Chabot Mathematics
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Chabot Mathematics. §7.6 Double Integrals. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] 7.5. Review §. Any QUESTIONS About §7.5 → Lagrange Multipliers Any QUESTIONS About HomeWork §7.5 → HW-8. Partial- Deriv↔Partial - Integ.

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Bruce mayer pe licensed electrical mechanical engineer bmayer chabotcollege edu

Chabot Mathematics

§7.6 DoubleIntegrals

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]



Review §

  • Any QUESTIONS About

    • §7.5 → Lagrange Multipliers

  • Any QUESTIONS About HomeWork

    • §7.5 → HW-8

Partial deriv partial integ


  • Extend the Concept of a “Partial” Operation to Integration.

  • Consider the mixed 2nd Partial

  • ReWrite the Partial in Lebniz Notation:

  • Now Let:

Partial deriv partial integ1


  • Thus with ∂z/∂y = u:

  • Now Multiply both sides by ∂x and Integrate

  • Integration with respect to the Partial Differential, ∂x, implies that y is held CONSTANT during the AntiDerivation

Partial deriv partial integ2


  • Performing The AntiDerivation while not including the Constant find:

  • Now Let:

  • Then substitute, then multiply by ∂x

Partial deriv partial integ3


  • Integrating find:

  • After AntiDerivation:

  • But ReCall:

  • Back Substituting find

  • By the Associative Property

Partial deriv partial integ4


  • Also ReCallClairaut’s Theorem:

  • This Order-Independence also Applies to Partial Integrals Which leads to the Final Statement of the Double Integral

    • C is the Constant of Integration

Area between curves

Area BETWEEN Curves

  • As before Find Area by adding Vertical Strips.

  • In this case for the Strip Shown:

    • Width = Δx

    • Height = ytop − ybot or

  • Then the strip area

Area between curves1

Area BETWEEN Curves

  • Note that for every CONSTANT xk, that y runs:

  • Now divide the Hgt into pieces Δy high

  • So then ΔA:

  • Then Astrip is simply the sum of all the small boxes

Area between curves2

Area BETWEEN Curves

  • Substitute:

  • Then

  • Next Add Up all the Strips to find the Total Area, A

Area between curves3

Area BETWEEN Curves

  • This Relation

  • Is simply a Riemann Sum

  • Then in the Limit

  • Find

Example area between curves

Example  Area Between Curves

  • Find the area of the region contained between the parabolas

Example area between curves1

Example  Area Between Curves

  • SOLUTION: Use Double Integration

Matlab code

% Bruce Mayer, PE

% MTH-16 • 22Feb14

% Area_Between_fcn_Graph_BlueGreen_BkGnd_Template_1306.m

% Ref: E. B. Magrab, S. Azarm, B. Balachandran, J. H. Duncan, K. E.

% Herhold, G. C. Gregory, "An Engineer's Guide to MATLAB", ISBN

% 978-0-13-199110-1, Pearson Higher Ed, 2011, pp294-295


clc; clear; clf; % Clear Figure Window


% The Function

xmin = -2; xmax = 2;

ymin = 0; ymax = 10;

x = linspace(xmin,xmax,500);

y1 = -x.^2 + 9;

y2 = x.^2 + 1;


plot(x,y1,'--', x,y2,'m','LineWidth', 5), axis([0 xmaxyminymax]),...

grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x) = -x^2 + 9 & x^2 + 1'),...

title(['\fontsize{16}MTH16 • Bruce Mayer, PE',]),...

legend('-x^2 + 9','x^2 + 1') %

display('Showing 2Fcn Plot; hit ANY KEY to Continue')

% "hold" = Retain current graph when adding new graphs

hold on

disp('Hit ANY KEY to show Fill')



xn = linspace(xmin, xmax, 500);

fill([xn,fliplr(xn)],[-xn.^2 + 9, fliplr(x.^2 + 1)],[.49 1 .63]), grid on

% alternate RGB triple: [.78 .4 .01]


