Higher Mathematics Surface Integral Problems, Higher Mathematics Surface Integral Problems?

You're kind of nowhere to go. Originally, the calculation was quite simple, and you had to calculate the ...... of the projection coincident part when you project to yozThe graduate school entrance examination will not be so specifying the projection surface of the question, it didn't come out last year anyway.

Dear, I'll give you a copy of King Li's book

Remember

1: Since it is directional, there is.

even zero odd multiple" property, contrary to the general situation.

When f(x) is an even function, if is symmetrical with respect to the corresponding faces, one part takes + and one part takes -

The result is f(x) -f(- x) = f(x) -f(x) = 0, and the two parts cancel each other out.

f(x) odd function, in the same case, one part takes + and one part takes -

The result is f(x) -f(- x) = f(x) +f(x) = 2f(x), and the integrals of both parts are equal and can be stacked.

2: 3-in-1 formula.

For is in the form z = z(x,y).

Normal n =

then pdydz+qdzdx+rdxdy

d) dxdy

When taking the front right side, take the + sign.

When removing the Left Rear side, take the - sign.

3: Gaussian formula.

(pdydz+qdzdx+rdxdy

p/∂x+∂q/∂y+∂r/∂z) dxdydz

and) pdydz+qdzdx+rdxdy

In the latter part (and), if the originally given surface cannot be enclosed into a closed space, the Gaussian formula cannot be used directly, and the area needs to be closed after making up a few surfaces, for example, if a number of (and) surfaces are added, the Gaussian formula can be used, and it should be noted that the corresponding integral of the surface (and) should be reduced in the end.

4: Dig a hole. If there is a singularity in the integrand on , the Gaussian formula cannot be used directly.

You need to fill in a small space r= that is large enough to include all the internal singularities, and then take the radius to tend to 0

When using the Gaussian formula, the corresponding integral of this part is also subtracted.

So there is =

5: Substitution. If the equation of the integrand f is on , you can preferentially substitute the equation of f into f.

For example, give the equation: x + y + z = a

then pdydz+qdzdx+rdxdy) (x +y +z )

pdydz+qdzdx+rdxdy)/a

1/a)∫∫pdydz+qdzdx+rdxdy

In this way, the situation of 4: can be avoided without digging a hole.

After removing the singularity, you can continue to fill the surface using the Gaussian formula.

Here is another way of thinking: the centroid method.

r = xds/(4πr^2) =xds = 4πr^3

therefore, ∫2rxds = 8πr^4

Looking at x-r as a whole t, it is clear that the integral surface can become.

t²+y²+z²=r²

Obviously, last year about t symmetry, at the same time, the integral function t is odd, so, 2rtds=0

i.e. 2r (x-r) ds=0

The curvature of the coordinates is not as good as you say. This question applies the Gaussian formula, gets.

Original = 0+y-z)DXDYDZ = 0, 3>dz <0, 2 >dt <0,1>(rsint-z)RDR

0, 3>dz <0, 2 >dt[(1 3)r 3sint-(1 2)zr 2]<0,1> branches.

0, 3>dz∫<0, 2π>[1/3)sint-(1/2)z]dt

0, 3>dz[-(1/3)cost-(1/2)zt]<0, 2π>

0, 3>(-z)dz = 2)[z^2]<0, 3> =9π/2

It's not a projective calculation, it's a Gaussian formula:

>pdydz+qdzdx+rdxdy = p/∂x+∂q/∂y+∂r/∂z)dxdydz

The first problem is the second type of surface division, the surface is a parabola, and the projection on each coordinate plane is two similar parabolas and horizontal lines, and a circle, respectively, calculate the plane integrals on these projection planes, and finally add them.

Of course, there is a second way, which is to use Gauss's formula

The original surface division is integrated with a circular plane (the center of the circle is at (0,2,0) and the radius is 1) to obtain the closed surface division, which can be transformed into a triple integral, and the volume of the projectile is exactly obtained.

That is, the final equal to the volume of the parabolic object minus the integral of a circular plane (parallel to the xoz plane, that is, the bottom surface of the projectile, which satisfies dy=0, y=2 at this time) (i.e., (6)dxdz = 6 circle area =6), the curve l of problem 2 is the boundary of a circular plane centered on the origin point (which is also the center of the sphere of a sphere with radius a), and the Stokes formula can be applied to integrate the closed curve and convert it into a surface integral.

p=y-4q=z+3

After r=x+1 finds each partial derivative, we get exactly the surface area, which is the area of the circle a 2

is a curve integral.

x 2+y 2 = 4, the circumference of the semicircle is 2 .

i = ∫(2+x^2+y^2)ds = ∫6ds = 6 · 2π = 12π。Select C.

According to the basic steps, replace with equivalent, and select D

Remember that the closed region enclosed in the xoy plane is dxy, then.

∑>f(x,y,z)ds = ∫∫f[x,y,z(x, y)]√1+(∂z/∂x)^2+(∂z/∂y)^2]dxdy

It is possible to convert the surface integral to a double integral solution.

Can be proved simply:

Draw two different paths from A to B1, 2 so that the area enclosed by the two paths is single-connected.

The integral goes from A via path 1 to b and back to a via path 2. This completes a cycle in which Green's formula is applied to this enclosed area and the integral value is zero. That is, the integral of A via path 1 to b + the integral of returning to a via path 2 = 0, so the integral of A via path 1 to b = - the integral of a via path 2 to a = the integral of A via path 2 to b. qed

Single-connected areas.

The integration of the first type of curve, which is equal to zero on a piecewise smooth closed curve, is only one of them.

Looking at the formula I wrote in **, you can also deduce this conclusion that has nothing to do with the path.