 Calculus Volume 3

# 6.6Surface Integrals

Calculus Volume 36.6 Surface Integrals

### Learning Objectives

• 6.6.1 Find the parametric representations of a cylinder, a cone, and a sphere.
• 6.6.2 Describe the surface integral of a scalar-valued function over a parametric surface.
• 6.6.3 Use a surface integral to calculate the area of a given surface.
• 6.6.4 Explain the meaning of an oriented surface, giving an example.
• 6.6.5 Describe the surface integral of a vector field.
• 6.6.6 Use surface integrals to solve applied problems.

We have seen that a line integral is an integral over a path in a plane or in space. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. We can extend the concept of a line integral to a surface integral to allow us to perform this integration.

Surface integrals are important for the same reasons that line integrals are important. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. In particular, surface integrals allow us to generalize Green’s theorem to higher dimensions, and they appear in some important theorems we discuss in later sections.

### Parametric Surfaces

A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. In this sense, surface integrals expand on our study of line integrals. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field.

However, before we can integrate over a surface, we need to consider the surface itself. Recall that to calculate a scalar or vector line integral over curve C, we first need to parameterize C. In a similar way, to calculate a surface integral over surface S, we need to parameterize S. That is, we need a working concept of a parameterized surface (or a parametric surface), in the same way that we already have a concept of a parameterized curve.

A parameterized surface is given by a description of the form

$r(u,v)=〈x(u,v),y(u,v),z(u,v)〉.r(u,v)=〈x(u,v),y(u,v),z(u,v)〉.$

Notice that this parameterization involves two parameters, u and v, because a surface is two-dimensional, and therefore two variables are needed to trace out the surface. The parameters u and v vary over a region called the parameter domain, or parameter space—the set of points in the uv-plane that can be substituted into r. Each choice of u and v in the parameter domain gives a point on the surface, just as each choice of a parameter t gives a point on a parameterized curve. The entire surface is created by making all possible choices of u and v over the parameter domain.

### Definition

Given a parameterization of surface $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉,r(u,v)=〈x(u,v),y(u,v),z(u,v)〉,$ the parameter domain of the parameterization is the set of points in the uv-plane that can be substituted into r.

### Example 6.58

#### Parameterizing a Cylinder

Describe surface S parameterized by

$r(u,v)=〈cosu,sinu,v〉,−∞

#### Analysis

Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to $0≤u≤π,0 then the surface would be a half-cylinder of height 6.

### Checkpoint6.47

Describe the surface with parameterization $r(u,v)=〈2cosu,2sinu,v〉,0≤u<2π,−∞

It follows from Example 6.58 that we can parameterize all cylinders of the form $x2+y2=R2.x2+y2=R2.$ If S is a cylinder given by equation $x2+y2=R2,x2+y2=R2,$ then a parameterization of S is

$r(u,v)=〈Rcosu,Rsinu,v〉,0≤u<2π,−∞

We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.

### Example 6.59

#### Describing a Surface

Describe surface S parameterized by

$r(u,v)=〈ucosv,usinv,u2〉,0≤u<∞,0≤v<2π.r(u,v)=〈ucosv,usinv,u2〉,0≤u<∞,0≤v<2π.$

### Checkpoint6.48

Describe the surface parameterized by $r(u,v)=〈ucosv,usinv,u〉,−∞

### Example 6.60

#### Finding a Parameterization

Give a parameterization of the cone $x2+y2=z2x2+y2=z2$ lying on or above the plane $z=−2.z=−2.$

### Checkpoint6.49

Give a parameterization for the portion of cone $x2+y2=z2x2+y2=z2$ lying in the first octant.

We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. To parameterize a sphere, it is easiest to use spherical coordinates. The sphere of radius $ρρ$ centered at the origin is given by the parameterization

$r(ϕ,θ)=〈ρcosθsinϕ,ρsinθsinϕ,ρcosϕ〉,0≤θ≤2π,0≤ϕ≤π.r(ϕ,θ)=〈ρcosθsinϕ,ρsinθsinϕ,ρcosϕ〉,0≤θ≤2π,0≤ϕ≤π.$

The idea of this parameterization is that as $ϕϕ$ sweeps downward from the positive z-axis, a circle of radius $ρsinϕρsinϕ$ is traced out by letting $θθ$ run from 0 to $2π.2π.$ To see this, let $ϕϕ$ be fixed. Then

$x2+y2=(ρcosθsinϕ)2+(ρsinθsinϕ)2=ρ2sin2ϕ(cos2θ+sin2θ)=ρ2sin2ϕ=(ρsinϕ)2.x2+y2=(ρcosθsinϕ)2+(ρsinθsinϕ)2=ρ2sin2ϕ(cos2θ+sin2θ)=ρ2sin2ϕ=(ρsinϕ)2.$

This results in the desired circle (Figure 6.61).

Figure 6.61 The sphere of radius $ρ ρ$ has parameterization $r ( ϕ , θ ) = 〈 ρ cos θ sin ϕ , ρ sin θ sin ϕ , ρ cos ϕ 〉 , r ( ϕ , θ ) = 〈 ρ cos θ sin ϕ , ρ sin θ sin ϕ , ρ cos ϕ 〉 ,$ $0 ≤ θ ≤ 2 π , 0 ≤ ϕ ≤ π . 0 ≤ θ ≤ 2 π , 0 ≤ ϕ ≤ π .$

Finally, to parameterize the graph of a two-variable function, we first let $z=f(x,y)z=f(x,y)$ be a function of two variables. The simplest parameterization of the graph of $ff$ is $r(x,y)=〈x,y,f(x,y)〉,r(x,y)=〈x,y,f(x,y)〉,$ where x and y vary over the domain of $ff$ (Figure 6.62). For example, the graph of $f(x,y)=x2yf(x,y)=x2y$ can be parameterized by $r(x,y)=〈x,y,x2y〉,r(x,y)=〈x,y,x2y〉,$ where the parameters x and y vary over the domain of $f.f.$ If we only care about a piece of the graph of $ff$—say, the piece of the graph over rectangle $[1,3]×[2,5][1,3]×[2,5]$—then we can restrict the parameter domain to give this piece of the surface:

