Surface and  

Volume Integrals 29.2 

Introduction Avector 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.

 ① be familiar with vector derivatives Prerequisites ② be familiar with surface and volume Before starting this Section you should ... integrals  

✓ be able to carry out operations involving Learning Outcomes integrations of vector fields. After completing this Section you should be able to ... 1. Surface 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 byn ˆ.

Example If S is the surface of the sphere x2 + y2 + z2 = a2 find the unit normaln ˆ

Solution 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 x2 + y2 + z2 i.e. the unit vector x y z nˆ = √ x i + √ y j + √ z k = i + j + k where a = x2 + y2 + z2 x2+y2+z2 x2+y2+z2 x2+y2+z2 a a a

z

y n

x

Example For the cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, find the unit normaln ˆ

Solution On the face given by x =0,the unit normal points in the negative x-direction. Hence the unit normal is −i. Similarly :- On the face x =1the unit normal is i On the face y =0the unit normal is −j On the face y =1the unit normal is j On the face z =0the unit normal is −k On the face z =1the unit normal is k

HELM (VERSION 1: April 14, 2004): Workbook Level 1 2 29.2: Surface and Volume Integrals dS and the unit normal The vector dS is a vector, an element of the surface with magnitude dudv and direction per- pendicular to the surface.

If the plane in question is the Oxy plane, then dS =ˆn du dv = k dx dy.

If the plane in question is not one of the three coordinate planes (Oxy, Oxz, Oyz), appro- priate adjustments must be made to express dS 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 =(0, 0, 0), A =(1, 0, 0),B =(1, 1, 1) and C =(0, 1, 1). Find a unit vectorn ˆ normal to the plane and an appropriate vector dS expressed in terms of dx and dy.

z

y C B

E D O x A

Solution −→ −→ 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 2sothe unit normal vector isn ˆ = √1 (−j + k)=− √1 j + √1 k. 2 2 2 The vector dS is therefore (− √1 j + √1 k) du dv where du and dv are increments in the plane 2 2 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)2 +(dz)2.However,√ as z = y, dz = dy and hence dv = 2dy. Thus, dS =(− √1 j + √1 k) dx 2 dy =(−j + j) dx dy. 2 √ 2 Note :- the factor of 2 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 . 2

Integrating a scalar field A function can be integrated over a surface in a manner similar to that shown in sections 29.1 and 29.2. Often, such integrals can be carried out with respect to an element containing the unit normal.

3 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals Example Evaluate the integral 1 2 dS A 1+x where S is the unit normal over the area A and A is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z =0.

Solution In this integral, S becomes k dx dy i.e. the unit normal times the surface element. Thus the integral is 1 1 1 k −1 1 2 dx dy = k tan x 0 dy y=0 x=0 1+x y=0 1 π 1 π 1 = k( − 0) dy = k dy y=0 4 0 4 y=0 π = k 4

2 2 2 2 Example Find S udS where u = r = x + y + z and S is the surface of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

Solution The unit cube has six faces and the normal vectorn ˆ 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 =0,the normaln ˆ = −i and the surface integral is 1 1 1 1 1 (02 + y2 + z2)(−i) dz dy = −i y2z + z3 dy 3 y=0 z=0 y=0 z=0 1 1 1 1 1 2 = −i y2 + dy = −i y3 + y = − i y=0 3 3 3 0 3 On the face x =1,the normaln ˆ = i and the surface integral is 1 1 1 1 1 (12 + y2 + z2)(i) dz dy = i z + y2z + z3 dy 3 y=0 z=0 y=0 z=0 1 4 1 4 1 5 = i y2 + dy = i y3 + y = i y=0 3 3 3 0 3 − 2 5 The net contribution from the faces x =0and x =1is 3 i + 3 i = i. Due to the symmetry of the scalar field u and the unit cube, the net contribution from the faces y =0and y =1isj while the net contribution from the faces z =0and z =1isk. The sum i.e. the surface integral S udS= i + j + k

HELM (VERSION 1: April 14, 2004): Workbook Level 1 4 29.2: Surface and Volume Integrals 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 dS.

