa (a + h ) rho a rho φ ρdρ 1 hπa 1. For F (x +y )i+(x +z )j +xzk and being the rectangle bounded by (1,, 1), (1,, 1), ( 1,, 1) and ( 1,, 1) find the integral F d. For F (x +y )i+(x +z )j +xzk and being the rectangle bounded by (1,, 1), (1,, 1), ( 1,, 1) and ( 1,, 1) (i.e. the same F and as in question 1), find the integral F d 3. Evaluate the integral φ d for φ x z sin y and being the rectangle bounded by (,, ), (1,, 1), (1,π,1) and (,π,). 4. Evaluate the integral ( F ) d where F xey i + ze y j and represents the unit square x 1, y Using spherical polar coordinates, evaluate the integral F d where F r cos θ ˆr and is the curved surface of the top half of the sphere r a. 1.) 8 3 HELM (VERION 1: April 14, 4): Workbook Level 1 1 .) 4 k 3 3.) π 3 4.) (e 1)j 5.) 13 HELM (VERION 1: April 14, 4): Workbook Level 1 πa 3. Volume integrals involving vectors Integrating a scalar function of a vector over a volume is essentially the same procedure as in ection 9.3. The volume element dv may be considered as dx dy dz. However, the scalar function may be the divergence of a vector functions. Example Integrate F over the unit cube x 1, y 1, z 1 where F is the vector function x yi +(x z)j +xz k. olution F x (x y)+ y (x z)+ z (xz )xy +4xz The integral is 3x dx (xy +4xz) dz dy dx xy +x] dy dx ] 1 3 x 3 xy +xy ] 1 dx xyz +xz ] 1 dy dx Key Point The volume integral of a scalar function (including the divergence of a vector) is a scalar 1. Evaluate FdV when F is the vector field yzi + xyj and V is V the unit cube x 1, y 1.. For the vector field F (x y + sin z)i +(xy + e z )j +(z + x y )k, find the integral FdV where V is the volume inside the tetrahedron V bounded by x,y,z and x + y + z Using spherical polar coordinates and the vector field F r ˆr+r sin θ ˆφ, evaluate the integral F dv over the sphere given by r a. V HELM (VERION 1: April 14, 4): Workbook Level 1 14 1.) 1.) 1 3.) 4πa 4 Integrating a vector function over a volume integral is similar but care should be taken with the various components. It may help to think in terms of a separate volume integral for each component. The vector function may be of the form f or F. Example Integrate the function F x i +j over the prism given by x 1, y, z (1 x). 15 HELM (VERION 1: April 14, 4): Workbook Level 1 z y 1 1 x olution The integral is 1 6 i +j x x i +j dz dy dx x (1 x)i + (1 x)j ] dy dx (x x 3 )i +(4 4x)j ] dx x zi +zj ] 1 x dy dx ( 3 x3 1 x4 )i +(4x x )j (x x 3 )i +( x)j ] dy dx ] 1 Example For F x yi + y j evaluate ( F)dV where V is the volume under V the plane z x + y +(and above z )for 1 x 1, 1 y 1. HELM (VERION 1: April 14, 4): Workbook Level 1 16 olution i j k F x y z x k x y y so ( F)dV V x 1 y 1 x 1 y 1 x 1 x 1 x 1 y 1 x+y+ ( x )k dz dy dx ( x )zk ] x+y+ dy dx x 3 x y x ] dy dx k x 3 y 1 x y x y ] 1 x 3 4x ] dx k y 1 dx k 1 x4 4 3 x3 ] 1 1 k 8 3 k z (1, 1, 4) y ( 1, 1, ) (1, 1, ) x ( 1, 1, ) (1, 1, ) Key Point The volume integral of a vector function (including the gradient of a scalar or the curl of a vector) is a vector 17 HELM (VERION 1: April 14, 4): Workbook Level 1 1. Evaluate the integral V F dv for the case where F xi + y j + zk and V being the cube 1 x 1, 1 y 1, 1 z 1.. For f x + yz, and V being the volume bounded by y,x + y 1 and x + y 1for 1 z 1, find the integral ( f) dv. V 3. Evaluate the integral ( F)dV for the case where F xzi V +(x3 + y 3 )j 4yk and V being the cube 1 x 1, 1 y 1, 1 z 1. 1.) 8 j 3.) k 3 3.) 3i +8k HELM (VERION 1: April 14, 4): Workbook Level 1 18

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