$r(x,y)=〈x,y,x2y〉,1≤x≤3,2≤y≤5.r(x,y)=〈x,y,x2y〉,1≤x≤3,2≤y≤5.$

Similarly, if S is a surface given by equation $x=g(y,z)x=g(y,z)$ or equation $y=h(x,z),y=h(x,z),$ then a parameterization of S is

$r(y,z)=〈g(y,z),y,z〉r(y,z)=〈g(y,z),y,z〉$ or $r(x,z)=〈x,h(x,z),z〉,r(x,z)=〈x,h(x,z),z〉,$ respectively. For example, the graph of paraboloid $2y=x2+z22y=x2+z2$ can be parameterized by $r(x,z)=〈x,x2+z22,z〉,0≤x<∞,0≤z<∞.r(x,z)=〈x,x2+z22,z〉,0≤x<∞,0≤z<∞.$ Notice that we do not need to vary over the entire domain of y because x and z are squared.

Figure 6.62 The simplest parameterization of the graph of a function is $r ( x , y ) = 〈 x , y , f ( x , y ) 〉 . r ( x , y ) = 〈 x , y , f ( x , y ) 〉 .$

Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization $r(t),a≤t≤br(t),a≤t≤b$ is regular if $r′(t)≠0r′(t)≠0$ for all t in $[a,b].[a,b].$ For a curve, this condition ensures that the image of r really is a curve, and not just a point. For example, consider curve parameterization $r(t)=〈1,2〉,0≤t≤5.r(t)=〈1,2〉,0≤t≤5.$ The image of this parameterization is simply point $(1,2),(1,2),$ which is not a curve. Notice also that $r′(t)=0.r′(t)=0.$ The fact that the derivative is zero indicates we are not actually looking at a curve.

Analogously, we would like a notion of regularity for surfaces so that a surface parameterization really does trace out a surface. To motivate the definition of regularity of a surface parameterization, consider parameterization

$r(u,v)=〈0,cosv,1〉,0≤u≤1,0≤v≤π.r(u,v)=〈0,cosv,1〉,0≤u≤1,0≤v≤π.$

Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure 6.63). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that $ru=〈0,0,0〉ru=〈0,0,0〉$ and $rv=〈0,−sinv,0〉,rv=〈0,−sinv,0〉,$ and the corresponding cross product is zero. The analog of the condition $r′(t)=0r′(t)=0$ is that $ru×rvru×rv$ is not zero for point $(u,v)(u,v)$ in the parameter domain, which is a regular parameterization.

Figure 6.63 The image of parameterization $r ( u , v ) = 〈 0 , cos v , 1 〉 , 0 ≤ u ≤ 1 , 0 ≤ v ≤ π r ( u , v ) = 〈 0 , cos v , 1 〉 , 0 ≤ u ≤ 1 , 0 ≤ v ≤ π$ is a line.

### Definition

Parameterization $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ is a regular parameterization if $ru×rvru×rv$ is not zero for point $(u,v)(u,v)$ in the parameter domain.

If parameterization r is regular, then the image of r is a two-dimensional object, as a surface should be. Throughout this chapter, parameterizations $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ are assumed to be regular.

Recall that curve parameterization $r(t),a≤t≤br(t),a≤t≤b$ is smooth if $r′(t)r′(t)$ is continuous and $r′(t)≠0r′(t)≠0$ for all t in $[a,b].[a,b].$ Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners.

### Definition

A surface parameterization $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ is smooth if vector $ru×rvru×rv$ is not zero for any choice of u and v in the parameter domain.

A surface may also be piecewise smooth if it has smooth faces but also has locations where the directional derivatives do not exist.

### Example 6.61

#### Identifying Smooth and Nonsmooth Surfaces

Which of the figures in Figure 6.64 is smooth?

Figure 6.64 (a) This surface is smooth. (b) This surface is piecewise smooth.

### Checkpoint6.50

Is the surface parameterization $r(u,v)=〈u2v,v+1,sinu〉,0≤u≤2,0≤v≤3r(u,v)=〈u2v,v+1,sinu〉,0≤u≤2,0≤v≤3$ smooth?

### Surface Area of a Parametric Surface

Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. The second step is to define the surface area of a parametric surface. The notation needed to develop this definition is used throughout the rest of this chapter.

Let S be a surface with parameterization $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ over some parameter domain D. We assume here and throughout that the surface parameterization $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ is continuously differentiable—meaning, each component function has continuous partial derivatives. Assume for the sake of simplicity that D is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Divide rectangle D into subrectangles $DijDij$ with horizontal width $ΔuΔu$ and vertical length $Δv.Δv.$ Suppose that i ranges from 1 to m and j ranges from 1 to n so that D is subdivided into mn rectangles. This division of D into subrectangles gives a corresponding division of surface S into pieces $Sij.Sij.$ Choose point $PijPij$ in each piece $Sij.Sij.$ Point $PijPij$ corresponds to point $(ui,vj)(ui,vj)$ in the parameter domain.

Note that we can form a grid with lines that are parallel to the u-axis and the v-axis in the uv-plane. These grid lines correspond to a set of grid curves on surface S that is parameterized by $r(u,v).r(u,v).$ Without loss of generality, we assume that $PijPij$ is located at the corner of two grid curves, as in Figure 6.65. If we think of r as a mapping from the uv-plane to $ℝ3,ℝ3,$ the grid curves are the image of the grid lines under r. To be precise, consider the grid lines that go through point $(ui,vj).(ui,vj).$ One line is given by $x=ui,y=v;x=ui,y=v;$ the other is given by $x=u,y=vj.x=u,y=vj.$ In the first grid line, the horizontal component is held constant, yielding a vertical line through $(ui,vj).(ui,vj).$ In the second grid line, the vertical component is held constant, yielding a horizontal line through $(ui,vj).(ui,vj).$ The corresponding grid curves are $r(ui,v)r(ui,v)$ and $r(u,vj),r(u,vj),$ and these curves intersect at point $Pij.Pij.$

Figure 6.65 Grid lines on a parameter domain correspond to grid curves on a surface.

