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Math 32B Review Sheet
- Tau Beta Pi - Boelter
- 1 Double Integrals Contents
- 1.1 Changing order of integration
- 1.2 Integrating over more general domains
- 2 Triple Integrals
- 3 Line Integrals
- 3.1 Vector Line Integrals
- 3.2 Conservative Vector Fields
- 4 Surface Integrals
- 4.1 Vector Surface Integrals
- 5 Fundamental Theorems of Multivariable Calculus
- 5.1 Green’s Theorem
- 5.2 Stokes’ Theorem
- 5.3 Divergence Theorem
1 Double Integrals
The basic premise of calculating a double integral is calculating the volume underneath a surface, just as the basic premise of calculating a single integral is finding the area under the line of a function. An example below shows a parabaloid where the function is z = x^2 + y^2.
Figure 1: The graph of the surface z = x^2 + y^2 is displayed. (Graph produced via Matlab)
We can see that many of the rules of calculating areas under curves also translate to calculate volumes under surfaces. For example, if we call the domain (− 10 , 10) × (− 10 , 10) as D (essentially x can take values from -10 to 10 and y can take values from -10 to 10 as shown in Figure 1), then the area under the paraboliod shown in Figure 1 is ¨
D
x^2 + y^2 dA
In general, calculating the (possibly signed) volume under the function z = f (x, y) is ¨
D
f (x, y) dA
As you might imagine, to calculate the volume ”over” the paraboloid instead, for this image, the volume between x^2 + y^2 and 200 is ¨
D
200 − (x^2 + y^2 ) dA
Double integrals seem to be fairly easy to visualize, so the hardest problem is usually computation, which has more nuances than in the single variable case. It is somewhat hard to show graphically, but the idea
1.1 Changing order of integration
Although the order of integration does not make a difference in the final value of the double integral, it can make a big difference in how easy it is to evaluate the integral analytically. For example
ˆ (^1)
0
0
xexy^ dxdy
Technically, we can integrate with respect to x first holding y constant using integration by parts. How- ever, it seems much easier to evaluate the integral with respect to y first instead.
ˆ (^2)
0
0
xexy^ dydx =
0
xexy^ ]^10 dx =
0
exdx = e^2 − 1
Sometimes, the integral might be impossible to evaluate analytically one way. Take ˆ (^2)
1
1
y
e
xy dydx
This integral is impossible to evaluate the way it is written. Calculating it the other way however, ˆ (^2)
1
1
y
e
xy dxdy =
1
1
y^2
(e
(^2) y − e
(^1) y ) dy = (e
(^1) y −
e
(^2) y )]^21 = e (^12) −
e +
e^2
1.2 Integrating over more general domains
While integrating over rectangles is quite useful, we have a much greater variety of domains that we did in single variable calculus. All domains of integration in single variable calculus had to be intervals of the form [a,b], but in 2-dimensions, we have rectangles, circles, ellipses, and so on which give much greater variety of domains to integrate over. Figure 2 from Rogawski sums up how to deal with the situation fairly well. If we make the rectangles very small so that they approximate the domain very well, we can see that the top value of the rectangle will be the top function g(x) and the bottom value the rectangles will reach will be the bottom function h(x). Therefore, the sum of the integrals of some function f (x, y) over this domain becomes
¨ f (x, y)dA =
ˆ (^) b
a
ˆ (^) g(x)
h(x)
f (x, y)dydx
Here, a and b are the smallest and largest values x can take respectively and still remain inside the domain. The function g(x) and h(x) take care of the values that y can take inside the domain. Notice that this integral eventually evaluates to a number, so swapping the integral in this case to somehow put g(x) and h(x) on the outside and integrate with respect to x first does not really make sense. Instead, we have to find the upper and lower bounds on x as a function of y and then use those as the bounds (more on this later). It would be easiest to see these integrals evaluated with an example:
ˆ (^1)
0
ˆ (^) x
0
1dydx
This domain is described by the picture in Figure 3. The double integral gives the area of the triangle, since it gives the volume under the surface of height 1, which gives the same numerical value as the area. Doing the integral, we get
ˆ (^1)
0
ˆ (^) x
0
1dydx =
0
y]x 0 dx =
0
xdx =
We know the triangle represents the bounds on the integral because y goes from 0 to x, which is the boundary y = x and x goes from 0 to 1. If we wanted to switch the bounds, we need the bounds on x as a function of y first and then the overall bounds on y. Looking at Figure 3, the largest x can become is 1. However, the smallest x can become is y. The values y can take also range between 0 and 1. Thus, we have
Figure 2: The graph of an arbitrary domain approximated using rectangles. (Graph adapted from Rogawski, 3rd edition, pg. 849)
0
ˆ (^) x
0
1dydx =
0
y
1dxdy =
0
x]^1 y dy =
0
1 − ydy =
2 Triple Integrals
After learning double integrals, triple integrals largely use the same tools, but simply add an extra step of integration. If double integrals are defined over an area domain, triple integrals are defined over a volume domain which means it is not quite possible to visually see the value being calculated, but it is still useful to see the domain graphically. We can evaluated triple integrals over boxes and more general domains just as before:
0
0
0
x^2 + y^2 + z^2 dxdydz =
0
0
x^3 + xy^2 + xz^2 ]^10 dydz =
0
0
Figure 4: The domain of integration, W, taken from Rogawski, 3rd edition, page 865.The domain is the volume found in between the surfaces which is over the first quadrant in the x-y plane.
