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

7.5

Review §

  • Any QUESTIONS About

    • §7.5 → Lagrange Multipliers

  • Any QUESTIONS About HomeWork

    • §7.5 → HW-8


Partial deriv partial integ

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

Partial-Deriv↔Partial-Integ

  • 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

Partial-Deriv↔Partial-Integ

  • Performing The AntiDerivation while not including the Constant find:

  • Now Let:

  • Then substitute, then multiply by ∂x


Partial deriv partial integ3

Partial-Deriv↔Partial-Integ

  • Integrating find:

  • After AntiDerivation:

  • But ReCall:

  • Back Substituting find

  • By the Associative Property


Partial deriv partial integ4

Partial-Deriv↔Partial-Integ

  • 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')

pause

%

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]

MATLAB code


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

Appendix

Do On

Wht/BlkBorad

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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