Now consider the vectors that are tangent to these grid curves. For grid curve $r(ui,v),r(ui,v),$ the tangent vector at $PijPij$ is

$tv(Pij)=rv(ui,vj)=〈xv(ui,vj),yv(ui,vj),zv(ui,vj)〉.tv(Pij)=rv(ui,vj)=〈xv(ui,vj),yv(ui,vj),zv(ui,vj)〉.$

For grid curve $r(u,vj),r(u,vj),$ the tangent vector at $PijPij$ is

$tu(Pij)=ru(ui,vj)=〈xu(ui,vj),yu(ui,vj),zu(ui,vj)〉.tu(Pij)=ru(ui,vj)=〈xu(ui,vj),yu(ui,vj),zu(ui,vj)〉.$

If vector $N=tu(Pij)×tv(Pij)N=tu(Pij)×tv(Pij)$ exists and is not zero, then the tangent plane at $PijPij$ exists (Figure 6.66). If piece $SijSij$ is small enough, then the tangent plane at point $PijPij$ is a good approximation of piece $Sij.Sij.$

Figure 6.66 If the cross product of vectors $t u t u$ and $t v t v$ exists, then there is a tangent plane.

The tangent plane at $PijPij$ contains vectors $tu(Pij)tu(Pij)$ and $tv(Pij),tv(Pij),$ and therefore the parallelogram spanned by $tu(Pij)tu(Pij)$ and $tv(Pij)tv(Pij)$ is in the tangent plane. Since the original rectangle in the uv-plane corresponding to $SijSij$ has width $ΔuΔu$ and length $Δv,Δv,$ the parallelogram that we use to approximate $SijSij$ is the parallelogram spanned by $Δutu(Pij)Δutu(Pij)$ and $Δvtv(Pij).Δvtv(Pij).$ In other words, we scale the tangent vectors by the constants $ΔuΔu$ and $ΔvΔv$ to match the scale of the original division of rectangles in the parameter domain. Therefore, the area of the parallelogram used to approximate the area of $SijSij$ is

$ΔSij≈‖(Δutu(Pij))×(Δvtv(Pij))‖=‖tu(Pij)×tv(Pij)‖ΔuΔv.ΔSij≈‖(Δutu(Pij))×(Δvtv(Pij))‖=‖tu(Pij)×tv(Pij)‖ΔuΔv.$

Varying point $PijPij$ over all pieces $SijSij$ and the previous approximation leads to the following definition of surface area of a parametric surface (Figure 6.67).

Figure 6.67 The parallelogram spanned by $t u t u$ and $t v t v$ approximates the piece of surface $S i j . S i j .$

### Definition

Let $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ with parameter domain D be a smooth parameterization of surface S. Furthermore, assume that S is traced out only once as $(u,v)(u,v)$ varies over D. The surface area of S is

$∬D‖tu×tv‖dA,∬D‖tu×tv‖dA,$
(6.18)

where $tu=〈∂x∂u,∂y∂u,∂z∂u〉tu=〈∂x∂u,∂y∂u,∂z∂u〉$ and $tv=〈∂x∂v,∂y∂v,∂z∂v〉.tv=〈∂x∂v,∂y∂v,∂z∂v〉.$

### Example 6.62

#### Calculating Surface Area

Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height h and radius r.

#### Analysis

The surface area of a right circular cone with radius r and height h is usually given as $πr2+πrh2+r2.πr2+πrh2+r2.$ The reason for this is that the circular base is included as part of the cone, and therefore the area of the base $πr2πr2$ is added to the lateral surface area $πrh2+r2πrh2+r2$ that we found.

### Checkpoint6.51

Find the surface area of the surface with parameterization $r(u,v)=〈u+v,u2,2v〉,0≤u≤3,0≤v≤2.r(u,v)=〈u+v,u2,2v〉,0≤u≤3,0≤v≤2.$

### Example 6.63

#### Calculating Surface Area

Show that the surface area of the sphere $x2+y2+z2=r2x2+y2+z2=r2$ is $4πr2.4πr2.$

### Checkpoint6.52

Show that the surface area of cylinder $x2+y2=r2,0≤z≤hx2+y2=r2,0≤z≤h$ is $2πrh.2πrh.$ Notice that this cylinder does not include the top and bottom circles.

In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. Let $y=f(x)≥0y=f(x)≥0$ be a positive single-variable function on the domain $a≤x≤ba≤x≤b$ and let S be the surface obtained by rotating $ff$ about the x-axis (Figure 6.69). Let $θθ$ be the angle of rotation. Then, S can be parameterized with parameters x and $θθ$ by

$r(x,θ)=〈x,f(x)cosθ,f(x)sinθ〉,a≤x≤b,0≤x<2π.r(x,θ)=〈x,f(x)cosθ,f(x)sinθ〉,a≤x≤b,0≤x<2π.$
Figure 6.69 We can parameterize a surface of revolution by $r ( x , θ ) = 〈 x , f ( x ) cos θ , f ( x ) sin θ 〉 , r ( x , θ ) = 〈 x , f ( x ) cos θ , f ( x ) sin θ 〉 ,$ $a ≤ x ≤ b , 0 ≤ x < 2 π . a ≤ x ≤ b , 0 ≤ x < 2 π .$

### Example 6.64

#### Calculating Surface Area

Find the area of the surface of revolution obtained by rotating $y=x2,0≤x≤by=x2,0≤x≤b$ about the x-axis (Figure 6.70).

Figure 6.70 A surface integral can be used to calculate the surface area of this solid of revolution.

### Checkpoint6.53

Use Equation 6.18 to find the area of the surface of revolution obtained by rotating curve $y=sinx,0≤x≤πy=sinx,0≤x≤π$ about the x-axis.

### Surface Integral of a Scalar-Valued Function

Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let’s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion.