then choosing a good order of integration would do wonders for easiness of the calculation. Since the volume is bounded by the two surfaces which give the bounds on z as functions of x and y, it makes sense to make z the inner bounds.
˚
W
xdV =
D
ˆ (^4) −x^2 −y^2
x^2 +3y^2
xdzdA
For the outer bounds, we need to see the largest values that x and y can take while still remaining inside the domain. This is given by the region where the two surfaces intersect, where
x^2 + 3y^2 = 4 − x^2 − y^2 ⇐⇒ x^2 + 2y^2 = 2 This domain gives the ellipse shown in Figure 5. The domain is fairly symmetric, so we can choose to either find the bounds for x in terms of y or vice versa. We choose x in terms of y because the bounds of y are simpler and the this will yield even polynomial orders of y after the x integration, which will be nice for cancelling out the square root. The bounds on x are then
0 ≤ x ≤
2 − 2 y^2 and the bounds on y are [0, 1]. Thus, the integral is ˚
W
xdV =
0
ˆ √ 2 − 2 y 2
0
ˆ (^4) −x (^2) −y 2
x^2 +3y^2
xdzdxdy =
0
ˆ √ 2 − 2 y 2
0
x
4 − 4 y^2 − 2 x^2
dxdy
0
ˆ √ 2 − 2 y 2
0
4 x
1 −y^2
− 2 x^3 dxdy =
0
2 x^2
1 −y^2
x^4
]√ 2 − 2 y 2
0
dy =
0
2 − 2 y^2
1 −y^2
2 − 2 y^2
dy
It is the not the point of this class to solve complicated single integrals, but in case you are interested, this simply involves quite a bit of expansion and polynomial integration.
Figure 5: The outer domain, D, which describes the bounds on x and y. (Produced via Matlab)
0
2 − 2 y^2
1 − y^2
2 − 2 y^2
dy =
0
4 − 8 y^2 + 4y^4
2 − 4 y^2 + 2y^4
dy =
0
2 − 4 y^2 + 2y^4 dy
= 2y −
y^3 +
y^5
] 1
0
We were fortunate enough to avoid trigonometric substitution in the last problem and as it happens, trigonometric substitution very rarely shows up in multiple integration due to our ability to make variable substitutions instead.
2.1 Change of Variables
The proof of change of variables is something done using linear algebra and so is left out of this class, but an example explains the concept fairly well. Take the parallelogram displayed in Figure 6 as the domain. We would like to calculate ¨
D
1dA
Computing this integral with respect to either x or y would be messy as either the upper bound or the lower bound would change twice, leading to the sum of three double integrals, something we would all like to avoid doing. Consider the 32A way of calculating the area given by two parallelograms. It was calculated using the magnitude of the cross product of the two vectors that made up the sides, in this case,
Figure 7: The transformation above defines the following square of sides 1 in (u,v) coordinates which is now the new domain, D’. (Produced via Matlab)
Then, the double integral
D f^ (x, y)dxdy^ can be found using ¨
D
f (x, y)dxdy =
D′
f (x(u, v), y(u, v)) |Jac(u, v)|dudv
where the |Jac(u, v)| is the (absolute value) of the determinant of the Jacobian. This idea can be generalized to 3-dimensions using the 3-dimensional Jacobian instead.
