How is the circled part of this double integral formula calculated?

1 makes a ray outward from the pole, and this ray intersects at two points, and the function of these two points is the range of r

2 There is also a case where the pole is in the region, then there is only one intersection point, so it is the distance from the function of that point to the pole (that is, the meaning of 0, the lower line is 0).

Teaching purpose: to be proficient in the calculation method of double integral.

Teaching focus: Calculate double integrals using Cartesian and polar coordinates.

Teaching difficulty: the finite problem of turning double integrals into quadratic integrals.

Teaching content: It is obviously impractical to use the definition of double integral to calculate double integral, and the calculation of double integral is realized by the calculation of two definite integrals (i.e., quadratic integral).

1. Calculate the double integral using Cartesian coordinates.

We will discuss the calculation of double integrals from a geometric point of view.

In the discussion, we assume that;

It is assumed that the integral region can be represented by the inequality , where , is continuous on .

According to the geometric meaning of the double integral, the value of is equal to the volume of the curved top cylinder with the surface as the bottom.

A point is arbitrarily taken on the interval to make a plane parallel to the surface, and the cross-section obtained by the plane section top cylinder is a curved trapezoidal with the interval as the bottom and the curve as the curve, and its area is.

In general, the area of the cross-section obtained by a planar truncated top cylinder that passes through any point above the interval and is parallel to the surface is.

Using the method of calculating the volume of a three-dimensional with a parallel cross-sectional area as known, the volume of the curved top cylinder is .

The calculation is as follows:

A general method for double integralization of cumulative integralsAccording to the principle above, the double integral is the accumulation of the integrand function on a two-dimensional integral-sensitive region; No matter which method of double integration is adopted, the key is to "accurately" represent the range of the two integration variables.

Once it is expressed, it can be easily written as a cumulative integral, and the calculation of the double integral is only left to calculate the two definite integrals.

The integral region of the two integral variables must be "exactly" represented by the range of these two variables, and whoever is first and who is last can do it, so that there must be two representations: taking Cartesian coordinates as an example, these two representations also ensure that the double integral must be converted into a progressive integral in two ways.

What is used here is the content of the double integral part of the high number, first draw a sketch of the integration area, then find the coordinates of each intersection and defeat point, and then try to integrate from left to right from top to bottom, or from top to bottom from left to right as much as possible, and then exchange the integration variables, as for how to draw the integration area, first draw a sketch, and then write out the area of the first integration variable y and the area of the second integration variable x, and then exchange the integration order, and convert the first variable y to the first integration variable x.

Here's a two-fold integral-inflection formula.

The back product is set first, and a line is drawn within the limit, and the upper limit is handed in first, and the upper limit is handed in later.

For this mantra application is this:

First of all, you have to make an integral region, and first make an integral for which, you need to draw a straight line parallel to him through the integral region, and then alternate the order of the integration, which should be determined according to the integration region, and then write the region d, and write another expression according to the sketch.

For circles where the center of the circle is not at the origin, use variable substitution, x=1+u, y=2+v, dxdy=dudv. Then you can use the polar coordinates to find the double integral.

Double integrals have a wide range of applications, such as calculating the area of the surface of the surface, the center of gravity of the plane sheet, the moment of inertia of the plane sheet, the gravitational force of the plane sheet to the particle, and so on. In addition, double integration is also widely used in real life, such as radio.

Definition of Double Integral:

Let the binary function z=f(x,y) be defined on the bounded closed region d, and arbitrarily divide region d into n subdomains δδi(i=1,2,3,..., n) and denote the area of the ith subdomain with δδi. On δδi take either point ( i, i), make and lim n n i = 1 (i, i)δδi)If the limit of this sum exists when the maximum value in the diameter of each subdomain tends to zero, then this limit is called the double product of the function f(x,y) over the region d, denoted by f(x,y)dδ, i.e.

auspicious chain f(x,y)dδ=lim 0( f( i, i)δδi)

In this case, f(x,y) is integrable over d, where f(x,y) is called the integrand, f(x,y)dδ is called the integrand expression, dδ is called the area element, d is called the integral field, and is called the double integral sign.

At the same time, double integration has a wide range of applications, which can be used to calculate the area of the surface, the center of gravity of the plane sheet, the moment of inertia of the plane sheet, the gravitational force of the plane sheet to the particle point, and so on. In addition, double integration is also widely used in real life, such as radio.