Let S be a piecewise smooth surface with parameterization $r(u,v)=〈x(u,v),y(u,v),z(u,v)〉r(u,v)=〈x(u,v),y(u,v),z(u,v)〉$ with parameter domain D and let $f(x,y,z)f(x,y,z)$ be a function with a domain that contains S. For now, assume the parameter domain D is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Divide rectangle D into subrectangles $DijDij$ with horizontal width $ΔuΔu$ and vertical length $Δv.Δv.$ Suppose that i ranges from 1 to m and j ranges from 1 to n so that D is subdivided into mn rectangles. This division of D into subrectangles gives a corresponding division of S into pieces $Sij.Sij.$ Choose point $PijPij$ in each piece $Sij,Sij,$ evaluate $PijPij$ at $ff$, and multiply by area $ΔSijΔSij$ to form the Riemann sum

$∑i=1m∑j=1nf(Pij)ΔSij.∑i=1m∑j=1nf(Pij)ΔSij.$

To define a surface integral of a scalar-valued function, we let the areas of the pieces of S shrink to zero by taking a limit.

### Definition

The surface integral of a scalar-valued function of $ff$ over a piecewise smooth surface S is

$∬Sf(x,y,z)dS=limm,n→∞∑i=1m∑j=1nf(Pij)ΔSij.∬Sf(x,y,z)dS=limm,n→∞∑i=1m∑j=1nf(Pij)ΔSij.$

Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.

The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.

Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas $ΔSijΔSij$ with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors $tutu$ and $tv:tv:$

$tu=〈∂x∂u,∂y∂u,∂z∂u〉andtv=〈∂x∂v,∂y∂v,∂z∂v〉.tu=〈∂x∂u,∂y∂u,∂z∂u〉andtv=〈∂x∂v,∂y∂v,∂z∂v〉.$

From the material we have already studied, we know that

$ΔSij≈‖tu(Pij)×tv(Pij)‖ΔuΔv.ΔSij≈‖tu(Pij)×tv(Pij)‖ΔuΔv.$

Therefore,

$∬Sf(x,y,z)dS≈limm,n→∞∑i=1m∑j=1nf(Pij)‖tu(Pij)×tv(Pij)‖ΔuΔv.∬Sf(x,y,z)dS≈limm,n→∞∑i=1m∑j=1nf(Pij)‖tu(Pij)×tv(Pij)‖ΔuΔv.$

This approximation becomes arbitrarily close to $limm,n→∞∑i=1m∑j=1nf(Pij)ΔSijlimm,n→∞∑i=1m∑j=1nf(Pij)ΔSij$ as we increase the number of pieces $SijSij$ by letting m and n go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

$∬Sf(x,y,z)dS=∬Df(r(u,v))‖tu×tv‖dA.∬Sf(x,y,z)dS=∬Df(r(u,v))‖tu×tv‖dA.$
(6.19)

Equation 6.19 allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to Equation 6.20 for line integrals:

$∬Cf(x,y,z)ds=∫abf(r(t))‖r′(t)‖dt.∬Cf(x,y,z)ds=∫abf(r(t))‖r′(t)‖dt.$

In this case, vector $tu×tvtu×tv$ is perpendicular to the surface, whereas vector $r′(t)r′(t)$ is tangent to the curve.

### Example 6.65

#### Calculating a Surface Integral

Calculate surface integral $∬S5dS,∬S5dS,$ where $SS$ is the surface with parameterization $r(u,v)=〈u,u2,v〉r(u,v)=〈u,u2,v〉$ for $0≤u≤20≤u≤2$ and $0≤v≤u.0≤v≤u.$

### Example 6.66

#### Calculating the Surface Integral of a Cylinder

Calculate surface integral $∬S(x+y2)dS,∬S(x+y2)dS,$ where S is cylinder $x2+y2=4,0≤z≤3x2+y2=4,0≤z≤3$ (Figure 6.71).

Figure 6.71 Integrating function $f ( x , y , z ) = x + y 2 f ( x , y , z ) = x + y 2$ over a cylinder.

### Checkpoint6.54

Calculate $∬S(x2−z)dS,∬S(x2−z)dS,$ where S is the surface with parameterization $r(u,v)=〈v,u2+v2,1〉,0≤u≤2,0≤v≤3.r(u,v)=〈v,u2+v2,1〉,0≤u≤2,0≤v≤3.$

### Example 6.67

#### Calculating the Surface Integral of a Piece of a Sphere

Calculate surface integral $∬Sf(x,y,z)dS,∬Sf(x,y,z)dS,$ where $f(x,y,z)=z2f(x,y,z)=z2$ and S is the surface that consists of the piece of sphere $x2+y2+z2=4x2+y2+z2=4$ that lies on or above plane $z=1z=1$ and the disk that is enclosed by intersection plane $z=1z=1$ and the given sphere (Figure 6.72).

Figure 6.72 Calculating a surface integral over surface S.

#### Analysis

In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.

### Checkpoint6.55

Calculate line integral $∬S(x−y)dS,∬S(x−y)dS,$ where S is cylinder $x2+y2=1,0≤z≤2,x2+y2=1,0≤z≤2,$ including the circular top and bottom.

Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface S and the density of the sheet at point $(x,y,z)(x,y,z)$ is $ρ(x,y,z),ρ(x,y,z),$ then mass m of the sheet is $m=∬Sρ(x,y,z)dS.m=∬Sρ(x,y,z)dS.$

### Example 6.68

#### Calculating the Mass of a Sheet

A flat sheet of metal has the shape of surface $z=1+x+2yz=1+x+2y$ that lies above rectangle $0≤x≤40≤x≤4$ and $0≤y≤2.0≤y≤2.$ If the density of the sheet is given by $ρ(x,y,z)=x2yz,ρ(x,y,z)=x2yz,$ what is the mass of the sheet?

### Checkpoint6.56

A piece of metal has a shape that is modeled by paraboloid $z=x2+y2,0≤z≤4,z=x2+y2,0≤z≤4,$ and the density of the metal is given by $ρ(x,y,z)=z+1.ρ(x,y,z)=z+1.$ Find the mass of the piece of metal.

### Orientation of a Surface

Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration.

On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation.