Example Find S fdS where f is the function 2x and S is the surface of the triangle bounded by (0, 0, 0), (0, 1, 1) and (1, 0, 1).

z √ Area 3 2 y

x 1 Area 2

Solution

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 i+ √1 j− √1 k. √ 3 3 3 3 1 As the√ area of the triangle is 2 and the area of its projection in the Oxy plane is 2 , the vector 3/2 dS = 1/2 n dy dx =(i + j + k) dy dx. Thus 1 1−x fdS = 2xdydx(i + j + k) S x=0 y=0 1 1−x = [2xy]y=0 dx (i + j + k) x=0 1 = (2x − 2x2) dx (i + j + k) x=0 2 1 1 = x2 − x3 (i + j + k)= (i + j + k) 3 0 3

The scalar function being integrated may be the divergence of a suitable vector function.

5 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals ∇· Example Find S( F )dS where F =2xi + yzj + xyk and S is the surface of the triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0).

Solution Note that ∇·F =2+z =2asz =0everywhere along S.Asthe triangle lies in the Oxy plane, the normal vector n = k and dS = k dy dx. Thus, 1 x (∇·F )dS = 2 dy dx S x=0 y=0 1 1 x 2 1 = [2y]0 dx = 2xdx= x 0 =1 0 0

1. Evaluate the integral S 4xdS where S represents the trapezium with vertices at (0, 0), (3, 0), (2, 1) and (0, 1). 2. Evaluate the integral S xydS where S is the triangle with vertices at (0, 0, 4), (0, 2, 0) and (1, 0, 0). 3. Find the integral S xyz dS where S is the surface of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. ∇· 2 2 4. Evaluate the integral S (x i + yzj + x yk) dS where S is the rectangle with vertices at (1, 0, 0), (1, 1, 0), (1, 1, 1) and (1, 0, 1).

Your solution 1.) (a) Find the vector dS (b) write the surface integral as a double integral (c) evaluate

this double integral

=0 x =0 y 3

k ) c ( yk d x d x 4 (b) k (a)

38

y 3 1

−

HELM (VERSION 1: April 14, 2004): Workbook Level 1 6 29.2: Surface and Volume Integrals Your solution

2.)

3 3 3

k + j + i

2 4 8

Your solution

3.)

4

) z + y + x ( 1

Your solution

4.)

2

i 5

7 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals Integrating a vector field

In a similar manner, a vector field may be integrated over a surface. Two common integrals are · × S F (r) dS and S F (r) dS which integrate to a scalar and a vector respectively. Again, when the unit normal dS is expressed appropriately, the expression will reduce to a double integral.

Example Evaluate the integral (x2yi + zj +(2x + y)k) · dS A where S is the unit normal over the area A and A is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z =0.

Solution On A, the unit normal is dx dy k so the integral becomes (x2yi + zj +(2x + y)k) · (k dx dy) A 1 1 1 2 1 = (2x + y) dx dy = x + xy x=0 dy y=0 x=0 y=0 1 1 1 3 = (1 + y) dy = y + y2 = y=0 2 0 2

· Example Evaluate A r dS where the surface A represents the surface of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 and r represents the vector xi + yj + zk .

HELM (VERSION 1: April 14, 2004): Workbook Level 1 8 29.2: Surface and Volume Integrals Solution The unit normal dS will be a constant vector on each face but will be different for each face. On the face x =0(left), dS = −dy dz i and the integral on this face is 1 1 1 1 (0i + yj + zk) · (− dy dz i)= 0 dy dz =0 z=0 y=0 z=0 y=0

Similarly on the face y =0(front), dS = −dx dz j and the integral on this face is 1 1 1 1 (xi +0j + zk) · (− dx dz j)= 0 dx dz =0 z=0 x=0 z=0 x=0 Furthermore on the face z =0(bottom), dS = −dx dy k and the integral on this face is 1 1 1 1 (xi + yj +0k) · (− dx dy k)= 0 dx dy =0 x=0 y=0 x=0 y=0 On these three faces, the contribution to the integral is zero. However, on the face x =1 (right), dS =+dy dz i and the integral on this face is 1 1 1 1 (1i + yj + zk) · (+ dy dz i)= 1 dy dz =1 z=0 y=0 z=0 y=0

(using the techniques of double integration from Workbook 27). Similarly, on the face y =1(back), dS =+dx dz j and the integral on this face is 1 1 1 1 (xi +1j + zk) · (+ dx dz j)= 1 dx dz =1 z=0 x=0 z=0 x=0 and finally,on the face z =1(top), dS =+dx dy k and the integral on this face is 1 1 1 1 (xi + yj +1k) · (+ dx dy k)= 1 dx dy =1 y=0 x=0 y=0 x=0 · Adding together the contributions from the various faces gives A r dS =0+0+0+1+1+1 = 3

2 2 2 × Example If F = x i + y j + z k,evaluate S F dS where S is the part of the plane z =0bounded by x = ±1, y = ±1.

