Lecture 45: Calculation of Double Integrals

45.1 Calculation of Double Integrals

Core Idea: Transform a double integral into the calculation of a definite integral in one variable, simplifying the calculation method;

Method Selection

Choose methods based on the following:
1. integrand;
2. Scoring area;

Suitable for polar coordinates

(1) Integrands suitable for polar coordinates calculations
Formula:
$$f(x^2+y^2),f(\sqrt{x^2+y^2}),f(\frac yx),f(\frac xy)$$
Cause:
$\sqrt{x^2+y^2}$ is more complex in the Cartesian coordinate system, but represents ρ in polar coordinates
$\frac yx$ In polar coordinates, it represents an angle
(2) Integration fields suitable for polar coordinates
$$x^{2}+y^{2}\leq R^{2};\quad\quad\quad r^{2}\leq x^{2}+y^{2}\leq R^{2};\quad\quad\quad\\x^{2}+y^{2}\leq2ax;\quad\quad\quad x^{2}+y^{2}\leq2by;$$
Note:
When the center is not at the origin, $x-x_0$ can be set to $\rho\sin\theta$, and similarly, $y-y_0$
If (1) and (2) have a conflict, (1) takes precedence

45.2 Calculation Using Cartesian Coordinate System

#####Theorem: Doubleintegrationcalculationbasedonrectangularcoordinatesystem

description:

(1) First Y Later X: $$\int\int_D{f(x,y)d\sigma = \int_{a}^{b}[\int_{y_{1}(x)}^{y_{2}(x)}f(x,y)dy]dx}$$

Area: $\begin{aligned}\varphi_1(x)&\leq y\leq\varphi_2(x)\cdot\\a&\leq x\leq b\end{aligned}$

$\varphi_1(x)\leq y\leq\varphi_2(x)$ represents the range of x values, which is the function of x with respect to y, and $\varphi_1(x)$ is the actual value of y equal to.

(2) First X Later Y: $$\iint_Df(x,y)\mathrm{d}\sigma=\int_c^ddy\int_{\psi_1(y)}^{\psi_2(y)}f(x,y)dx$$

Area: $\begin{aligned}\Psi(y)&\leq X\leq\Psi_2(y)\\c&\leq y\leq d\end{aligned}$

Method: Set the points limit;

When Y comes first, then X:
dy upper and lower integral limits ->Draw a ray from bottom to top, with the lower end of the ray being the dy lower limit of integration and the upper end of the ray being the upper limit of the integral of dy;
dx upper and lower limits of integration -> Observe the graph to see the range of x values;
When X comes first, then Y:
The upper and lower limits of dx integration ->Draw a ray from left to right, with the left end of the ray as the dx lower limit and the upper end as the dx lower limit;
dy upper and lower limits of integration -> Observe the image and see therangeof y;

Explanation

Concept:
Both can be used; choose whichever is more convenient for calculation;
First Y X:
Concept:
$\int_{a}^{b}[\int_{y_{1}(x)}^{y_{2}(x)}f(x,y)dy]dx$ or $\int_a^b\mathbf{d}x\int_{y_1(x)}^{y_2(x)}f(x,y)\operatorname{d}y.$
Reasoning:
1. First, find the cross-sectional area: $S(x)=\int_{y_1(x)}^{y_2(x)}f(x,y)\mathsf{d}y.$
2. Then, for the curved edge body area, solve: $V=\int_a^bS(x)\operatorname{d}x$
3. Merging the two: $\int_{a}^{b}[\int_{y_{1}(x)}^{y_{2}(x)}f(x,y)dy]dx$
Illustration:
2D:
3D:
First X Y:
Premise: $(\sigma)=\{(x,y)\mid x_1(y)\leq x\leq x_2(y),c\leq y\leq d\}$
Illustration:
Convert to two single-invariant integrals with x before y;
Formula: $\begin{aligned}\iint_{(\sigma)}f(x,y)\operatorname{d}\sigma&=\int_c^d[\int_{x_1(y)}^{x_2(y)}f(x,y)dx]\operatorname{d}y\\\\&=\int_c^d\operatorname{d}\left.y\right]_{x_1(y)}^{x_2(y)}f(x,y)\operatorname{d}x.\end{aligned}$
Non-X and non-Y regions
Illustration
Concept:
A complex double integral of a non-X and non-Y figure;
Method:
Canbe converted by partitioning to: sum of multiple X-types and multiple Y-types;

45.3 Calculation Using Polar Coordinates

#####Theorem: Doubleintegralcalculationbasedonpolarcoordinates

description: $$\text{ First }\rho\text{ Afterwards }\theta\quad\iint_Df (x, y)\mathrm{d}\sigma=\int_\alpha^\beta d\theta\int_{\varphi_1 (\theta)}^{\varphi_2 (\theta)}f (\rho\cos\theta,\rho\sin\theta)\rho d\rho$$