Let S be a smooth surface. For any point $(x,y,z)(x,y,z)$ on S, we can identify two unit normal vectors $NN$ and $−N.−N.$ If it is possible to choose a unit normal vector N at every point $(x,y,z)(x,y,z)$ on S so that N varies continuously over S, then S is “orientable.” Such a choice of unit normal vector at each point gives the orientation of a surface S. If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the “negative” side and the side of the surface at which the water flows away is the “positive” side. Informally, a choice of orientation gives S an “outer” side and an “inner” side (or an “upward” side and a “downward” side), just as a choice of orientation of a curve gives the curve “forward” and “backward” directions.

Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the positive orientation of the closed surface (Figure 6.74). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.

Figure 6.74 An oriented sphere with positive orientation.

A portion of the graph of any smooth function $z=f(x,y)z=f(x,y)$ is also orientable. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points “below” the surface at each point. To get such an orientation, we parameterize the graph of $ff$ in the standard way: $r(x,y)=〈x,y,f(x,y)〉,r(x,y)=〈x,y,f(x,y)〉,$ where x and y vary over the domain of $f.f.$ Then, $tx=〈1,0,fx〉tx=〈1,0,fx〉$ and $ty=〈0,1,fy〉,ty=〈0,1,fy〉,$ and therefore the cross product $tx×tytx×ty$ (which is normal to the surface at any point on the surface) is $〈−fx,−fy,1〉.〈−fx,−fy,1〉.$ Since the z component of this vector is one, the corresponding unit normal vector points “upward,” and the upward side of the surface is chosen to be the “positive” side.

Let S be a smooth orientable surface with parameterization $r(u,v).r(u,v).$ For each point $r(a,b)r(a,b)$ on the surface, vectors $tutu$ and $tvtv$ lie in the tangent plane at that point. Vector $tu×tvtu×tv$ is normal to the tangent plane at $r(a,b)r(a,b)$ and is therefore normal to S at that point. Therefore, the choice of unit normal vector

$N=tu×tv‖tu×tv‖N=tu×tv‖tu×tv‖$

gives an orientation of surface S.

### Example 6.69

#### Choosing an Orientation

Give an orientation of cylinder $x2+y2=r2,0≤z≤h.x2+y2=r2,0≤z≤h.$

### Checkpoint6.57

Give the “upward” orientation of the graph of $f(x,y)=xy.f(x,y)=xy.$

Since every curve has a “forward” and “backward” direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. Hence, it is possible to think of every curve as an oriented curve. This is not the case with surfaces, however. Some surfaces cannot be oriented; such surfaces are called nonorientable. Essentially, a surface can be oriented if the surface has an “inner” side and an “outer” side, or an “upward” side and a “downward” side. Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side.

The classic example of a nonorientable surface is the Möbius strip. To create a Möbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure 6.76). Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Therefore, the strip really only has one side.

Figure 6.76 The construction of a Möbius strip.

Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve.

### Surface Integral of a Vector Field

With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field. The definition is analogous to the definition of the flux of a vector field along a plane curve. Recall that if F is a two-dimensional vector field and C is a plane curve, then the definition of the flux of F along C involved chopping C into small pieces, choosing a point inside each piece, and calculating $F·NF·N$ at the point (where N is the unit normal vector at the point). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface S into small pieces, choose a point in the small (two-dimensional) piece, and calculate $F·NF·N$ at the point.

To place this definition in a real-world setting, let S be an oriented surface with unit normal vector N. Let v be a velocity field of a fluid flowing through S, and suppose the fluid has density $ρ(x,y,z).ρ(x,y,z).$ Imagine the fluid flows through S, but S is completely permeable so that it does not impede the fluid flow (Figure 6.77). The mass flux of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area. How could we calculate the mass flux of the fluid across S?

Figure 6.77 Fluid flows across a completely permeable surface S.

The rate of flow, measured in mass per unit time per unit area, is $ρN.ρN.$ To calculate the mass flux across S, chop S into small pieces $Sij.Sij.$ If $SijSij$ is small enough, then it can be approximated by a tangent plane at some point P in $Sij.Sij.$ Therefore, the unit normal vector at P can be used to approximate $N(x,y,z)N(x,y,z)$ across the entire piece $Sij,Sij,$ because the normal vector to a plane does not change as we move across the plane. The component of the vector $ρvρv$ at P in the direction of N is $ρv·Nρv·N$ at P. Since $SijSij$ is small, the dot product $ρv·Nρv·N$ changes very little as we vary across $Sij,Sij,$ and therefore $ρv·Nρv·N$ can be taken as approximately constant across $Sij.Sij.$ To approximate the mass of fluid per unit time flowing across $SijSij$ (and not just locally at point P), we need to multiply $(ρv·N)(P)(ρv·N)(P)$ by the area of $Sij.Sij.$ Therefore, the mass of fluid per unit time flowing across $SijSij$ in the direction of N can be approximated by $(ρv·N)ΔSij,(ρv·N)ΔSij,$ where N, $ρ,ρ,$ and v are all evaluated at P (Figure 6.78). This is analogous to the flux of two-dimensional vector field F across plane curve C, in which we approximated flux across a small piece of C with the expression $(F·N)Δs.(F·N)Δs.$ To approximate the mass flux across S, form the sum $∑i=1m∑j=1n(ρv·N)ΔSij.∑i=1m∑j=1n(ρv·N)ΔSij.$ As pieces $SijSij$ get smaller, the sum $∑i=1m∑j=1n(ρv·N)ΔSij∑i=1m∑j=1n(ρv·N)ΔSij$ gets arbitrarily close to the mass flux. Therefore, the mass flux is

$∬sρv·NdS=limm,n→∞∑i=1m∑j=1n(ρv·N)ΔSij.∬sρv·NdS=limm,n→∞∑i=1m∑j=1n(ρv·N)ΔSij.$

This is a surface integral of a vector field. Letting the vector field $ρvρv$ be an arbitrary vector field F leads to the following definition.

Figure 6.78 The mass of fluid per unit time flowing across $S i j S i j$ in the direction of N can be approximated by $( ρ v · N ) Δ S i j . ( ρ v · N ) Δ S i j .$

### Definition

Let F be a continuous vector field with a domain that contains oriented surface S with unit normal vector N. The surface integral of F over S is

$∬SF·dS=∬SF·NdS.∬SF·dS=∬SF·NdS.$
(6.20)

Notice the parallel between this definition and the definition of vector line integral $∫CF·Nds.∫CF·Nds.$ A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Integral $∬SF·NdS∬SF·NdS$ is called the flux of F across S, just as integral $∫CF·Nds∫CF·Nds$ is the flux of F across curve C. A surface integral over a vector field is also called a flux integral.

