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14.2aE: Double Integrals Part 1 (Exercises) - Mathematics


In exercises 1 and 2, use the midpoint rule with (m = 4) and (n = 2) to estimate the volume of the solid bounded by the surface (z = f(x,y)), the vertical planes (x = 1), (x = 2), (y = 1), and (y = 2), and the horizontal plane (x = 0).

1) (f(x,y) = 4x + 2y + 8xy)

Answer:
(27)

2) (f(x,y) = 16x^2 + frac{y}{2})

In exercises 3 and 4, estimate the volume of the solid under the surface (z = f(x,y)) and above the rectangular region R by using a Riemann sum with (m = n = 2) and the sample points to be the lower left corners of the subrectangles of the partition.

3) (f(x,y) = sin x - cos y), (R = [0, pi] imes [0, pi])

Answer:
(0)

4) (f(x,y) = cos x + cos y), (R = [0, pi] imes [0, frac{pi}{2}])

5) Use the midpoint rule with (m = n = 2) to estimate (iint_R f(x,y) ,dA), where the values of the function f on (R = [8,10] imes [9,11]) are given in the following table.

(y)
(x)99.51010.511
89.856.755.6
8.59.44.585.43.4
98.74.665.53.4
9.56.764.55.46.7
106.86.45.55.76.8
Answer:
(21.3)

6) The values of the function (f) on the rectangle (R = [0,2] imes [7,9]) are given in the following table. Estimate the double integral (iint_R f(x,y),dA) by using a Riemann sum with (m = n = 2). Select the sample points to be the upper right corners of the subsquares of R.

(y_0 = 7)(y_1 = 8)(y_2 = 9)
(x_0 = 0)10.2210.219.85
(x_1 = 1)6.739.759.63
(x_2 = 2)5.627.838.21

7) The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.

  1. Estimate the volume of water in the swimming pool by using a Riemann sum with (m = n = 2). Select the sample points using the midpoint rule on (R = [0,4] imes [0,4]).
  2. Find the average depth of the swimming pool.
    (y)
    (x)01234
    011.522.53
    111.522.53
    211.51.52.53
    3111.522.5
    41111.52
Answer:
a. 28 ( ext{ft}^3)
b. 1.75 ft.

8) The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.

  1. Estimate the volume of the hole by using a Riemann sum with (m = n = 3) and the sample points to be the upper left corners of the subsquares of (R).
  2. Find the average depth of the hole.
    (y)
    (x)0123
    066.56.46
    16.577.56.5
    26.56.76.56
    366.555.6

9) The level curves (f(x,y) = k) of the function (f) are given in the following graph, where (k) is a constant.

  1. Apply the midpoint rule with (m = n = 2) to estimate the double integral (iint_R f(x,y),dA), where (R = [0.2,1] imes [0,0.8]).
  2. Estimate the average value of the function (f) on (R).

Answer:
a. 0.112
b. (f_{ave} ≃ 0.175); here (f(0.4,0.2) ≃ 0.1), (f(0.2,0.6) ≃− 0.2), (f(0.8,0.2) ≃ 0.6), and (f(0.8,0.6) ≃ 0.2)

10) The level curves (f(x,y) = k) of the function (f) are given in the following graph, where (k) is a constant.

  1. Apply the midpoint rule with (m = n = 2) to estimate the double integral (iint_R f(x,y),dA), where (R = [0.1,0.5] imes [0.1,0.5]).
  2. Estimate the average value of the function f on (R).

11) The solid lying under the surface (z = sqrt{4 - y^2}) and above the rectangular region( R = [0,2] imes [0,2]) is illustrated in the following graph. Evaluate the double integral (iint_Rf(x,y)), where (f(x,y) = sqrt{4 - y^2}) by finding the volume of the corresponding solid.

Answer:
(2pi)

12) The solid lying under the plane (z = y + 4) and above the rectangular region (R = [0,2] imes [0,4]) is illustrated in the following graph. Evaluate the double integral (iint_R f(x,y),dA), where (f(x,y) = y + 4), by finding the volume of the corresponding solid.

In the exercises 13 - 20, calculate the integrals by reversing the order of integration.