x y z
g(u, v, w) h(u, v, w) k(u, v, w)
(^) =⇒ Jac =
∂x ∂u
∂x ∂v
∂x ∂y ∂w ∂u
∂y ∂v
∂y ∂z ∂w ∂u
∂z ∂v
∂z ∂w
Then, the integral becomes ˚
D
f (x, y, z)dV (x, y, z) =
D′
f (x(u, v, w), y(u, v, w), z(u, v, w)) ∗ |Jac(u, v, w)| dV (u, v, w)
It is again easiest to see this with an example, such as the area of an ellipse. Let us take a general ellipse given by the equation
( x a
y b
If we take the entire ellipse as the domain, we have to solve the following integral to get the area. ¨
D
1dA
Instead of solving this by expressing x in terms of y or vice versa, let us instead make the substitution u = xa and v = ya. This turns the (u,v) domain into a circle of radius 1. In terms of x and y, x = au and y = bv. This transformation is shown in Figure 8.
Figure 8: The defined transformation from an ellipse in (x,y) coordinates to the disk in (u,v) coordinates.
The determinant of the Jacobian is ∣ ∣ ∣ ∣
a 0 0 b
∣ =^ ab
Then, the integral becomes ¨
D
1dA(x, y) =
D′
abdA(u, v) = ab
D′
dA(u, v) = πab
Similarly, try showing that the volume of an ellipsoid which is surrounded by the surface
( (^) x a
( (^) y b
( (^) z c
= 1 is πabc. It is very hard to know what kind of substitution will make your life easier, but there are a few common ones that are frequently useful. The one we just performed was a scaling of the axes. Another one, in 2-dimensions, is a change to polar coordinates, given by the transformation x = r cos(θ) and y = r sin(θ). This transformation has the Jacobian matrix determinant
∣ ∣ ∣ ∣
∂x ∂r
∂x ∂y ∂θ ∂r
∂y ∂θ
cos(θ) −r sin(θ) sin(θ) r cos(θ)
∣ =^ r
In 3-dimensions, we can define this same polar coordinate transformation while keeping the z-coordinate and thus using cylindrical coordinates; x = r cos(θ), y = r sin(θ) and z = z.
∣ ∣ ∣ ∣ ∣ ∣ ∂x ∂u
∂x ∂v
∂x ∂y ∂w ∂u
∂y ∂v
∂y ∂z ∂w ∂u
∂z ∂v
∂z ∂w
cos(θ) −r sin(θ) 0 sin(θ) r cos(θ) 0 0 0 1
= r
Finally, another transformation that is common is using spherical coordinates, x = ρ cos(θ) sin(φ), y = ρ sin(θ) sin(φ), and z = ρ sin(φ). The physical interpretation of the symbols ρ, θ, and φ in this case can be seen in Figure 9. The Jacobian in this case is ∣ ∣ ∣ ∣ ∣ ∣ ∂x ∂u
∂x ∂v
∂x ∂y ∂w ∂u
∂y ∂v
∂y ∂z ∂w ∂u
∂z ∂v
∂z ∂w
cos(θ) sin(φ) −ρ sin(θ) sin(φ) ρ cos(θ) cos(φ) sin(θ) sin(φ) ρ cos(θ) sin(φ) ρ sin(θ) cos(φ) cos(φ) 0 −ρ sin(φ)
= −ρ^2 sin(φ)
In this case, the evaluation of the determinant is negative. However, this is simply a choice of the ordering of the coordinates since swapping two rows in a determinant flips the sign. Thus, we will take the absolute value of the determinant and get |Jac| = ρ^2 sin(φ).
3 Line Integrals
Line integrals are a very special type of integral unlike the other ones we have seen before. Figure 10 presents the essential idea of how we view an integral.
Figure 10: The physical intuition of an integral.