Just as with vector line integrals, surface integral $∬SF·NdS∬SF·NdS$ is easier to compute after surface S has been parameterized. Let $r(u,v)r(u,v)$ be a parameterization of S with parameter domain D. Then, the unit normal vector is given by $N=tu×tv‖tu×tv‖N=tu×tv‖tu×tv‖$ and, from Equation 6.20, we have

$∬SF·NdS=∬SF·NdS=∬SF·tu×tv‖tu×tv‖dS=∬D(F(r(u,v))·tu×tv‖tu×tv‖)‖tu×tv‖dA=∬D(F(r(u,v))·(tu×tv))dA.∬SF·NdS=∬SF·NdS=∬SF·tu×tv‖tu×tv‖dS=∬D(F(r(u,v))·tu×tv‖tu×tv‖)‖tu×tv‖dA=∬D(F(r(u,v))·(tu×tv))dA.$

Therefore, to compute a surface integral over a vector field we can use the equation

$∬SF·NdS=∬D(F(r(u,v))·(tu×tv))dA.∬SF·NdS=∬D(F(r(u,v))·(tu×tv))dA.$
(6.21)

### Example 6.70

#### Calculating a Surface Integral

Calculate the surface integral $∬SF·NdS,∬SF·NdS,$ where $F=〈−y,x,0〉F=〈−y,x,0〉$ and S is the surface with parameterization $r(u,v)=〈u,v2−u,u+v〉,0≤u<3,0≤v≤4.r(u,v)=〈u,v2−u,u+v〉,0≤u<3,0≤v≤4.$

### Checkpoint6.58

Calculate surface integral $∬SF·dS,∬SF·dS,$ where $F=〈0,−z,y〉F=〈0,−z,y〉$ and S is the portion of the unit sphere in the first octant with outward orientation.

### Example 6.71

#### Calculating Mass Flow Rate

Let $v(x,y,z)=〈2x,2y,z〉v(x,y,z)=〈2x,2y,z〉$ represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m3. Let S be hemisphere $x2+y2+z2=9x2+y2+z2=9$ with $z≥0z≥0$ such that S is oriented outward. Find the mass flow rate of the fluid across S.

### Checkpoint6.59

Let $v(x,y,z)=〈x2+y2,z,4y〉v(x,y,z)=〈x2+y2,z,4y〉$ m/sec represent a velocity field of a fluid with constant density 100 kg/m3. Let S be the half-cylinder $r(u,v)=〈cosu,sinu,v〉,0≤u≤π,0≤v≤2r(u,v)=〈cosu,sinu,v〉,0≤u≤π,0≤v≤2$ oriented outward. Calculate the mass flux of the fluid across S.

In Example 6.70, we computed the mass flux, which is the rate of mass flow per unit area. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral $∫∫Sv•NdS,∫∫Sv•NdS,$ which leaves out the density. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization of the flow rate of a fluid with velocity v across a surface S:

$Flow rate of fluid acrossS=∫∫Sv•dS.Flow rate of fluid acrossS=∫∫Sv•dS.$

To compute the flow rate of the fluid in Example 6.68, we simply remove the density constant, which gives a flow rate of $90πm3/sec.90πm3/sec.$

Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.

In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point $(x,y,z)(x,y,z)$ in an object is $T(x,y,z).T(x,y,z).$ Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field $F=−k∇T,F=−k∇T,$ where the constant k is the thermal conductivity of the substance from which the object is made (this constant is determined experimentally). The rate of heat flow across surface S in the object is given by the flux integral

$∬SF·dS=∬S−k∇T·dS.∬SF·dS=∬S−k∇T·dS.$

### Example 6.72

#### Calculating Heat Flow

A cast-iron solid cylinder is given by inequalities $x2+y2≤1,x2+y2≤1,$ $1≤z≤4.1≤z≤4.$ The temperature at point $(x,y,z)(x,y,z)$ in a region containing the cylinder is $T(x,y,z)=(x2+y2)z.T(x,y,z)=(x2+y2)z.$ Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.

### Checkpoint6.60

A cast-iron solid ball is given by inequality $x2+y2+z2≤1.x2+y2+z2≤1.$ The temperature at a point in a region containing the ball is $T(x,y,z)=13(x2+y2+z2).T(x,y,z)=13(x2+y2+z2).$ Find the heat flow across the boundary of the solid if this boundary is oriented outward.

### Section 6.6 Exercises

For the following exercises, determine whether the statements are true or false.

269 .

If surface S is given by ${(x,y,z):0≤x≤1,0≤y≤1,z=10},{(x,y,z):0≤x≤1,0≤y≤1,z=10},$ then $∬Sf(x,y,z)dS=∫01∫01f(x,y,10)dxdy.∬Sf(x,y,z)dS=∫01∫01f(x,y,10)dxdy.$

270 .

If surface S is given by ${(x,y,z):0≤x≤1,0≤y≤1,z=x},{(x,y,z):0≤x≤1,0≤y≤1,z=x},$ then $∬Sf(x,y,z)dS=∫01∫01f(x,y,x)dxdy.∬Sf(x,y,z)dS=∫01∫01f(x,y,x)dxdy.$

271 .

Surface $r=〈vcosu,vsinu,v2〉,for0≤u≤π,0≤v≤2,r=〈vcosu,vsinu,v2〉,for0≤u≤π,0≤v≤2,$ is the same as surface $r=〈vcos2u,vsin2u,v〉,r=〈vcos2u,vsin2u,v〉,$ for $0≤u≤π2,0≤v≤4.0≤u≤π2,0≤v≤4.$

272 .

Given the standard parameterization of a sphere, normal vectors $tu×tvtu×tv$ are outward normal vectors.

For the following exercises, find parametric descriptions for the following surfaces.

273 .

Plane $3x−2y+z=23x−2y+z=2$

274 .