In fact, it is to use the derivative formula of the variable limit integral, since the integral arctan[cos(3x+5 root number)]dx is actually a function of y on the 0 to the root number y, you might as well make it f(y), according to the derivative formula of the variable limit integral, the derivative of the integral f(y)dy on 0 to t is 2tf(t), so the formula obtained by the first standby double integral to t contains the factor 2t, because f(y) is the integral arctan[cos(3x+5 root number)]dx,f(t) on the integral arctan[cos(3x+5 root number)] dx,f(t In fact, it is to replace all the ys with t to get the second line, and by the limit number, t 0, the third line is respected.

Replace t in the second integral with x, write it down, and multiply it by the derivative of x (in this case, multiply by 1).

Double integrals. The spatial integral of a binary function, similar to a definite integral, is the limit of a particular form of sum. The essence is to find the volume of the curved top cylinder.

Reintegration has a wide range of applications, such as calculating the area of a surface, the center of gravity of a flat sheet, etc. The double integral of a planar region can be generalized to the integration on a (directional) surface in a high-dimensional space, called a surface integral.

There are several ways to calculate double integrals, the most commonly used of which are the commutation method and the hierarchical integration method.

1. Substitution method:

For double integrals.

f(x,y)dxdy

It can be split into two one-fold integrals:

f(x,y)dxdy= f(u,v)dudv, where u=u(x,y),v=v(x,y) are the functions of the two variables u, v, satisfying u=u(x,y) and v=v(x,y).

2. Hierarchical integration method:

For double integrals.

f(x,y)dxdy

It can be broken down into two one-fold integrals:

f(x,y)dxdy=∫∫f(x,y)dxdy+∫∫f(x,y)dydx

That is, first find the integral of x, and then find the integral of y.

Hope mine can help you out!

Summary. Hello

How to calculate the double integral is not complicated, but there must be a detailed process, and the drawings will be attached.

Hello solution one, I hope to write in detail, I have a poor foundation.

Hello This question is the first sedan chair god to draw the picture, the title already knows the image of x 1 and y imitation sail reed siege city, do you understand this step?

After the image is drawn, there are two methods, and the first solution is to use the x-axis. In this remorse painting stool good image, 1 x 2. Therefore, the upper and lower limits of points are 2 and 1.

It is the first half of the game, and the second half of the integral looks at the value of the y-axis range, and now a vertical line is drawn in the middle of the image, passing through the upper and lower sides of the image of the figure of the bridge, which is x y 2. This is the upper and lower limit of the second credit.

If I want the detailed process of this integral, I will only weigh it single, involving two letters (xy) and I don't know how to calculate it.

For this calculation, first calculate the difference of the integral value in the back, and don't need to look at the front of DX first, first calculate XYDY. This is to find the integral of y, so I put x out and put it in front of dx, which becomes ydy, the original function of y is 1 2y squared, and the upper and lower limits of [2 squares of y] are 2 and x

Solve it (2 minus the square of x Song Xiao 2), and it becomes the royal cherry selling x (2 minus 2 squares of x) and bringing X in to make fun of the town is the step written in the book.

2e^-(2x+y)

2e -2x * e -y first to x integral.

Think of e -y as a constant.

2e -2x is integrated at [0,x] to give 1-e -2x, which is now f(x,y)=(1-e -2x).

e ydy to y integral.

1-e -2x) as a constant.

e -y at [0,y].

Points are scored (1-e-y).

So the final answer is f(x,y)=(1-e-2x)—and (1-e-y)

Contractual exponent: [a,b] denotes a definite integral on [a,b].

y 2,y]e (x round comma chain y)dx=ye (x y)|[y^2,y]=ey-ye^y

Original = [1 2,1](ey-ye y)dy(e 2)y 2-(y-1)e y|[1/2,1](3e/8)-(e)/2

Hope it helps! Orange Sun.

You get x y upside down, and you see f(u,v) is the integral of x and v is the integral of y. It should be like this:

If this is the integral, the result is the opposite:

In fact, the integral interval is reversed, and the integral interval of x is reversed from the integral interval of y.

My own understanding ha.

I don't know, I'm sorry I couldn't help you.