When using calculus to find the area of a circle, there is a problem that I don t understand, so hel

It's not a simple addition of ordinates. There is also the integral formula, z(x,y)=x 2+y 2, so also consider the integral, you are only talking about the definition domain of y, and the definition domain of x.

I think you should understand the definition of calculus again. The domain of y is [0,1], while the domain of x is [0, (1-y 2) 2)].These two limits are multiplied and then the integrals are found.

The core idea of calculus is to divide and solve, and to find the area is to divide it into an infinite number of small rectangles, and then add these small areas. It's not just the addition of coordinates.

And don't forget the lim too.

Rectangles are used when differentiation is introduced, e.g. y is the length x is the width.

Then make the width smaller, and the area of the rectangle will become smaller and smaller, and finally add it together.

Rather than simply adding the value of y together.

s=2 upper limit 1 lower bound (-1) (1-x2)dx let x=sina, then (1-x2)=cosa because -1 x 1,0 (1-x2) 1

So the range of values for a is [-pi 2,pi 2], so s=2 upper limit (pi 2) lower bound (-pi 2) cosadsina 2 upper limit (pi 2) lower limit (-pi 2) cos 2 (a) dacos 2 (a) da, let f=cosa, dg=cosada=dsina, then df=-sinada, g=sina.

cos 2(a)da=cosasina+ sin2(a)dacosasina+ 1-cos2(a)dacosasina+a- cos 2(a)da, so cos 2(a)da=1 2(cosasina+a)=1 4sin(2a)+a2.

s={2[1 4sin(2a)+a 2]} upper limit (pi 2) lower bound (-pi 2) pi

Take x 2 + y 2 = r 2 as an example:Just figure out the quadrant before the first volt.

Then multiply by 4s 4= (0 to r) (r 2-x 2)dx so that x=rcosa

r^2-x^2)=rsina

dx=-rsinada

So s4= (2 to 0)rsina*(-rsina)da-r 2 ( 2 to 0)(sina) 2da r 2 4

So repentance which s = r 2.

Calculus Note:

The content mainly includes limits, differential calculus, integral science and their applications. Differential calculus includes the operation of finding derivatives and is a set of theories about the rate of change. It makes functions, velocities, accelerations.

and the slope of the curve, etc., can be discussed with a common set of symbols. Integralism, including the operation of finding integrals, provides a general set of methods for defining and calculating area, volume, etc.

Pair circlesDefinite integralsFinding the area will not be zero.

A definite integral can be used to find an area, but a definite integral is not equal to an area. Because the definite integral can be negative, but the area is positive. So when the curve of the integration is divided into the x-axis, the divisions (over 0 and less than 0) are calculated separately, and then the positive integration plus the negative integration is the absolute value of the lesion tolerance.

An equal area is the number of dimensions that represent a two-dimensional shape or shape or plane layer in a plane.

Surface area. is a simulator on a two-dimensional surface of a three-dimensional object.

This area can be understood as the amount of material with a given thickness of acres, and this area is necessary to form a model of shape. A one-answer function can have indefinite integrals, no definite integrals, definite integrals, or no indefinite integrals.

A continuous function.

There must be a definite integral and an indeterminate integral, if there is only one finite discontinuity point, then the definite integral exists, if there is a hop discontinuity point, then the original function cannot exist, i.e., the indefinite integral cannot exist.

Definite integrals

The definite integral is the area enclosed by the image in the interval a,b of the function f(x). That is, the area of the graph enclosed by y=0,x=a,x=b,y=f(x). This shape is called a curved trapezoid, with the exception of a curved triangle.

The definite integral is to divide the image a and b of the function in a certain interval into n parts, divide them into an infinite number of rectangles with a straight line parallel to the y-axis, and then find the sum of the areas of all these rectangles when n +.

Definite integrals and indefinite integrals seem to be incompatible, but they are intrinsically closely related due to the support of a mathematically important theory. It may seem impossible to infinitely subdivide a graph and add it up, but thanks to this theory, it can be translated into computational integrals.

The above content reference: Encyclopedia - Definite Integral.

To find the area of the circle, we can divide the circle into many small sectors and stitch these sectors together to form a polygon that approximates the circle. When the number of sides of a polygon increases infinitely, the perimeter of the polygon approaches the perimeter of the circle, and the area of the polygon approaches the area of the circle.

Suppose the circle is divided into n sectors, then the central angle of each sector is frac, and the area of the sector is fracr 2 sin( frac). Add the area of the fan to get the area of the polygon that approximates the circle: s n=n times fracr 2 sin( frac.

When n approaches infinity, frac approaches 0, sin( frac) approaches frac, and thus the area of the polygon of the approximated circle approaches the area of the circle. Namely:

s=\lim_n\times\fracr^2\sin（\frac）=\lim_n\times\fracr^2\frac=\pi r^

Therefore, the area of the circle is pi r 2, where pi is pi and r is the radius of the round paulownia. <>

1. Establish a coordinate system with the center of the circle as the origin.

2. Divide the circle into strips along the y-axis, let the radius of the circle be the liquid front r, and the width of the strip circle is dy, then the area of the strip (rectangle) is 2 (r 2-y 2)dy.

3. Integrate this equation, the lower limit is -r, the upper limit is r, and the area of the circle can be calculated as r 2.

1. Establish a coordinate system, taking the center of the liquid front circle as the origin to establish a coordinate system.

2. Divide the circle into strips along the y-axis, let the radius of the circle be r, and the width of the strip circle is dy at any y distance from the x-axis, then the area of the strip (rectangle) is 2 (r 2-y 2)dy.

3. Integrate this formula, the lower limit is -r, the upper limit is r, and the area of the circle can be calculated as r 2.

1. Establish a coordinate system, take the center of the circle as the origin, and establish a coordinate system.

2. Divide the circle of the section shed into liquid front strips along the y-axis, and let the radius of the circle be r, and the width of the strip circle is dy at any y point from the x-axis, then the surface of the strip (rectangle) is 2 (r 2-y 2)dy.

3. Integrate this equation, the lower limit is -r, the upper limit is r, and the area of the circle can be calculated as r 2.

Take x bai2+y 2=r 2 as an example.

Just count the first quadrant and multiply it by 4

s 4 = (0 to r) dao(r in 2-x 2) dx let x=rcosa

r^2-x^2)=rsina

dx=-rsinada

So S 4= ( Capacity 2 to 0)rsina*(-rsina)da=-r 2 ( 2 to 0)(sina) 2da=-r 2 ( 2 to 0)(1-cos2a) 2da=-r 2 4 ( 2 to 0)(1-cos2a)d2a=-r 2 4(2a-sin2a)( 2 to 0) = r 2 4

So s= r 2

Establish a coordinate system, with the center of the circle as the origin, and establish a coordinate system.

Divide the circle into strips along the y-axis, let the radius of the circle be back r, and at any y point from the x-axis, the width of the strip circle is dy, then the area of the strip (rectangle) is.

2 (r 2-y 2)dy, integrating this equation with a lower bound of -r and an upper limit of r, the area of the circle can be calculated as r 2

On a circle of 1 4, the integration comes out 4 times. You can find this at a glance when you look for a book.