Double integral math problems, double integration problems?

Here's how to solve it. 1) Note that the integral region is symmetrical with respect to the y-axis, while the integrand is odd with respect to x, so the integral is 0

2) Let the integrand be f(x,y), then f(0,0) = 1, and consecutively at (0,0) points.

For a disk d(r) of radius r, by the integral median theorem, the double integral = 1 (pi * r 2) *area of d(r)) *f (x1, y1) = f (x1, y1), where (x1, y1) is a point in the disk d(r).

Note that when r tends to 0, (x1,y1) tends to (0, 0), the limit of the continuity of the function f is f(0,0), that is, 1

1) Note that the integral region is symmetrical with respect to the y-axis, while the integrand is odd with respect to x, so the integral is 0

2) Let the integrand be f(x,y), then f(0,0) = 1, and consecutively at (0,0) points.

For a disk d(r) of radius r, by the integral median theorem, the double integral = 1 (pi * r 2) *area of d(r)) *f (x1, y1) = f (x1, y1), where (x1, y1) is a point in the disk d(r).

Note that when r tends to 0, (x1,y1) tends to (0, 0), the limit of the continuity of the function f is f(0,0), that is, 1

Here's how to do it, please make the first to check the ginseng test:

If it helps, the reeds are destroyed.

The previous piece is the odd function about y on the interval of the guan chaos, and the integral is 0, which is the basic property of the definite integral.

The latter one is replaced by polar coordinates, dxdy = rdrdrou, and then becomes d(r 2)drou

You said "it shouldn't be" and you are probably missing the above 1 2

A double integral is the spatial integral of a binary function, similar to a definite integral, and is the limit of a particular form of sum. The essence of punching and bending is to find the volume of the cylinder at the top of the curve. Reintegration has a wide range of applications, such as the area of a surface and the center of gravity of a flat sheet can be calculated by using scattered premature boredom.

The double integral of the planar region can be generalized to the integration on the (directed) surface leakage in the high-dimensional space, which is called the surface integral.

When the double integral is converted into a progressive integral, the integral of x on the symmetric interval occurs. Using the property of the odd function to integrate on a symmetric interval, the integral is 0

If you can't understand that x is an odd function, you can see y as a constant, and if you can't understand it, you can see y as 1, such as z=xy, which is z=x, which is an odd function, and z=x 2*y, which is z=x 2, which is an even function, and discuss what function is x, which has nothing to do with y, and what function about y is it, which has nothing to do with x.

Since x is an odd function, and y is regarded as a constant, and when the integral region is symmetric with respect to the y-axis, its integral can be understood in the same way as a definite integral, y=sin x, on to , the area above and below the x-axis is equal, the algebraic sum is 0, and the definite integral is 0. In the same way with the double integral, z=y*sin x, on to , z is symmetrical about the origin in space, so the volumes above and below the xoy plane are equal, and the algebraic sum is 0.

If the integrand is odd with respect to y, and the integral region is symmetric with respect to the x-axis, then its integral is 0. Similarly.

The method is shown in the figure below, 139

Check it yourself again, please check it carefully, and I wish you a happy study:

Draw a sketch first, and then look at the symmetry of the branches.