13) (displaystyle int_{-1}^1left(int_{-2}^2 (2x + 3y + 5),dx ight) space dy)

Answer:
(40)

14) (displaystyle int_0^2left(int_0^1 (x + 2e^y + 3),dx ight) space dy)

15) (displaystyle int_1^{27}left(int_1^2 (sqrt[3]{x} + sqrt[3]{y}),dy ight) space dx)

Answer:
(frac{81}{2} + 39sqrt[3]{2})

16) (displaystyle int_1^{16}left(int_1^8 (sqrt[4]{x} + 2sqrt[3]{y}),dy ight) space dx)

17) (displaystyle int_{ln 2}^{ln 3}left(int_0^1 e^{x+y},dy ight) space dx)

Answer:
(e - 1)

18) (displaystyle int_0^2left(int_0^1 3^{x+y},dy ight) space dx)

19) (displaystyle int_1^6left(int_2^9 frac{sqrt{y}}{y^2},dy ight) space dx)

Answer:
(15 - frac{10sqrt{2}}{9})

20) (displaystyle int_1^9 left(int_4^2 frac{sqrt{x}}{y^2},dy ight),dx)

In exercises 21 - 34, evaluate the iterated integrals by choosing the order of integration.

21) (displaystyle int_0^{pi} int_0^{pi/2} sin(2x)cos(3y),dx space dy)

Answer:
(0)

22) (displaystyle int_{pi/12}^{pi/8}int_{pi/4}^{pi/3} [cot x + an(2y)],dx space dy)

23) (displaystyle int_1^e int_1^e left[frac{1}{x}sin(ln x) + frac{1}{y}cos (ln y) ight] ,dx space dy)

Answer:
((e − 1)(1 + sin 1 − cos 1))

24) (displaystyle int_1^e int_1^e frac{sin(ln x)cos (ln y)}{xy} ,dx space dy)

25) (displaystyle int_1^2 int_1^2 left(frac{ln y}{x} + frac{x}{2y + 1} ight),dy space dx)

Answer:
(frac{3}{4}ln left(frac{5}{3} ight) + 2b space ln^2 2 - ln 2)

26) (displaystyle int_1^e int_1^2 x^2 ln(x),dy space dx)

27) (displaystyle int_1^{sqrt{3}} int_1^2 y space arctan left(frac{1}{x} ight) ,dy space dx)

Answer:
(frac{1}{8}[(2sqrt{3} - 3) pi + 6 space ln 2])

28) (displaystyle int_0^1 int_0^{1/2} (arcsin x + arcsin y),dy space dx)

29) (displaystyle int_0^1 int_0^2 xe^{x+4y},dy space dx)

Answer:
(frac{1}{4}e^4 (e^4 - 1))

30) (displaystyle int_1^2 int_0^1 xe^{x-y},dy space dx)

31) (displaystyle int_1^e int_1^e left(frac{ln y}{sqrt{y}} + frac{ln x}{sqrt{x}} ight),dy space dx)

Answer:
(4(e - 1)(2 - sqrt{e}))

32) (displaystyle int_1^e int_1^e left(frac{x space ln y}{sqrt{y}} + frac{y space ln x}{sqrt{x}} ight),dy space dx)

33) (displaystyle int_0^1 int_1^2 left(frac{x}{x^2 + y^2} ight),dy space dx)

Answer:
(-frac{pi}{4} + ln left(frac{5}{4} ight) - frac{1}{2} ln 2 + arctan 2)

34) (displaystyle int_0^1 int_1^2 frac{y}{x + y^2},dy space dx)

In exercises 35 - 38, find the average value of the function over the given rectangles.

35)(f(x,y) = −x +2y), (R = [0,1] imes [0,1])

Answer:
(frac{1}{2})

36) (f(x,y) = x^4 + 2y^3), (R = [1,2] imes [2,3])

37) (f(x,y) = sinh x + sinh y), (R = [0,1] imes [0,2])

Answer:
(frac{1}{2}(2 space cosh 1 + cosh 2 - 3)).

38) (f(x,y) = arctan(xy)), (R = [0,1] imes [0,1])

39) Let (f) and (g) be two continuous functions such that (0 leq m_1 leq f(x) leq M_1) for any (x ∈ [a,b]) and (0 leq m_2 leq g(y) leq M_2) for any( y ∈ [c,d]). Show that the following inequality is true:

[m_1m_2(b-a)(c-d) leq int_a^b int_c^d f(x) g(y),dy dx leq M_1M_2 (b-a)(c-d).]