The integral finds the area, A, between an arbitrary function f(x) and the x axis between the x values of a and b. We sum up over dx, the differential element that describes how to change from a to b. In the line integral case, we instead have a surface z = f (x, y) defined over a curve in the two dimensional case. The curve is in the x-y plane and the surface is above the x-y plane and denotes a height for every point along the curve. To find the area of the curtain that now drops from the surface to the curve, we sum up the areas of the rectangles, which have individual areas
∆A = f (x, y)∆s where s describes the length of the rectangle, which is the arc-length of the segment of the curve and f(x,y) is the value of the surface which is also the height of the curve. Then, to find the area, we would need take the limit as all the ∆s become small, which is the same as computing the integral over the curve, c. ˆ
c
f (x, y)ds
This area is defined independent of a parametrization given to the curve since it is simply an intrinsic property of the curve and the surface over it. However, this integral is in practice computing by expressing both x and y in terms of a parameter and thus computing the integral as such. Let the parametrization be ~r(t) = 〈x(t), y(t)〉 and the beginning and ending points of the integral be the points on the curve given by ~r(t 0 ) and ~r(t 1 ) respectively. Then,
ˆ
c
f (x, y)ds =
ˆ (^) t 1
t 0
f (x(t), y(t))
ds dt
dt
Since s described the arc length of the curve,
ds dt
= ||~c(t)|| =
dx dt
dy dt
c
f (x, y)ds =
ˆ (^) t 1
t 0
f (x(t), y(t))
dx dt
dy dt
dt
Again, integrating the function f (x, y) = 1 will give you the length of the curve. As an example, let us integrate the function f (x, y) = x along the path defined by the parabola given by y = x^2 from the point
〈 0 , 0 〉 to the point 〈 1 , 1 〉. A parametrization of this curve can simply be given by
〈x(t), y(t)〉 = 〈t, t^2 〉
Then,
ds dt
1 + 4t^2
The function in terms of the parameter t is
f (x(t), y(t)) = t Thus, we have ˆ
c
f (x, y)ds =
0
t
1 + 4t^2 dt =
1
udu =
[
u (^32)
] 5
1
(^32) − 1 12 This idea extends nicely to three dimensions, which is beyond our ability to visualize, but still can be computed.
ˆ
c
f (x, y, z)ds =
ˆ (^) t 1
t 0
f (x(t), y(t), z(t))
ds dt
dt =
ˆ (^) t 1
t 0
f (x(t), y(t), z(t))
dx dt
dy dt
dz dt
dt
As an example, consider the function f (x, y, z) = x^2 z over the path ~r(t) = 〈et,
2 t, e−t〉 from t = 0 to t = 1. Then, we have
ˆ
c
f (x, y, z)ds =
0
et
e^2 t^ + 2 + e−^2 tdt =
0
et
et^ + e−t
dt =
0
e^2 t^ + 1dt =
e^2 t^ + t
] 1
0
e^2 + 1 2 The dsdt term in the line integral makes most integrals of this form impossible to compute analytically, but there is a special class of integrands that do not have this problem that are common; vector line integrals.
3.1 Vector Line Integrals
Before discussing a vector line integral itself, we first need to have a concept of a vector field. A vector field is a function that takes possibly multiple inputs and returns outputs as a vector. An example is F^ ~ (x, y) = 〈x^2 + y^2 , xy〉. In general, it is of the form F~ (x, y) = 〈f (x, y), g(x, y)〉, or in three dimensions, F^ ~ (x, y, z) = 〈f (x, y, z), g(x, y, z), h(x, y, z)〉.
In two dimensions, this is equivalent to taking every point inside the domain inside the x-y plane and attaching a 2-dimensional vector to it. An example vector field, F~ (x, y) = 〈x, y〉 is shown in Figure 11. Another common example is the vector field given by F~ (x, y) = 〈y, −x〉, which is Figure 12. Finally, a more complex vector field, F~ (x, y) = 〈y, sin(x)〉, is shown in Figure 13. Notice how as x approaches 0, the arrows become horizontal as sin(0) = 0 and as y approaches 0, the arrows becomes vertical.The arrows also flip from left to right moving right on the x=axis just as sin(x) flips from negative to positive.
Finally, a three-dimensional example is rarely drawn, but it is instructive to see at least one, shown in Figure 14 of F~ (x, y, z) = 〈x, y, z〉. A vector line integral is that which calculates ˆ
c
F~ · Tˆ
ds
Figure 12: The vector field F~ (x, y) = 〈y, −x〉 is displayed.