9 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals Solution Here dS = dx dyk and hence i j k F × dS = x2 y2 z2 = y2 dx dyi − x2 dx dyj and 00 dx dy × 1 1 2 − 1 1 2 S F dS = y=−1 x=−1 y dx dy i y=−1 x=−1 x dx dy j The integral 1 1 1 2 2 1 y dx dy = y x x=−1 dy − − − y= 1 x= 1 y= 1 1 2 1 4 = 2y2 dy = y3 = y=−1 3 −1 3 1 1 2 4 Similarly y=−1 x=−1 x dx dy = 3 . × 4 − 4 Thus S F dS = 3 i 3 j

Key Point · 1. An integral of the form S F (r) dS evaluates to a scalar. × 2. An integral of the form S F (r) dS evaluates to a vector.

The vector function involved may be the gradient of a scalar or the curl of a vector. ∇ 2 Example Integrate S( φ).dS where φ = x +2yz and S is the area between y =0 and y = x2 for 0 ≤ x ≤ 1 and z =0.

y

x

HELM (VERSION 1: April 14, 2004): Workbook Level 1 10 29.2: Surface and Volume Integrals Solution Here ∇φ =2xi +2zj +2yk and dS = k dy dx.Thus(∇φ).dS =2ydydxand 1 x2 (∇φ).dS = 2ydydx S x=0 y=0 1 2 1 2 x 4 = y y=0 dx = x dx x=0 x=0 1 1 1 = x5 = 5 0 5

· Forintegrals of the form S F dS, 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 Section 28.3), find the integral · 2 S F (r) dS for F = ρzρˆ + z sin φzˆ and S being the complete surface (in- cluding ends) of the cylinder ρ ≤ a,0≤ z ≤ 1.

z

y z =1

x ρ =1

Solution · The integral S F (r) dS must be evaluated separately for the curved surface and the ends. For the curved surface, dS =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 · dS = a2 zdφdzand a 2π F (r) · dS = a2zdφdz S z=0 φ=0 a 1 a =2πa2 zdz=2πa2 z2 = πa4 z=0 2 0

11 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals Solution On the bottom, z =0soF =0and the contribution to the integral is zero. On the top, z =1and dS =ˆz rdrdφand F · dS = ρz sin2 φdφdρ= ρh sin2 φdφdρand a 2π F (r) · dS = ρh sin2 φdφdρ S rho=0 φ=0 a 1 = hπ ρdρ= hπa2 rho=0 2 · 4 1 2 2 2 h So S F (r) dS = πa + 2 hπa = πa (a + 2 )

1. For F =(x2 +y2)i+(x2 +z2)j +2xzk and S being the rectangle bounded − − − − · by (1, 0, 1), (1, 0, 1), ( 1, 0, 1) and ( 1, 0, 1) find the integral S F dS 2. For F =(x2 +y2)i+(x2 +z2)j +2xzk and S being the rectangle bounded by (1, 0, 1), (1, 0, −1), (−1, 0, −1) and (−1, 0, 1) (i.e. the same F and S × as in question 1), find the integral S F dS ∇ · 2 3. Evaluate the integral S φ dS for φ = x z sin y and S being the rectangle bounded by (0, 0, 0), (1, 0, 1), (1,π,1) and (0,π,0). ∇× × y y 4. Evaluate the integral S( F ) dS where F = xe i + ze j and S represents the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. · 5. Using spherical polar coordinates, evaluate the integral S F dS where F = r cos θ rˆ and S is the curved surface of the top half of the sphere r = a.

Your solution

1.)

3 8

HELM (VERSION 1: April 14, 2004): Workbook Level 1 12 29.2: Surface and Volume Integrals Your solution

2.)

3

k 4

Your solution

3.)

3

− π

Your solution

4.)

j 1) e ( −

Your solution 5.)