Volume under a surface

Volume Under a Surface

  • Use Long Strips to find the Area under a Curve (AuC) by Riemann Summation

  • Use Long Boxesto find the VolumeUnder a Surface(VUS) by Riemann Summation

Vus by double integral

VUS by Double Integral

Example vol under surf

Example  Vol under Surf

  • Find the volume under the Surface described by

  • Over the Domain

  • See Plot at Right

Example vol under surf1

Example  Vol under Surf

  • SOLUTION: Find Vol by Double Integral

Example vol under surf2

Example  Vol under Surf

  • Completing the Reduction

Vus for nonrectangular region

VUS for NonRectangular Region

  • If the Base Region, R, for a Volume Integral is NonRectangular and can be described by InEqualities

  • Then by adding up all the long boxes

  • If R described by

  • Then:

Example nonrectangular vus

Example  NonRectangular VUS

  • Find the volume under the surface

  • Over the Region Bounded by

  • SOLUTION: First, visualize the limits of integration using a graph of the Base PlaneIntegration Region:

Example nonrectangular vus1

Example  NonRectangular VUS

  • The outer limits of integration need to be numerical (no variables), but the Inner limits can contain expressions in x (or y) as in the definition.

  • In this case, choose the inner limits to be with respect to y, then find the limits of the y values in terms of x

Example nonrectangular vus2

Example  NonRectangular VUS

  • Each y-value in the region is restricted by the constant height 0 at the top, at the bottom by the Line:

  • Thus the Double Integral (so far):

  • In Simplified Notation

Example nonrectangular vus3

Example  NonRectangular VUS

  • Now, Because the outer integral needs to contain only numbers values, consider only the absolute limits on the x-values in the figure:

    • a MINimum of 0 and a MAXimum of 5

  • Thus the Completed Double Integral

Example nonrectangular vus4

Example  NonRectangular VUS

  • Complete the Mathematical Reduction

Example nonrectangular vus5

Example  NonRectangular VUS

  • Complete the Mathematical Reduction

  • The volume contained underneath the surface and over the triangular region in the XY plane is approximately 69.8 cubic units.

Example nonrectangular vus6

Example  NonRectangular VUS

  • Verify Constrained VUS by MuPad

V := int((int(x+E^(x+2*y), y=x-5..0)), x=0..5)

Vnum = float(V)

Average value

Average Value

  • Recall from Section 5.4 that the average value of a function f of one variable defined on an interval [a, b] is

  • Similarly, the average value of a function f of two variables defined on a rectangle R to be:

Example average sales

Example  Average Sales

  • Weekly sales of a new product depend on its price p in dollars per item and time t in weeks after its release, can be Modeled by:

    • Where S is measured in k-units sold

  • Find the average weekly sales of the product during the first six weeks after release and when the product’s price varies between 15 – t and 25 – t.

Example average sales1

Example  Average Sales

  • SOLUTION: first find the area of the region of integration as shown below

  • Note that The price Constraints producea Parallelogram-likeRegion

  • By the ParallelogramArea Formula

Example average sales2

Example  Average Sales

  • Proceed with the Double Integration

Example average sales3

Example  Average Sales

  • Continue the Double Integration

Example average sales4

Example  Average Sales

  • Complete the Double Integration

  • The average weekly sales is 21,900 units over the time and pricing constraints given.

Whiteboard work

WhiteBoard Work

  • Problems From §7.6

    • P7.6-89 → Exposure to Disease

All done for today

All Done for Today

Volume byRiemannSum

Bruce mayer pe licensed electrical mechanical engineer bmayer chabotcollege edu

Chabot Mathematics


Do On


Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]

7 3 learning goals

§7.3 Learning Goals

  • Define and compute double integrals over rectangular and NONrectangular regions in the xy plane

  • Use double integrals in problems involving

    • Area

    • Volume,

    • Average Value

    • Population Density

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