F^ ~ (~r(t)) = 〈cos(t), sin(t)〉 d~r dt
= 〈− sin(t), cos(t)〉 ˆ
c
F^ ~ · d~r =
ˆ (^2) π
0
〈cos(t), sin(t)〉 · 〈− sin(t), cos(t)〉dt =
ˆ (^2) π
0
0 dt = 0
This answer somewhat makes sense because we have constructed a path that starts at a point, 〈 1 , 0 〉 and ends at the same point. However, this is actually not always the case, as we can see integrating the vector field F~ = 〈y, −x〉. We instead get
F^ ~ (~r(t)) = 〈sin(t), − cos(t)〉 ˆ (^2) π
0
〈sin(t), − cos(t)〉 · 〈− sin(t), cos(t)〉dt =
ˆ (^2) π
0
−1dt = − 2 π
Therefore, the vector field determines whether line integrals around closed paths are 0. To denote a closed path, the following notation is common. ˛
c
F^ ~ · d~r
3.2 Conservative Vector Fields
When a vector field integrates such that its line integral is 0 for all closed loops, we call it a conservative vector field. This happens when the vector field is the gradient of some function f(x,y,z) because, from 32A,
df (~r(t)) = ∇~f ·
d~r dt
dt
Figure 13: The vector field F~ (x, y) = 〈−y, x〉 is displayed.
˛
c
F^ ~ · d~r =
ˆ (^) t 1
t 0
∇^ ~f · d~r dt
dt =
ˆ (^) t 1
t 0
df (~r(t)) = f (~r(t))
]t 1 t 0
The last expression is 0 because the closed line integral implies that ~r(t 0 ) = ~r(t 1 ). A useful way of testing whether a vector field is the gradient of some function is checking the curl of the function, which checks all mixed derivatives. From 32A, we have
∂ ∂y
( (^) ∂f ∂x
∂x
( (^) ∂f ∂y
so, for 2-dimensions, if F~ = 〈F 1 (x, y), F 2 (x, y)〉 we must check ∣ ∣ ∣ ∣
∂ ∂x
∂y ∂θ F 1 F 2
∣ =^
∂F 2
∂x
∂F 1
∂y
For 3-dimensions, F~ = 〈F 1 , F 2 , F 3 〉 we must check
∇ ×^ ~ F~ =
ˆi ˆj ˆk ∂ ∂x
∂ ∂y
∂ ∂z F 1 F 2 F 3
∂F 3
∂y
∂F 2
∂z
ˆi +
∂F 1
∂z
∂F 3
∂x
ˆj +
∂F 2
∂x
∂F 1
∂y
ˆi = 〈 0 , 0 , 0 〉
If this is the case, then we can find the function f itself and not bother with calculating the line integral directly. For example, let us calculate the line integral of F~ = 〈y, x, z^3 〉 over the path ~r(t) = 〈t, t^2 , t^3 〉 from t = 0 to t = 1. In checking whether the field is conservative, we obtain
∣ ∣ ∣ ∣ ∣ ∣ ˆi ˆj ˆk ∂ ∂x
∂ ∂y
∂ ∂z y x z^3
i +
j +
i = 〈 0 , 0 , 0 〉
To find the field itself, let 〈 ∂f ∂x
∂f ∂y
∂f ∂z
= 〈y, x, z^3 〉
f (x, y, z) = xy +
z^4 4
+ C
The final C is truly a constant independent of x, y, and z. It will cancel out in the difference when doing the integral. Thus, for any path starting at 〈 0 , 0 , 0 〉 and ending at 〈 1 , 1 , 1 〉, the line integral for this vector field is
ˆ
c
F^ ~ · d~r = f (1, 1 , 1) − f (0, 0 , 0) =
4 Surface Integrals
We can also think about what it means to define an integral over a surface just as we did for a line. We do not have a visual representation in this case because surfaces are already 3-dimensional objects, but we can imagine a function that gives a value for every (x,y,z) coordinate on a surface. Then, we would somehow want to compute
∑ f (x, y, z)∆S
for all small surface area elements we divide the surface into and take the limit as ∆S goes to 0 to get ¨
S
f (x, y, z)dS
The ds in the line integral case denoted the differential arc length element and so dS in this case must represent the differential surface area. From the example in Figure 6, we remember that the area of a parallelogram is given by the magnitude of the cross product of the vectors that make up the sides. Since the surface area can be approximated as parallelograms that become increasingly accurate as the surface area is divided up more and more, we can find dS as local cross product of the tangent vectors.