In exercises 40 - 43, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

40) (frac{1}{e^2} leq iint_R e^{-x^2 - y^2} space dA leq 1), where (R = [0,1] imes [0,1])

41) (frac{pi^2}{144} leq iint_R sin x cos y space dA leq frac{pi^2}{48}), where (R = left[ frac{pi}{6}, frac{pi}{3} ight] imes left[ frac{pi}{6}, frac{pi}{3} ight])

42) (0 leq iint_R e^{-y}space cos x space dA leq frac{pi}{2}), where (R = left[0, frac{pi}{2} ight] imes left[0, frac{pi}{2} ight])

43) (0 leq iint_R (ln x)(ln y) ,dA leq (e - 1)^2), where (R = [1, e] imes [1, e] )

44) Let (f) and (g) be two continuous functions such that (0 leq m_1 leq f(x) leq M_1) for any (x ∈ [a,b]) and (0 leq m_2 leq g(y) leq M_2) for any (y ∈ [c,d]). Show that the following inequality is true:

[(m_1 + m_2) (b - a)(c - d) leq int_a^b int_c^d |f(x) + g(y)| space dy space dx leq (M_1 + M_2)(b - a)(c - d)]

In exercises 45 - 48, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

45) (frac{2}{e} leq iint_R (e^{-x^2} + e^{-y^2}) ,dA leq 2), where (R = [0,1] imes [0,1])

46) (frac{pi^2}{36}iint_R (sin x + cos y),dA leq frac{pi^2 sqrt{3}}{36}), where (R = [frac{pi}{6}, frac{pi}{3}] imes [frac{pi}{6}, frac{pi}{3}])

47) (frac{pi}{2}e^{-pi/2} leq iint_R (cos x + e^{-y}),dA leq pi), where (R = [0, frac{pi}{2}] imes [0, frac{pi}{2}])

48) (frac{1}{e} leq iint_R (e^{-y} - ln x) ,dA leq 2), where (R = [0, 1] imes [0, 1])

In exercises 49 - 50, the function (f) is given in terms of double integrals.

  1. Determine the explicit form of the function (f).
  2. Find the volume of the solid under the surface (z = f(x,y)) and above the region (R).
  3. Find the average value of the function (f) on (R).
  4. Use a computer algebra system (CAS) to plot (z = f(x,y)) and (z = f_{ave}) in the same system of coordinates.

49) [T] (f(x,y) = int_0^y int_0^x (xs + yt) ds space dt), where ((x,y) in R = [0,1] imes [0,1])

Answer:

a. (f(x,y) = frac{1}{2} xy (x^2 + y^2));
b. (V = int_0^1 int_0^1 f(x,y),dx space dy = frac{1}{8});
c. (f_{ave} = frac{1}{8});

d.

50) [T] (f(x,y) = int_0^x int_0^y [cos(s) + cos(t)] , dt space ds), where ((x,y) in R = [0,3] imes [0,3])

51) Show that if (f) and (g) are continuous on ([a,b]) and ([c,d]), respectively, then

(displaystyle int_a^b int_c^d |f(x) + g(y)| dy space dx = (d - c) int_a^b f(x),dx)

(displaystyle + int_a^b int_c^d g(y),dy space dx = (b - a) int_c^d g(y),dy + int_c^d int_a^b f(x),dx space dy).

52) Show that (displaystyle int_a^b int_c^d yf(x) + xg(y),dy space dx = frac{1}{2} (d^2 - c^2) left(int_a^b f(x),dx ight) + frac{1}{2} (b^2 - a^2) left(int_c^d g(y),dy ight)).

53) [T] Consider the function (f(x,y) = e^{-x^2-y^2}), where ((x,y) in R = [−1,1] imes [−1,1]).

  1. Use the midpoint rule with (m = n = 2,4,..., 10) to estimate the double integral (I = iint_R e^{-x^2 - y^2} dA). Round your answers to the nearest hundredths.
  2. For (m = n = 2), find the average value of f over the region R. Round your answer to the nearest hundredths.
  3. Use a CAS to graph in the same coordinate system the solid whose volume is given by (iint_R e^{-x^2-y^2} dA) and the plane (z = f_{ave}).
Answer:

a. For (m = n = 2), (I = 4e^{-0.5} approx 2.43)
b. (f_{ave} = e^{-0.5} simeq 0.61);

c.