Paraboloid $z=x2+y2,z=x2+y2,$ for $0≤z≤9.0≤z≤9.$

275 .

Plane $2x−4y+3z=162x−4y+3z=16$

276 .

The frustum of cone $z2=x2+y2,for2≤z≤8z2=x2+y2,for2≤z≤8$

277 .

The portion of cylinder $x2+y2=9x2+y2=9$ in the first octant, for $0≤z≤30≤z≤3$ 278 .

A cone with base radius r and height h, where r and h are positive constants

For the following exercises, use a computer algebra system to approximate the area of the following surfaces using a parametric description of the surface.

279 .

[T] Half cylinder ${(r,θ,z):r=4,0≤θ≤π,0≤z≤7}{(r,θ,z):r=4,0≤θ≤π,0≤z≤7}$

280 .

[T] Plane $z=10−x−yz=10−x−y$ above square $|x|≤2,|y|≤2|x|≤2,|y|≤2$

For the following exercises, let S be the hemisphere $x2+y2+z2=4,x2+y2+z2=4,$ with $z≥0,z≥0,$ and evaluate each surface integral.

281 .

$∬ S z d S ∬ S z d S$

282 .

$∬ S ( x − 2 y ) d S ∬ S ( x − 2 y ) d S$

283 .

$∬ S ( x 2 + y 2 ) z d S ∬ S ( x 2 + y 2 ) z d S$

For the following exercises, evaluate $∫∫SF·NdS∫∫SF·NdS$ for vector field F, where N is an upward pointing normal vector to surface S.

284 .

$F(x,y,z)=xi+2yj−3zk,F(x,y,z)=xi+2yj−3zk,$ and S is that part of plane $15x−12y+3z=615x−12y+3z=6$ that lies above unit square $0≤x≤1,0≤y≤1.0≤x≤1,0≤y≤1.$

285 .

$F(x,y,z)=xi+yj,F(x,y,z)=xi+yj,$ and S is hemisphere $z=1−x2−y2.z=1−x2−y2.$

286 .

$F(x,y,z)=x2i+y2j+z2k,F(x,y,z)=x2i+y2j+z2k,$ and S is the portion of plane $z=y+1z=y+1$ that lies inside cylinder $x2+y2=1.x2+y2=1.$ For the following exercises, approximate the mass of the homogeneous lamina that has the shape of given surface S. Round to four decimal places.

287 .

[T] S is surface $z=4−x−2y,withz≥0,x≥0,y≥0;ξ=x.z=4−x−2y,withz≥0,x≥0,y≥0;ξ=x.$

288 .

[T] S is surface $z=x2+y2,withz≤1;ξ=z.z=x2+y2,withz≤1;ξ=z.$

289 .

[T] S is surface $x2+y2+x2=5,withz≥1;ξ=θ2.x2+y2+x2=5,withz≥1;ξ=θ2.$

290 .

Evaluate $∬S(y2zi+y3j+xzk)·dS,∬S(y2zi+y3j+xzk)·dS,$ where S is the surface of cube $−1≤x≤1,−1≤y≤1,and0≤z≤2.−1≤x≤1,−1≤y≤1,and0≤z≤2.$ Assume an outward pointing normal.

291 .

Evaluate surface integral $∬SgdS,∬SgdS,$ where $g(x,y,z)=xz+2x2−3xyg(x,y,z)=xz+2x2−3xy$ and S is the portion of plane $2x−3y+z=62x−3y+z=6$ that lies over unit square R: $0≤x≤1,0≤y≤1.0≤x≤1,0≤y≤1.$

292 .

Evaluate $∬S(x+y+z)dS,∬S(x+y+z)dS,$ where S is the surface defined parametrically by $R(u,v)=(2u+v)i+(u−2v)j+(u+3v)kR(u,v)=(2u+v)i+(u−2v)j+(u+3v)k$ for $0≤u≤1,and0≤v≤2.0≤u≤1,and0≤v≤2.$ 293 .

[T] Evaluate $∬S(x−y2+z)dS,∬S(x−y2+z)dS,$ where S is the surface defined by $R(u,v)=u2i+vj+uk,0≤u≤1,0≤v≤1.R(u,v)=u2i+vj+uk,0≤u≤1,0≤v≤1.$ 294 .

[T] Evaluate $∫∫S(x2+y2-z)dS∫∫S(x2+y2-z)dS$ where $SS$ is the surface defined by $R(u,v)=ui−u2j+vk,0≤u≤2,0≤v≤1.R(u,v)=ui−u2j+vk,0≤u≤2,0≤v≤1.$

295 .

Evaluate $∬S(x2+y2)dS,∬S(x2+y2)dS,$ where S is the surface bounded above hemisphere $z=1−x2−y2,z=1−x2−y2,$ and below by plane $z=0.z=0.$

296 .

Evaluate $∬S(x2+y2+z2)dS,∬S(x2+y2+z2)dS,$ where S is the portion of plane $z=x+1z=x+1$ that lies inside cylinder $x2+y2=1.x2+y2=1.$

297 .

[T] Evaluate $∬Sx2zdS,∬Sx2zdS,$ where S is the portion of cone $z2=x2+y2z2=x2+y2$ that lies between planes $z=1z=1$ and $z=4.z=4.$ 298 .

[T] Evaluate $∬S(xz/y)dS,∬S(xz/y)dS,$ where S is the portion of cylinder $x=y2x=y2$ that lies in the first octant between planes $z=0,z=5,y=1,z=0,z=5,y=1,$ and $y=4.y=4.$ 299 .

[T] Evaluate $∬S(z+y)dS,∬S(z+y)dS,$ where S is the part of the graph of $z=1−x2z=1−x2$ in the first octant between the xz-plane and plane $y=3.y=3.$ 300 .

Evaluate $∬SxyzdS∬SxyzdS$ if S is the part of plane $z=x+yz=x+y$ that lies over the triangular region in the xy-plane with vertices (0, 0, 0), (1, 0, 0), and (0, 2, 0).

301 .

