Ask for two advanced math problems and two advanced math problems

Wait, I'll draw you a picture.

The second problem itself requires the integral of the area enclosed by x=0 x=1 y=0 y=1.

That is, the integral of a + b + c + d.

But when you get to 1 (x 2+1)-0 (x 2+0), you lose 0 (x 2+0), which is c+d, so your 4 is actually a+b

After changing the order of points.

You accumulate. 1 (y 2+1)-0 (y 2+0) loses 0 (y 2+0), which is b+d.

So the -4 you accumulate is a+c

Because the concrete form of your function f(x,y) = (x 2-y 2) (x 2+y 2) 2

So f(y,x)=-f(x,y) because f(x,y) does not =f(y,x) so the graph of this function is not symmetrical with respect to y=x, so b does not =c, so the a+b you find is of course not =a+c.

Exactly here b = 4-a

c=-π/4-a

1.Drawing the integration range, you can see that the upper and lower bounds of the integration have nothing to do with x or y, so no matter where f(x) is, the result is the same whether it is integrated first for x or first for y.

2.Based on the above explanation, B should be chosen for the second question

Let p=axy 3-y 2cosx, q=1+bysinx+3x 2y 2

The partial derivative of p to y is 3axy 2-2ycos, and the partial derivative of q to x is bycosx+6xy 2

The above two equations are equal, 3ax=6x, -2y=by

a=2,b=-2

This is a direct credit.

The product of the first pair of x results is ax 2y 3 2-y 2sinx, and the second pair of the product of y results is y+by 2sinx 2+x 2y 3 comparison coefficient a=2 b=-2

This function is 2x 2y 3-y 2sinx+y+c

Perpendicular to the y-axis, equivalent to parallel to the surface xoz

a+ b=(3 +2 ,5 + 8 -4 ) a+ b is parallel to xoz, so y=0, i.e. 5 + =0 =(-2,2,1) =(2,1,2) ·=-2x2+2x1+1x2=0

The answer to the first question is: (under the root number) 2, and the answer to the second question is: 1 16; with polar coordinates.

For the first question, use e (-x 2)dx* e (-y 2)dy= e (-x 2-y 2)dxdy

where e (-x 2)dx= e (-y 2)dy

The second problem can be found directly, first to find the integral of y, and then to the integral of x.