54) [T] Consider the function (f(x,y) = sin (x^2) space cos (y^2)), where ((x,y in R = [−1,1] imes [−1,1]).

  1. Use the midpoint rule with (m = n = 2,4,..., 10) to estimate the double integral (I = iint_R sin (x^2) cos (y^2) space dA). Round your answers to the nearest hundredths.
  2. For (m = n = 2), find the average value of (f)over the region R. Round your answer to the nearest hundredths.
  3. Use a CAS to graph in the same coordinate system the solid whose volume is given by (iint_R sin(x^2) cos(y^2) space dA) and the plane (z = f_{ave}).

In exercises 55 - 56, the functions (f_n) are given, where (n geq 1) is a natural number.

  1. Find the volume of the solids (S_n) under the surfaces (z = f_n(x,y)) and above the region (R).
  2. Determine the limit of the volumes of the solids (S_n) as (n) increases without bound.

55) (f(x,y) = x^n + y^n + xy, space (x,y) in R = [0,1] imes [0,1])

Answer:
a. (frac{2}{n + 1} + frac{1}{4})
b. (frac{1}{4})

56) (f(x,y) = frac{1}{x^n} + frac{1}{y^n}, space (x,y) in R = [1,2] imes [1,2])

57) Show that the average value of a function (f) on a rectangular region (R = [a,b] imes [c,d]) is (f_{ave} approx frac{1}{mn} sum_{i=1}^m sum_{j=1}^n f(x_{ij}^*,y_{ij}^*)),where ((x_{ij}^*,y_{ij}^*)) are the sample points of the partition of (R), where (1 leq i leq m) and (1 leq j leq n).

58) Use the midpoint rule with (m = n) to show that the average value of a function (f) on a rectangular region (R = [a,b] imes [c,d]) is approximated by

[f_{ave} approx frac{1}{n^2} sum_{i,j =1}^n f left(frac{1}{2} (x_{i=1} + x_i), space frac{1}{2} (y_{j=1} + y_j) ight).]

59) An isotherm map is a chart connecting points having the same temperature at a given time for a given period of time. Use the preceding exercise and apply the midpoint rule with (m = n = 2) to find the average temperature over the region given in the following figure.

Answer:
(56.5^{circ}) F; here (f(x_1^*,y_1^*) = 71, space f(x_2^*, y_1^*) = 72, space f(x_2^*,y_1^*) = 40, space f(x_2^*,y_2^*) = 43), where (x_i^*) and (y_j^*) are the midpoints of the subintervals of the partitions of ([a,b]) and ([c,d]), respectively.

Problem Set 7

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14.2aE: Double Integrals Part 1 (Exercises) - Mathematics

14. Use a double integral to determine the volume of the region bounded by (z = 6 - 5) and the planes (y = 2x), (y = 2),(x = 0) and the (xy)-plane.

Show All Steps Hide All Steps

Let’s first get a sketch of the solid that we’re working with. If you are not good at visualizing these types of solids in your head these graphs can be invaluable in helping to get the integral set up.

The surface is sketched with a traditional set of axes and well as a “box frame” set of axes. Sometimes it is easier to see what is going on with the surface when both sketches are present.

The upper surface (the orange surface) is the graph of (z = 6 - 5). The blue plane is the graph of (y = 2) which is nothing more than the plane parallel to the (xz)-plane at (y = 2). The red plane is the graph of (y = 2x) and this is simply the plane that is perpendicular to the (xy)-plane and goes through the line (y = 2x) in the (xy)-plane. The surface given by (x = 0) is simply the (yz)-plane (i.e the back of the solid) and is not shown and the (xy)-plane is the bottom of the surface and again is not shown in the sketch.

In this section the only method that we have for determining the volume of a solid is to find the volume under a surface. In this case it is hopefully clear that we are looking for the surface that is under (z = 6 - 5) and is above the region (D) in the (xy)-plane defined by where the other three planes intersect it. In other words, the region (D) is the regoin in the (xy)-plane that is bounded by (y = 2), (y = 2x) and (x = 0).

The integral for the volume is then,

where (D) is sketched below.

This integral can be integrated in any order so let’s integrate with respect to (y) first to avoid fractions in the limits (which we’d get with one if we integrated with respect to (x) first). Here are the limits for our integral.