Area: $\begin{aligned}\varphi_1(0)&\leq p\leq\varphi_2(0)\\\alpha&\leq\theta\leq\beta.\end{aligned}$

Explanation

Note:
When integrating with $\rho$, $\theta$ in the expression can be regarded as a constant. Similarly, when integrating with $\theta$;
Supplement:
The contents of a function can be divided into two integrals for calculation;
$\iint_D\frac{x\sin (\pi\sqrt{x^2+y^2})}{x+y}dxdy=\int_0^{\frac\pi 2}\frac{\cos\theta}{\cos\theta+\sin\theta}d\theta\cdot\int_1^2\rho\sin (\pi\rho) d\rho$
$\Delta\sigma=\frac12[(\rho+\Delta\rho)^2\Delta\theta-\rho^2\Delta\theta]$ = $\rho\Delta\rho\Delta\theta+\frac12(\Lambda\rho)^2\Delta\theta.$

45.4 Calculation Using Symmetry and Parity

45.4.1 Parity

Concepts: When Y is symmetrical, look at X; when about X, look at Y;

Property One: If the integral D relation Y axial, then the function is evenly X:

If the function with respect to X is even, it doubles; If the function with respect to X is odd, it is 0;
$$\iint\limits_{D}f(x,y)d\sigma=\begin{cases}2\iint\limits_{D_{x\geq0}}f(x,y)\mathrm{d}\sigma;&f(-x,y)=f(x,y)\\0;&f(-x,y)=-f(x,y)\end{cases}$$

Property Two: If the integral D relation X axiosymmetric, then the function has parity with respect to Y:

$$\iint\limits_{D}f(x,y)d\sigma=\begin{cases}2\iint\limits_{D_{y_{z_0}}}f(x,y)\mathrm{d}\sigma&f(x,-y)=f(x,y)\\0&f(x,-y)=-f(x,y)\end{cases}$$

45.4.2 Symmetry

#####Theorem: Variablesymmetryofdoubleintegrals

description: $$\text{ If }D\text{ About }y=x\text{ Symmetric, then }\quad\iint_Df(x,y)\mathrm{d}\sigma=\iint_Df(y,x)\mathrm{d}\sigma $$

Explanation

The integration field is the same, where the independent variable is swapped;
Because the point of $(x,y)$ symmetric about y=x is $(y,x)$

Inference: A more generalized conclusion

$$\int\int_{D(x,y)} f(x,y)\mathrm{d}x\mathrm{d}y=\int\int_{D(u,v)} f(u,v)\mathrm{d}u\mathrm{d}v=\int\int_{D(y,x)} f(y,x)\mathrm{d}y\mathrm{d}x$$

45.5 Frequently Tested Question Types

Question Type: Cumulativepointsexchangeorderorcalculation

PART 1: Problem-solving methods

Problem-solving steps: Example $\text{ Exchange cumulative points }\int_0^1dx\int_{x^2}^{2-x}f(x,y)dy\text{ The order }$

Step 1: Draw the domain
Step 2: After drawing the domain, define the domain in another order
Swap order and reline:
$$\int_0^1dy\int_0^{\sqrt{y}}f(x,y)dx+\int_1^2dy\int_0^{2-y}f(x,y)dx$$
Additional note: If it's hard to calculate after swapping order, consider using polar coordinates;

Question Type: Cumulative integrals in polar coordinates

Example:
$$\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{0}^{\cos\theta}f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho $$
Steps:
(1) Drawing area;
(2) Draw the limit, convert the polar coordinate equation into a rectangular coordinate equation;

Question Type: Calculate cumulative points

PART 2: Typical Example Problems

PART 3: Key Points Review

Question Type: Calculationofdoubleintegrals

PART 1: Problem-solving methods

Question Type: Calculation of double integrals

First, draw the graph based on the function of D
First, observe the form and analyze:
1. Can parity, symmetry, -> simplify the integral to be used;
2. Observe the formula, whether Cartesian or polar coordinate systems are appropriate;
Calculation:
Using the rules for calculating double integrals, calculating;

Question Type: Double Integrals and Inequalities

Basic idea: The integrand of the integrand is larger;

PART 2: Typical Example Problems

Example: Let $D=\{(x,y)|x^2+y^2\leq1\}$, then $\iint_D(x^2-y)dxdy=$

Analysis
Directly calculating the double integral of $x^2-y$ is difficult to compute, so consider analyzing parity and symmetry;
Analysis
From parity, it can be inferred: $\int\int ydxdy=0$
From symmetry, it can be inferred: $\text{ Original form }=\int\int y^2dxdy=\int\int x^2dxdy=\frac{1}{2}\int\int(x^{2}+y^{2})db$