13 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals

92 ufc n oueIntegrals Volume and Surface 29.2:

14 EM(ESO :Arl1,20) okokLvl1 Level Workbook 2004): 14, April 1: (VERSION HELM

V

. a r by given sphere the r ove dV F vlaeteintegral the evaluate ≤ ∇·

F field vector the and coordinates polar spherical Using 3. , φ θ sin r + ˆ r r =

ˆ 2 2

=1. z + y + x and =0 z =0, y =0, x by ounded b

V

stevlm nietetetrahedron the inside volume the is V where dV F integral ∇·

F field vector the r Fo 2. n the find , k ) x + z +( j ) e + xy +( i ) z sin + y x =(

y 2 z 2 2

1. y 0 1, x 0 cube unit the

≤ ≤ ≤ ≤

V

is V and xyj + yzi field vector the is F when dV F Evaluate 1. ∇·

h oueitga fasaa ucin(nldn h iegneo etr sascalar a is vector) a of divergence the (including function scalar a of integral volume The

oint P Key

0 =0 x 2 2

x = = xdx 3 =

2

3 3

1

1

=0 y =0 x =0 x

0

= xy = dx dy ] x +2 xy [2 dx xy +2

2

1

1 1 1

=0 z =0 y =0 x =0 y =0 x

0

xz +2 xyz 2 = dx dy dz ) xz +4 xy (2 ydx dy

2

1

1 1 1 1 1

h nerlis integral The

∂z ∂y ∂x

F xz (2 )+ z x ( x ( xz +4 xy )=2 )+ y =

∇· −

2 2 ∂ ∂ ∂

Solution

. k xz +2 j ) z x +( yi x function vector the is

− 2 2

Example F where 1 z 0 1, y 0 1, x 0 cube unit the r ove F Integrate ≤ ≤ ≤ ≤ ≤ ≤ ∇·

ucinmyb h iegneo etrfunctions. vector a of divergence the be may function

eto 93 h oueelement volume The 29.3. Section a ecniee as considered be may dV r h scalar the er, v owe .H dz dy dx

nertn clrfnto favco vravlm sesnilytesm rcdr sin as procedure same the essentially is volume a over vector a of function scalar a Integrating .Vlm nerl novn vectors involving integrals Volume 2.

πa3 Your solution

1.)

2 1

Your solution

2.)

20 1

Your solution

3.)

πa 4 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 = x2i +2j over the prism given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ (1 − x).

15 HELM (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals z

y

1 2

x 1

Solution The integral is 1 2 1−x 1 2 − x2i +2j dz dy dx = x2zi +2zj 1 x dy dx z=0 x=0 y=0 z=0 x=0 y=0 1 2 1 2 = x2(1 − x)i + 2(1 − x)j dy dx = (x2 − x3)i +(2− 2x)j dy dx x=0 y=0 x=0 y=0 1 2 1 1 = (2x2 − 2x3)i +(4− 4x)j dx = ( x3 − x4)i +(4x − 2x2)j x=0 3 2 0 1 = i +2j 6

2 2 ∇× Example For F = x yi + y j evaluate V ( F )dV where V is the volume under the plane z = x + y +2(and above z =0)for −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.

HELM (VERSION 1: April 14, 2004): Workbook Level 1 16 29.2: Surface and Volume Integrals Solution i j k ∂ ∂ ∂ ∇×F = = −x2k ∂x ∂y ∂z x2yy2 0 so 1 1 x+y+2 (∇×F )dV = (−x2)k dz dy dx V x=−1 y=−1 z=0 1 1 − 2 x+y+2 = ( x )zk z=0 dy dx x=−1 y=−1 1 1 = −x3 − x2y − 2x2 dy dx k − − x= 1 y= 1 1 1 1 = −x3y − x2y2 − 2x2y dx k − 2 − x= 1 y= 1 1 1 4 1 8 = −2x3 − 0 − 4x2 dx k = − x4 − x3 k = − k x=−1 2 3 −1 3

z

(1, 1, 4) y

(−1, 1, 0) (1, 1, 0)

x

(−1, −1, 0) (1, −1, 0)

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 (VERSION 1: April 14, 2004): Workbook Level 1 29.2: Surface and Volume Integrals 2 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.

2. For f = x2 + yz, and V being the volume bounded by y =0,x + y =1 − − ≤ ≤ ∇ and x + y =1for 1 z 1, find the integral V ( f) dV . ∇× 3 3. Evaluate the integral V ( F )dV for the case where F = xzi +(x + y3)j − 4yk and V being the cube −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, −1 ≤ z ≤ 1.

Your solution

1.)

3

j 8

Your solution

2.)

3

k 2

Your solution

3.)

k +8 i 32 −

HELM (VERSION 1: April 14, 2004): Workbook Level 1 18 29.2: Surface and Volume Integrals