[egin0 le x le 1 2x le y le 2end]

The integral for the volume is then,

Now all we need to do is evaluate the integral. Here is the (y) integration.

Note that in doing the (y) integration we acknowledged that the whole integrand contained no (y)’s and so could be considered a constant and so would just be multiplied by (y). We could also have done each term individually but sometimes it is just as easy or even easier to do what we’ve done here. Of course, we then had to multiply out the integrand for the next step but it wasn’t too bad.


14.2aE: Double Integrals Part 1 (Exercises) - Mathematics

13. Use a double integral to determine the volume of the region that is between the (xy)‑plane and(fleft( ight) = 2 + cos left( <> ight)) and is above the triangle with vertices (left( <0,0> ight)), (left( <6,0> ight)) and (left( <6,2> ight)).

Show All Steps Hide All Steps

Let’s first get a sketch of the function and the triangle that lies under it.

The surface is sketched with a traditional set of axes and well as a “box frame” set of axes. Sometimes it is easier to see what is going on with the surface when both sketches are present.

The greenish triangle underneath the surface is the triangle referenced in the problem statement.

Now, the volume we are after is given by the following integral,

where (D) is the triangle referenced in the problem statement.

So, in order to evaluate the integral we’ll need a sketch of (D) so we can determine an order of integration as well as limits for the integrals.

The region (D) can easily be described for either order of integration. However, it should be pretty clear that the integral can’t be integrated with respect to (x) first and so we’ll need to integrate with respect to (y) first.

Here are the limits for the integral with this order.

[egin 0 le x le 6 0 le y le displaystyle frac<1><3>xend]

The integral for the volume is then,

Now all we need to do is evaluate the integral. Here is the (y) integration.

Finally, the (x) integration and hence the volume is,

[V = left. <3> + frac<1><6>sin left( <> ight),> ight)> ight|_0^6 = equire box[2pt,border:1px solid black]<<12 + frac<1><6>sin left( <36> ight) = 11.8347>>]

Don’t forget to have your calculator set to radians if you are converting to decimals!


What rules do I have to follow?

There are rules to keep in mind. First, the function . f(x). must be continuous during the the interval in question. This means that between . a. and . b. the graph of the function cannot have any breaks (where it does not exist), holes (where it does not exist at a single point) or jumps (where the function exists at two separate . y. -values for a single . x. -value). Second, the interval must be closed, which means that both limits must be constants (real numbers only, no infinity allowed).

When it comes to solving a problem using Part 1 of the Fundamental Theorem, we can use the chart below to help us figure out how to do it.


Math H53: Honors Multivariable Calculus

Instructor Office hours: Regular office hours: 4:30-5:30 on Tuesday and 2-3:30 on Thursday. Check Bcourses "Syllabus" for Zoom ID. Always feel free to send me questions or ask for alternative office hours.

Final exam: Check UC Berkeley final exam schedule

Prerequisites: Math 1B or equivalent.

Text: The primary texts for this course are Vector Calculus by Michael Corral ([Co]) and Notes on Multivariable Calculus by Cain and Herod ([CH]). Students should feel free to consult other books for additional exercises and/or alternative presentations of the material. Wikipedia also has lots of great articles on the topics at hand. Students are expected to read the relevant sections of the notes, as the lectures are meant to complement the notes, not replace it, and we have a lot of material to cover.

Grading: Your homework grade (hw) will be the average of all homeworks, with the lowest dropped. Your exam grade (exams) will be computed based on the maximum of the following three schemes: (0.2)MT1 + (0.2)MT2 + (0.4)F (0.2)MT1 + (0.6)F (0.2)MT2 + (0.6)F. Finally, your total grade will be calculated as the maximum of: (0.2)hw + (0.8)exams, (0.3)hw + (0.7)exams.

Website: For now, the only website is this page, http://math.berkeley.edu/

dcorwin/mathh53s21.html. I will use bcourses for solutions and other non-public information, such as book excerpts, exams, and my phone number.