Find the mass of a lamina of density $ξ(x,y,z)=zξ(x,y,z)=z$ in the shape of hemisphere $z=(a2−x2−y2)1/2.z=(a2−x2−y2)1/2.$

302 .

Compute $∫∫SF·NdS,∫∫SF·NdS,$ where $F(x,y,z)=xi−5yj+4zkF(x,y,z)=xi−5yj+4zk$ and N is an outward normal vector S, where S is the union of two squares $S1:x=0,0≤y≤1,0≤z≤1S1:x=0,0≤y≤1,0≤z≤1$ and $S2:z=1,0≤x≤1,0≤y≤1.S2:z=1,0≤x≤1,0≤y≤1.$ 303 .

Compute $∫∫SF·NdS,∫∫SF·NdS,$ where $F(x,y,z)=xyi+zj+(x+y)kF(x,y,z)=xyi+zj+(x+y)k$ and N is an upward pointing normal vector S, where S is the triangular region cut off from plane $x+y+z=1x+y+z=1$ by the positive coordinate axes.

304 .

Compute $∫∫SF·NdS,∫∫SF·NdS,$ where $F(x,y,z)=2yzi+(tan−1(xz))j+exykF(x,y,z)=2yzi+(tan−1(xz))j+exyk$ and N is an outward normal vector S, where S is the surface of sphere $x2+y2+z2=1.x2+y2+z2=1.$

305 .

Compute $∫∫SF·NdS,∫∫SF·NdS,$ where $F(x,y,z)=xyzi+xyzj+xyzkF(x,y,z)=xyzi+xyzj+xyzk$ and N is an outward normal vector S, where S is the surface of the five faces of the unit cube $0≤x≤1,0≤y≤1,0≤z≤10≤x≤1,0≤y≤1,0≤z≤1$ missing $z=0.z=0.$

For the following exercises, express the surface integral as an iterated double integral by using a projection on S on the yz-plane.

306 .

$∬Sxy2z3dS;∬Sxy2z3dS;$ S is the first-octant portion of plane $2x+3y+4z=12.2x+3y+4z=12.$

307 .

$∬S(x2−2y+z)dS;∬S(x2−2y+z)dS;$ S is the portion of the graph of $4x+y=84x+y=8$ bounded by the coordinate planes and plane $z=6.z=6.$

For the following exercises, express the surface integral as an iterated double integral by using a projection on S on the xz-plane

308 .

$∬Sxy2z3dS;∬Sxy2z3dS;$ S is the first-octant portion of plane $2x+3y+4z=12.2x+3y+4z=12.$

309 .

$∬S(x2−2y+z)dS;∬S(x2−2y+z)dS;$ S is the portion of the graph of $4x+y=84x+y=8$ bounded by the coordinate planes and plane $z=6.z=6.$

310 .

Evaluate surface integral $∬SyzdS,∬SyzdS,$ where S is the first-octant part of plane $x+y+z=λ,x+y+z=λ,$ where $λλ$ is a positive constant.

311 .

Evaluate surface integral $∬S(x2z+y2z)dS,∬S(x2z+y2z)dS,$ where S is hemisphere $x2+y2+z2=a2,z≥0.x2+y2+z2=a2,z≥0.$

312 .

Evaluate surface integral $∬SzdA,∬SzdA,$ where S is surface $z=x2+y2,0≤z≤2.z=x2+y2,0≤z≤2.$

313 .

Evaluate surface integral $∬Sx2yzdS,∬Sx2yzdS,$ where S is the part of plane $z=1+2x+3yz=1+2x+3y$ that lies above rectangle $0≤x≤3and0≤y≤2.0≤x≤3and0≤y≤2.$

314 .

Evaluate surface integral $∬SyzdS,∬SyzdS,$ where S is plane $x+y+z=1x+y+z=1$ that lies in the first octant.

315 .

Evaluate surface integral $∬SyzdS,∬SyzdS,$ where S is the part of plane $z=y+3z=y+3$ that lies inside cylinder $x2+y2=1.x2+y2=1.$

For the following exercises, use geometric reasoning to evaluate the given surface integrals.

316 .

$∬Sx2+y2+z2dS,∬Sx2+y2+z2dS,$ where S is surface $x2+y2+z2=4,z≥0x2+y2+z2=4,z≥0$

317 .

$∬S(xi+yj)·dS,∬S(xi+yj)·dS,$ where S is surface $x2+y2=4,1≤z≤3,x2+y2=4,1≤z≤3,$ oriented with unit normal vectors pointing outward

318 .

$∬S(zk)·dS,∬S(zk)·dS,$ where S is disc $x2+y2≤9x2+y2≤9$ on plane $z=4,z=4,$ oriented with unit normal vectors pointing upward

319 .

A lamina has the shape of a portion of sphere $x2+y2+z2=a2x2+y2+z2=a2$ that lies within cone $z=x2+y2.z=x2+y2.$ Let S be the spherical shell centered at the origin with radius a, and let C be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the z-axis. Determine the mass of the lamina if $δ(x,y,z)=x2y2z.δ(x,y,z)=x2y2z.$ 320 .

A lamina has the shape of a portion of sphere $x2+y2+z2=a2x2+y2+z2=a2$ that lies within cone $z=x2+y2.z=x2+y2.$ Let S be the spherical shell centered at the origin with radius a, and let C be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the z-axis. Suppose the vertex angle of the cone is $ϕ0,with0≤ϕ0<π2.ϕ0,with0≤ϕ0<π2.$ Determine the mass of that portion of the shape enclosed in the intersection of S and C. Assume $δ(x,y,z)=x2y2z.δ(x,y,z)=x2y2z.$ 321 .

A paper cup has the shape of an inverted right circular cone of height 6 in. and radius of top 3 in. If the cup is full of water weighing $62.5lb/ft3,62.5lb/ft3,$ find the total force exerted by the water on the inside surface of the cup.

For the following exercises, the heat flow vector field for conducting objects i $F=−k∇T,whereT(x,y,z)F=−k∇T,whereT(x,y,z)$ is the temperature in the object and $k>0k>0$ is a constant that depends on the material. Find the outward flux of F across the following surfaces S for the given temperature distributions and assume $k=1.k=1.$

322 .