The transformation of polar coordinates into rectangular coordinates by double integration 1. Let x = RCOs θ y = rsin θ, why is the corresponding rectangular coordinate equation of R = 2cos θ (x-1) ^ 2 + y ^ 2 = 1 2. 2cosθ< r < 2 0

The transformation of polar coordinates into rectangular coordinates by double integration 1. Let x = RCOs θ y = rsin θ, why is the corresponding rectangular coordinate equation of R = 2cos θ (x-1) ^ 2 + y ^ 2 = 1 2. 2cosθ< r < 2 0

eight

Does it have geometric significance to transform polar double integral into quadratic integral

After the transformation, we can integrate from the radius and angle values to become the integral of the square region

Determination of the upper and lower limits of the parameters in polar coordinates of double integrals
How to determine the upper and lower limits of two integral parameters x = RCOs θ, y = rsin θ in polar coordinates of double integral?
Let me put a question to you. I don't know how to take the corresponding range for the upper and lower limits of the parameters R and θ after setting the parameters x = RCOs θ and y = rsin θ
∫ (x + y) d θ, where d = {(x, y) / x ^ 2 + y ^ 2

"The origin is in the graph, and the tangent can't be drawn through the origin. How can we take the upper and lower limits of θ?"
The origin is in the graph, theta angle range is 0 to 2 π, R is a function related to theta, which is the polar coordinate equation of the integral field

How to find the upper and lower limits of polar double integral R

Starting from the pole, a ray is introduced to pass through the integral region. The first ray passes through is the lower limit, and the second ray passes through is the upper limit

What is the area of a circle?

Radius x radius x pie
Pai = 22 / 7
This one is less used
Normally,
Pai = 3.14159265359
Make it simple
Pai = 3.14

When we use calculus to calculate the area of a circle, we have a problem,
If the radius of the circle is 1, then its area should be π. If the circle is drawn in the plane rectangular coordinate system, if the first quadrant part is 1 / 4 circle, then its area should be 1 / 4 π, less than 1. But if we want to use the integral to calculate the area of the 1 / 4 circle, we should add up all the ordinates of the first quadrant. The maximum ordinate of the first quadrant is 1, and the minimum is 0. After adding up, it must be greater than 1. What's the matter?

It's not a simple addition of ordinates. There's also the integral formula, Z (x, y) = x ^ 2 + y ^ 2, so we have to consider the integral. What you're talking about is only the domain of Y and the domain of X

Given the square area formula, prove the circle area formula s = R ^ 2 (related to calculus)

The specific measures are as follows:
Let's assume that the center of the circle is at the origin, and the distance between the moving point P (x, y) and the origin is a fixed value R. we know that the trajectory of the moving point P must be a circle with radius R. let's find the area of the circle
Change P (x, y) to polar coordinates
Let x = rcost, y = rsint
Since the center of the circle is inside the circle, then the parameter t ∈ (0,2 π)
According to the integral formula of the curve:
A=1/2[∫xdy+ydx ]
=1/2[∫RcostdRsint - RsintdRcost]
=1/2[∫R^2 cost^2dt+R^2sint^2dt]
=1/2R^2∫1dt
=1/2R^2 t
Then the function value of lower limit 0 is subtracted from the function value of upper limit 2 π
The results are as follows
A=πR^2
So we can see that the area of a circle with radius r must be π R ^ 2
The proof is complete

How to calculate the area of a circle? OK, 50 points

S＝πr²
S = area
π=3.1415926
R = radius
The length of a rectangle is equal to half the circumference of a circle
I.e. = = π R
The width of a rectangle is equal to the radius r of a circle
Because the area of a rectangle is length x width
So the area of the circle is π R × r = π R & sup2;
(3) the area formula of a circle is derived from the transformation of a circle into a rectangle. Students, can we convert a circle into other figures to derive the area formula of a circle?
4. Sum up the area formula of circle
S＝πr²

Area formula of ellipse? Area volume formula of sphere? Volume formula of cone?

The area formula of ellipse? S = π × a × B / 4 (where a and B are the length of major axis and minor axis of ellipse respectively)
The area volume formula of sphere? S = 4 π R ^ 2 v = 4 / 3 π R ^ 3
The volume formula of cone? V = 1 / 3 * π R ^ 2 h

Which is the derivation of the area formula and perimeter formula of a circle? Is one of them an axiom or a definition?
How to determine that π is a constant?

1. Cut the circle into several equal parts along the radius (the more the better) (into several sectors)
2. Divide the sector into two parts equally, and put them together to form a figure similar to a rectangle. (the more, the closer to a rectangle)
3. The area of a rectangle is equal to the length multiplied by the width. The length of the rectangle is half of the circumference of the circle (2pr). Therefore, the length is pr (the symbol of PI, I will not type it, expressed by P), and the width is the radius of the circle R. therefore, the calculation formula of the area of the circle is s = pr. r = PR2 (square)
Derivation of circumference
Find several circular objects, measure their perimeter and diameter respectively, and calculate the ratio of perimeter and diameter. Through experiments and statistics, we can know that the perimeter of a circle is always more than three times the diameter. Then, the ratio of perimeter and diameter of any circle is a fixed number (circumference ratio). Because the perimeter of a circle is always Π times the diameter, when we know the diameter or radius of a circle, We can calculate its perimeter, that is, C = Πd, C = 2 Π R
The derivation of circle area is as follows
Draw a circle on the cardboard, divide the circle into several equal parts, cut it open, and then use these similar isosceles triangle pieces of paper to put together to form an approximate parallelogram. If the score is more, each part will be thinner. The figure will be closer to the rectangle. The length of the rectangle is equal to half of the circumference of the circle, that is, R, and the width is equal to the radius of the circle, Because the area of the rectangle is length × width, the area of the garden is R × r = R & # 178, that is s = Πr & # 178;