  • Homework will be assigned regularly (see the syllabus) and due at 11pm on Gradescope. I grant extensions in reasonable circumstances, but you must talk to me as early as possible. The longer you wait, the less flexible I will be.
  • You may work together to figure out homework problems, but you must write up your solutions in your own words in order to receive credit. In particular, please do not copy answers from the internet or solution manuals. Since a major purpose of the homework is to prepare you for the exams, I encourage you to give each problem an honest shot by yourself (say, at least thirty minutes) before discussing it with others. Another useful practice is if you're stuck on a problem, come to office hours and ask for a hint. The more you figure out on your own, the better your understanding of the material, and the better you'll do both on the exams and in your future endeavours that might require abstract algebra.
  • You may cite any results from the notes, unless otherwise stated.
  • The usual expectations and procedures for academic integrity at UC Berkeley apply. Cheating on an exam will result in a failing grade and will be reported to the University Office of Student Conduction. Please don't put me through this.
  • Please let me know sooner rather than later if you need any accommodations related to the Disabled Student Program (DSP). I am more than happy to make arrangements, but it really helps if you tell me earlier rather than later.
  • Per university guidelines, it is your responsibility to notify the instructor in writing by the end of the second week of classes (January 31) of any scheduling conflicts due to religious observance or extracurricular activities, and to propose a resolution for those conflicts.

Additional resources (will be on Bcourses when needed):

  • Calculus: Early Transcendentals by James Stewart, denoted [S]
  • Calculus Volume II by Tom Apostol, denoted [A]
  • div grad curl and all that by H. M. Schey, denoted [dgcaat]
  • Line Integrals and Green's Theorem by Jeremy Orloff, denoted [O]

Course Overview: Outlined below is the rough course schedule. Depending on how the class progresses it may be subject to minor changes over the course of the semester.


Practice Final Exam

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Test: Double And Triple Integrals - 2

The value of dxdy is


[change the order]

The volume of an object expressed in spherical coordinates is given by

Q. The value of the integral is

Consider the shaded triangular region P shown in the figure. What is xy dxdy?

The intercept form of equation is


where, a is x intercept and b is y intercept.
So Given, equation is
implies x+2y=2
implies x=2(1 - y)
Limit of x is form 0 to 2(1 - y ) and limit of y is from 0 to 1.
Therefore,

= 1/6

Changing the order of integration in the double integral leads to, then the value of q is

Given, I =
Equation of the line is x = Ay Starting val ue of x is zero and final value of x is 8 but we want the result of first integration in terms of y, so that we can integrate the result with respect to y to get the final result. So, limits are from 0 to 4y.
And integration limit for y is from 0 to 2.

So,

Hence, r=0, s=2, p=0, q=4y

dxdy over the positive quadrant of the circle x 2 + y 2 = 1 is given by

Let
Changing to polar coordinates let x=r cos&theta, y=r sin&theta


Therefore,
= 1/14

The volume of the cylinder x 2 + y 2 = a 2 bounded below by z = 0 and bounded above by z = h is given by

The equation of the cylinder is
x 2 + y 2 = a 2
The equation of surface CDE is z = h
So, the required volume is


