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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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Chabot Mathematics

§7.6 DoubleIntegrals

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]

7.5

- Any QUESTIONS About
- §7.5 → Lagrange Multipliers

- Any QUESTIONS About HomeWork
- §7.5 → HW-8

- Extend the Concept of a “Partial” Operation to Integration.
- Consider the mixed 2nd Partial
- ReWrite the Partial in Lebniz Notation:
- Now Let:

- 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

- Performing The AntiDerivation while not including the Constant find:
- Now Let:
- Then substitute, then multiply by ∂x

- Integrating find:
- After AntiDerivation:
- But ReCall:
- Back Substituting find
- By the Associative Property

- 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

- 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

- 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

- Substitute:
- Then
- Next Add Up all the Strips to find the Total Area, A

- This Relation
- Is simply a Riemann Sum
- Then in the Limit

- Find

- Find the area of the region contained between the parabolas

- SOLUTION: Use Double Integration

% 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]

- 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

- Find the volume under the Surface described by
- Over the Domain
- See Plot at Right

- SOLUTION: Find Vol by Double Integral

- Completing the Reduction

- 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:

- 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:

- 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

- 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

- 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

- Complete the Mathematical Reduction

- 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.

- Verify Constrained VUS by MuPad

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

Vnum = float(V)

- 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:

- 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.

- 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

- Proceed with the Double Integration

- Continue the Double Integration

- Complete the Double Integration
- The average weekly sales is 21,900 units over the time and pricing constraints given.

- Problems From §7.6
- P7.6-89 → Exposure to Disease

Volume byRiemannSum

Chabot Mathematics

Appendix

Do On

Wht/BlkBorad

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]

–

- 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