Find the volume of solid surrounded by x ^ 2 + y ^ 2 under the sign of cone z = root, x ^ 2 + y ^ 2 = 1 on cylinder and z = 0 on plane

Find the volume of solid surrounded by x ^ 2 + y ^ 2 under the sign of cone z = root, x ^ 2 + y ^ 2 = 1 on cylinder and z = 0 on plane

V =∫dt∫r*rdr =2π/3.

Find the volume of the solid surrounded by x ^ 2 + y ^ 2 and z = 2-x ^ 2-y ^ 2 under z = root of the cone

http://hi.baidu.com/522597089/album/item/d33979029fbb74761c9583ac.html#

∫∫ (x ^ 2 + y ^ 2) dzdx + (Z-1) DXDY how to integrate with Gauss formula?
Trajectory: the cone surface is x ^ 2 + y ^ 2 = Z ^ 2 (0

Integral surface is not closed, complement plane ∑ 1: z = 1, X & # 178; + Y & # 178; ≤ 1 upper side
The combination of the two surfaces is a closed surface
∫∫ (x²+y²)dzdx+(z-1)dxdy
=Which cone is the domain of ∫ ∫ (2Y + 1) dxdydz integral
Since the cone is symmetric with respect to the xoz plane and 2Y in the integrand is an odd function, the integral result is 0
=∫∫∫ 1 dxdydz
The integrand function is 1, and the result of integration is the volume of the region. The volume of the cone is: (1 / 3) π
=(1/3)π
Now subtract the integral of the plane to be complemented
∫∫(Σ1) (x²+y²)dzdx+(z-1)dxdy
=The integral region of ∫ - 1 DXDY is: X & # 178; + Y & # 178; ≤ 1
=-π
So the final result of this problem is: (1 / 3) π - (- π) = (4 / 3) π
If you don't understand, please ask. If you can solve the problem, please click "select as satisfactory answer" below

Let ∑ be the upper side of the surface z = x ^ 2 + y ^ 2 (Z ≤ 1), and find the surface integral ∫ (x + Z ^ 2) dydz zdxdy
appeal

Two double integrals, if their integrands are equal and always greater than 0, and the integral region is in the same image
Two double integrals, if their integrands are equal and always greater than 0, and the integral region is within one quadrant, is it true that the larger the area of the integral region, the greater the value of the double integral?

Not necessarily
It's possible that area a is a large area full of bushes
Area B is a small giant fir forest
However, if the small integral area is contained in the large integral area
Then it must be right

How to understand that the integrand is Max {XY, 1}?

F (x) = (XY-1 + | XY-1 |) / 2; this situation is easy to understand!

The function f (x) is defined on (0, + ∞), f (2) = 0; when x > 1, f (x)

1. F (x) is a decreasing function in the domain of definition
It is proved that f (XY) = f (x) + F (y) - 1,
Let y be a constant a larger than 1 but infinitely close to 1
Then when ax > x, f (a) 1, f (x) 2
So just find the abscissa a, 0 of the intersection a

If f (XY, x + y) = x ^ 2 + XY + y ^ 2, then DF (x, y) =?

Because f (XY, x + y) = x ^ 2 + y ^ 2 + xy = (x + y) ^ 2-xy
So the function f (XY, x + y) = x ^ 2 + y ^ 2 + XY can be transformed into f (x, y) = y ^ 2-x

On double integral integrand
Why double integral, as long as the interval is symmetric, the integrand function is y, which is an odd function, directly get 0, and the integrand function is x, which is an even high number? How can we see that? And how can we distinguish the parity of the integrand function?

It's like this
① If the integral region of the integrand is symmetric with respect to x, and there is an odd function of Y in the integrand, then the value of the odd function of Y is 0
② If the integral region of the integrand is symmetric with respect to y, and there is an odd function about X in the integrand, then the value of the odd function about X is 0
③ If its integral region is not symmetric with respect to X or Y, but symmetric with respect to a certain line, it is easy to integrate as long as the integrand is reduced to the form containing the line

Single choice
20. When the integrand is a constant C and the integrand is an ellipse, the value of the double integral ()
A: It's the area of this ellipse
B: It's the volume of the cylinder with the bottom of the ellipse and the height of C
C: Is the length of this ellipse
D: It's the volume of a sphere with this ellipse as the bottom and the z-axis as C

B