Let x = a sin &theta
implies dx = a cos&theta d&theta
So, volume V

dxdydz is equal to


14.2aE: Double Integrals Part 1 (Exercises) - Mathematics

Lecture Description

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Course Index

  1. Summation Notation
  2. The Definite Integral: Understanding the Definition
  3. Approximating a Definite Integral Using Rectangles
  4. Trapezoidal Rule to Approximate a Definite Integral
  5. Simpson's Rule to Approximate a Definite Integral
  6. Simpson's Rule and Error Bounds
  7. Calculating a Definite Integral Using Riemann Sums (Part 1)
  8. Calculating a Definite Integral Using Riemann Sums (Part 2)
  9. Basic Integration Formulas
  10. Basic Antiderivate Examples: Indefinite Integral
  11. More Basic Integration Problems
  12. Basic Definite Integral Example
  13. Indefinite Integral: U-substitution
  14. Definite Integral: U-substitution
  15. More Integration Using U-Substitution (Part 1)
  16. More Integration Using U-Substitution (Part 2)
  17. Integration Involving Inverse Trigonometric Functions
  18. Integration By Parts: Indefinite Integral
  19. Integration By Parts: Definite Integral
  20. Indefinite/Definite Integral Examples
  21. Integration By Parts: Using IBP's Twice
  22. Integration By Parts: A "Loopy" Example
  23. Trigonometric Integrals: Part 1 of 6
  24. Trigonometric Integrals: Part 2 of 6
  25. Trigonometric Integrals: Part 3 of 6
  26. Trigonometric Integrals: Part 4 of 6
  27. Trigonometric Integrals: Part 5 of 6
  28. Trigonometric Integrals: Part 6 of 6
  29. Trigonometric Substitution (Part 1)
  30. Trigonometric Substitution (Part 2)
  31. Trigonometric Substitution (Part 3)
  32. Trigonometric Substitution (Part 4)
  33. Trigonometric Substitution (Part 5)
  34. Partial Fractions: Decomposing a Rational Function
  35. Partial Fractions: Coefficients of a Partial Fraction Decomposition
  36. Partial Fractions: Problem
  37. Partial Fractions: Problem Using a Rationalizing Substitution
  38. Calculating Double Integrals Over Rectangular Regions
  39. Calculating Double Integrals Over General Regions
  40. Reversing the Order of Integration (Part 1)
  41. Reversing the Order of Integration (Part 2)
  42. Finding Areas in Polar Coordinates
  43. Double Integral Using Polar Coordinates (Part 1)
  44. Double Integral Using Polar Coordinates (Part 2)
  45. Double Integral Using Polar Coordinates (Part 3)
  46. Triple Integrals
  47. Triple Integrals in Spherical Coordinates
  48. Line Integrals
  49. Solving First Order Linear Differential Equations
  50. Finding Centroids/Centers of Mass (Part 1)
  51. Finding Centroids/Centers of Mass (Part 2)
  52. Improper Integrals: Introduction
  53. Improper Integrals: Using L'Hospitals Rule
  54. Improper Integrals: Infinity in the Upper and Lower Limit
  55. Improper Integrals: Infinite Discontinuity at an Endpoint
  56. Improper Integrals: Infinite Discontinuity in the Middle of the Interval
  57. Volumes of Revolution: Disk/Washer Method & Rotating Regions About a Horizontal Line
  58. Volumes of Revolution: Disk/Washer Method & Rotating Regions About a Vertical Line
  59. Volumes of Revolution: Disk/Washer Method (cont.)
  60. Work Problems: Finding the Work To Empty a Tank Full of Water

Course Description


In this course, Calculus Instructor Patrick gives 60 video lectures on Integral Calculus. Some of the topics covered are: Indefinite Integrals, Definite Integrals, Trigonometric Integrals, Trigonometric Substitution, Partial Fractions, Double Integrals, Triple Integrals, Polar Coordinates, Spherical Coordinates, Line Integrals, Centroids/Centers of Mass, Improper Integrals, Volumes of Revolution, Work, and many more.


14.2aE: Double Integrals Part 1 (Exercises) - Mathematics

In this section we are going to investigate the relationship between certain kinds of line integrals (on closed paths) and double integrals.

Let’s start off with a simple (recall that this means that it doesn’t cross itself) closed curve (C) and let (D) be the region enclosed by the curve. Here is a sketch of such a curve and region.

First, notice that because the curve is simple and closed there are no holes in the region (D). Also notice that a direction has been put on the curve. We will use the convention here that the curve (C) has a positive orientation if it is traced out in a counter-clockwise direction. Another way to think of a positive orientation (that will cover much more general curves as well see later) is that as we traverse the path following the positive orientation the region (D) must always be on the left.

Given curves/regions such as this we have the following theorem.

Green’s Theorem

Let (C) be a positively oriented, piecewise smooth, simple, closed curve and let (D) be the region enclosed by the curve. If (P) and (Q) have continuous first order partial derivatives on (D) then,

Before working some examples there are some alternate notations that we need to acknowledge. When working with a line integral in which the path satisfies the condition of Green’s Theorem we will often denote the line integral as,

Both of these notations do assume that (C) satisfies the conditions of Green’s Theorem so be careful in using them.

Also, sometimes the curve (C) is not thought of as a separate curve but instead as the boundary of some region (D) and in these cases you may see (C) denoted as (partial D).

Let’s work a couple of examples.

Let’s first sketch (C) and (D) for this case to make sure that the conditions of Green’s Theorem are met for (C) and will need the sketch of (D) to evaluate the double integral.

So, the curve does satisfy the conditions of Green’s Theorem and we can see that the following inequalities will define the region enclosed.

[0 le x le 1hspace<0.5in>0 le y le 2x]

We can identify (P) and (Q) from the line integral. Here they are.