Suppose ~r(t, s) = 〈x(t, s), y(t, s), z(t, s)〉 describes the surface. An example of this would be how ~r(x, y) = 〈x, y, x^2 + y^2 〉 describes the paraboloid z = x^2 + y^2. Then, the surface area element is given by the magnitude of the cross product of the tangent vectors multiplied by area element in terms of (t,s) coordinates, or
dS =
∂~r ∂t
×
∂~r ∂s
∣dA(t, s) ¨
S
f (x, y, z)dS =
f (x(t, s), y(t, s), z(t, s))
∂~r ∂t
×
∂~r ∂s
∣dA(t, s)
As an example, let f (x, y, z) = x + y + z. Let us integrate this function over the surface of the cone z^2 = x^2 + y^2 from z = 0 to z = 4. The natural way to express the cone is in cylindrical coordinates, since we already have the equation z = r. Therefore, it seems natural to use x = r cos(θ) and y = r sin(θ) as our two parameters. The bounds on theta are [0, 2 π] in that case and for r, we have 0 ≤ r = z ≤ 4, so the bounds on r are [0, 4]. Therefore, we have our parametrization ~r(r, θ) = 〈r cos(θ), r sin(θ), r〉 and our r and θ bounds as specified above. Now we must calculate the cross product.
∂~r ∂r = 〈cos(θ), sin(θ), 1 〉
∂~r ∂θ
= 〈−r sin(θ), r cos(θ), 0 〉
∂~r ∂r
×
∂~r ∂θ
ˆi ˆj ˆk cos(θ) sin(θ) 1 −r sin(θ) r cos(θ) 0
= 〈−r cos(θ), −r sin(θ), r〉
∂~r ∂r
×
∂~r ∂θ
2 r
f (x(r, θ), y(r, θ), z(r, θ)) =
r cos(θ)
r sin(θ)
1 + sin(θ) + cos(θ)
S
f (x, y, z)dS =
D
r
1 + sin(θ) + cos(θ)
2 rdA(r, θ) =
ˆ (^2) π
0
0
r^2
1 + sin(θ) + cos(θ)
drdθ
0
r^2 dr
)( (^) ˆ (^2) π
0
1 + sin(θ) + cos(θ)
dθ
· 2 π =
128 π
4.1 Vector Surface Integrals
Just as we previously defined a vector line integrals as taking the integrand as a vector field dotted with the unit tangent vector, we accomplish a similar task with a surface integral. It does not make sense to calculate the amount of a vector field is ”parallel” to a surface, but it does make sense to compute how much of a vector field is normal to a surface, or F~ · nˆ. Then, a vector surface integral is ¨
S
F~ · ˆn
dS
Again, this quantity is independent of parametrization, but the parametrization greatly simplifies its calculation since ~n = ∂~∂tr × ∂~∂sr is already perpendicular to the surface as it is found through the cross product of the tangent vectors. Then, ¨
S
F~ · ˆn
dS =
S
F~ · ˆn
||~n||dA(t, s) =
S
F^ ~ · ~ndA(t, s)
As a shorthand, to declare that this value is independent of parametrization, it is often written as ¨
S
F^ ~ · dS~
A physical intuition that can be given to this type of integral is thinking of the vector field F~ as a field that describes the velocity of water at every point and S as a surface in the water. The value of
S F~ · dS~
describes how much water passes through the surface. As an example, let F~ = 〈y, x, z〉 and let the plane x + 2y + z = 1 be the surface inside the first octant with an upward pointing normal. A parametrization that can be given to this surface is ~r = 〈x, y, 1 − x − 2 y. Then,
∂~r ∂x
∂~r ∂y
∂~r ∂x
×
∂~r ∂y
ˆi ˆj ˆk 1 0 − 1 0 1 − 2
This is the normal vector we would have gotten from just taking the coefficients in front of the variables as usual from a plane, but that shortcut only works if the plane equation is expressed in the form ax+by+cz = 1.
F^ ~ (~r(x, y)) = 〈y, x, 1 − x − 2 y〉 ¨
S
F^ ~ · dS~ =
D
〈y, x, 1 − x − 2 y〉 · 〈 1 , 2 , 1 〉dA(x, y) =
D
1 + x − ydA(x, y)
The domain in this case is the area in the positive x-y plane where 0 ≤ 1 − x − 2 y or where y ≤ 1 − 2 x. Thus, the x bounds are [0, 1] and the y bounds are [0, 1 − 2 x].
F^ ~ (~r(x, y)) = 〈y, x, 1 − x − 2 y〉 ¨
S
F^ ~ · dS~ =
0
ˆ 1 − 2 x
0
(1 + x − y)dydx =
0
(1 − x^2 ) −
(1 − x)^2 dx =
0
(1 + x)^2 dx =
(1 + x)^3 3
] 1
0
