Play x? +y？ +z？ =a？ Substitute into the integral formula,

Get the original formula =a? ∫ds，

Where the curve integral of arc length ∫ds = curve length.

The second question,

Play x? +y？ =a？ Substitute into the integral formula,

Get the original formula =a? ∫dx+0dy，

In terms of Green's formula,

The area enclosed by l is d,

Let the function P= 1 and Q=0,

P'y=Q'x=0，

Then the original formula =-∫ ∫ [d] q 'x-p 'y dxdy = 0 is obtained from Green's formula.

The third question,

It can be divided into left and right sections, which can be directly calculated by calculation formula.

The following is calculated by Green's formula.

Therefore, a straight line segment from point (1, 0) to point (-1, 0) is added, which is recorded as L 1.

Note that the area enclosed by l and L 1 is d.

Let the function P=Q= 1, then P'y=Q'x=0.

Then the original formula =∫L…+∫L 1…-∫L 1…

Green's formula is used to get =0 of the first two integrals.

So the original formula =-∫L 1…

Because when L 1:y=0: Y = 0, and the range of x is 1 to-1, dy=0,

So the original formula =-∫ [1 to-1] dx = 2.