So, using Green’s Theorem the line integral becomes,

Okay, a circle will satisfy the conditions of Green’s Theorem since it is closed and simple and so there really isn’t a reason to sketch it.

Let’s first identify (P) and (Q) from the line integral.

Be careful with the minus sign on (Q)!

Now, using Green’s theorem on the line integral gives,

where (D) is a disk of radius 2 centered at the origin.

Since (D) is a disk it seems like the best way to do this integral is to use polar coordinates. Here is the evaluation of the integral.

So, Green’s theorem, as stated, will not work on regions that have holes in them. However, many regions do have holes in them. So, let’s see how we can deal with those kinds of regions.

Let’s start with the following region. Even though this region doesn’t have any holes in it the arguments that we’re going to go through will be similar to those that we’d need for regions with holes in them, except it will be a little easier to deal with and write down.

The region (D) will be ( cup ) and recall that the symbol ( cup ) is called the union and means that (D) consists of both (>) and (). The boundary of (>) is ( cup ) while the boundary of () is ( cup left( < - > ight)) and notice that both of these boundaries are positively oriented. As we traverse each boundary the corresponding region is always on the left. Finally, also note that we can think of the whole boundary, (C), as,

[C = left( <cup > ight) cup left( <cup left( < - > ight)> ight) = cup ]

since both () and ( - ) will “cancel” each other out.

Now, let’s start with the following double integral and use a basic property of double integrals to break it up.

Next, use Green’s theorem on each of these and again use the fact that we can break up line integrals into separate line integrals for each portion of the boundary.

Next, we’ll use the fact that,

Recall that changing the orientation of a curve with line integrals with respect to (x) and/or (y) will simply change the sign on the integral. Using this fact we get,

Finally, put the line integrals back together and we get,

So, what did we learn from this? If you think about it this was just a lot of work and all we got out of it was the result from Green’s Theorem which we already knew to be true. What this exercise has shown us is that if we break a region up as we did above then the portion of the line integral on the pieces of the curve that are in the middle of the region (each of which are in the opposite direction) will cancel out. This idea will help us in dealing with regions that have holes in them.

To see this let’s look at a ring.

Notice that both of the curves are oriented positively since the region (D) is on the left side as we traverse the curve in the indicated direction. Note as well that the curve () seems to violate the original definition of positive orientation. We originally said that a curve had a positive orientation if it was traversed in a counter-clockwise direction. However, this was only for regions that do not have holes. For the boundary of the hole this definition won’t work and we need to resort to the second definition that we gave above.

Now, since this region has a hole in it we will apparently not be able to use Green’s Theorem on any line integral with the curve (C = cup ). However, if we cut the disk in half and rename all the various portions of the curves we get the following sketch.

The boundary of the upper portion ((>))of the disk is ( cup cup cup ) and the boundary on the lower portion (())of the disk is ( cup cup left( < - > ight) cup left( < - > ight)). Also notice that we can use Green’s Theorem on each of these new regions since they don’t have any holes in them. This means that we can do the following,

Now, we can break up the line integrals into line integrals on each piece of the boundary. Also recall from the work above that boundaries that have the same curve, but opposite direction will cancel. Doing this gives,

But at this point we can add the line integrals back up as follows,

The end result of all of this is that we could have just used Green’s Theorem on the disk from the start even though there is a hole in it. This will be true in general for regions that have holes in them.

Let’s take a look at an example.

Notice that this is the same line integral as we looked at in the second example and only the curve has changed. In this case the region (D) will now be the region between these two circles and that will only change the limits in the double integral so we’ll not put in some of the details here.

Here is the work for this integral.

We will close out this section with an interesting application of Green’s Theorem. Recall that we can determine the area of a region (D) with the following double integral.

Let’s think of this double integral as the result of using Green’s Theorem. In other words, let’s assume that

and see if we can get some functions (P) and (Q) that will satisfy this.

There are many functions that will satisfy this. Here are some of the more common functions.

Then, if we use Green’s Theorem in reverse we see that the area of the region (D) can also be computed by evaluating any of the following line integrals.

where (C) is the boundary of the region (D).

Let’s take a quick look at an example of this.

We can use either of the integrals above, but the third one is probably the easiest. So,

where (C) is the circle of radius (a). So, to do this we’ll need a parameterization of (C). This is,

[x = acos thspace<0.25in>y = asin thspace<0.25in>0 le t le 2pi ]