Surface z x 2 y 2 The area of the surface that is cross sectioned by the plane z 1 z 2

The definite integral of 2pi*sqrt(z)dz on 1,2 is calculated to be 4 3 *pi(2*sqrt(2)-1).

s= ∫∫1+zx²+zy²dxdy

∫√1+x²+y²)dxdy

dθ∫(1+ρ²dρ (0《θ《2π ，0《ρ《1）

Formula for indefinite integrals.

1. A dx = ax + c, a and c are constants.

2. x a dx = [x (a + 1)] (a + 1) +c, where a is a constant and a ≠ 1

3、∫ 1/x dx = ln|x| +c

4. A x DX = (1 LNA)a x + C, where A > 0 and A ≠ 1

5、∫ e^x dx = e^x + c

6、∫ cosx dx = sinx + c

7、∫ sinx dx = - cosx + c

8、∫ cotx dx = ln|sinx| +c = - ln|cscx| +c

The details are as follows:

A surface is the trajectory of a straight line or curve under certain constraints. This moving straight line or curve is called the bus bar of the surface; A busbar at any position on the surface is called a prime line. The constraints on the movement of the bus bar are called the constraints of the motion.

In constraints, the straight line or curve that controls the motion of the bus bar is called a traverse; The plane that controls the movement of the busbar is called the guide plane.

When the moving line moves according to a certain law, the surface formed is called a regular surface; When the moving line moves irregularly, the surface formed is called an irregular surface. The busbars that form the surface can be straight or curved. If the surface is formed by linear motion, it is called a straight surface (such as a cylindrical surface, a conical surface, and so on).

Surfaces formed by curvilinear motion are called curved surfaces (e.g., spheres, toruses, etc.). Two consecutive straight lines of a straight plane are parallel to or intersect each other (i.e., they are located on the same plane), and this kind of surface can form a plane without deformation, which is a deployable surface. A surface that intersects two consecutive straight lines (that is, they are not on the same plane) is a non-expandable surface.

The representation of a surface is similar to that of a plane, and the most basic requirement is that the projection of the geometric elements of the surface should be decided, such as busbars, wires, guide surfaces, etc. In addition, in order to clearly represent a surface, it is generally necessary to draw the outline line of the surface to determine the extent of the surface.

z = 2-x 2-y 2,x and. y

is perfectly symmetrical, then it can be treated as a two-dimensional plane.

i.e. consider. z=2-x 2 when. z=0

Time. x=±√2

Find the area above xoy of this surface. =πx^2

Solution: Let the area be s. By z=2-x -y zx=-2x, zy=

2y s=∫∫

ds=∫∫dxy)

1+z²x+z²y)

dxdy= (upper limit 2, lower limit 0).

d (upper limit 2, lower limit 0).

1+4r²)

rdr=13π/3

Summary. Hello, glad to answer for you. <>

The cone z=(x 2+y 2) is planed z=2, and the surface area of the cut part is . As you can see from the title, the surface is a revolved surface with a bus bar of z=x +y. The surface is cut in two by the plane z=2, with the upper part forming the top of the cone and the lower part being the required surface.

Therefore, it is only necessary to specify the area of this lower part of the surface. Using the properties of the revolved surface, the lower surface can be projected onto the plane z=0 and expressed as the polar equation of the contour line: r= 2sin where theta can be valued in the range of [0, 2].

By calculating the area of the element for one of the arcs and accumulating it, the integral formula is obtained: therefore, the area of the lower part of the surface is pi.

Find the surface area of the cone z=(x 2+y 2) cut off by the plane z=2.

In no. Hello, glad to answer for you. <>

The cone z=(x 2+y 2) is planed z=2, and the surface area of the cut part is . As can be seen from the title, this surface is a rotating surface in the state of Eggplant, and its bus bar is z=x +y. The surface is cut in two by the plane z=2, with the upper part forming the top of the cone and the lower part being the required surface.

Therefore, it is only necessary to specify the area of this lower part of the surface. Using the properties of the revolved surface, the lower surface can be projected onto the plane z=0 and expressed as the polar equation for the contour wiring line: r= 2sin where theta can be valued in the range [0, 2].

By calculating and adding up the area of the quivering element for one of the arcs, the integral formula is obtained: therefore, the area of the lower part of the surface is pi.

Write about the specific process.

It's the cut part, not the bottom part.

I knew it turned out to be 4 root number 2 vultures, but it didn't turn out to be right.

Can you help with this?

The equation of the cone becomes z= [3(x 2+y 2)], p= z x=6x = 3x (x 2+y 2), q= z y= 3y (x 2+y 2), (1+p 2+q 2)= 1+(3x 2+3y 2) (x 2+y 2)]=2, the cone x 2+y 2=1 3z 2 The curve that is altered by the plane x+y+z=2 is an ellipse

3(x 2+y 2)=[2-(x+y)] 2, orthogonal transformation.

x=(u+v) 2, and (-u+v) 2, the above equation becomes.

3(U 2+V 2)=[2- 2V] 2,3U 2+V 2+4 2V+8=12,U 2 4+(V+2 2) 2 12=1, and its major semi-axis a=2 3;The minor semi-axis b=2 and the area is ab=4 3 , so the surface area is found = 1+p 2+q 2)dxdy=8 3 .

Summary. When intercepting a surface, we need to find the intersection equation of the two surfaces so that they are on the same plane. Here, the intersection equation for surface z=x 2+y 2 is x 2+y 2 and the intersection equation for cylinder x 2+y 2=4 is x 2+y 2=4.

From this, we can see that x 2+y 2 = x 2+y 2 = 4 is the intersection equation of two surfaces, which are in the same plane. Bringing this equation into z=x 2+y 2, we get z = 4. This is the surface that is intercepted, and this surface is a cylindrical surface with a height of 4.

Since a cylindrical surface is formed by the same circle extending in a specific direction, we can know that in this case, the resulting surface is a cylindrical surface.

1.Find the surface z=x 2+y 2 and be truncated by the cylinder x 2+y 2=4.

Hello, I'm glad to answer your questions! The surface z=x 2+y 2 intersects the column balance surface x 2+y 2=4, and its intersection line is x 2+y 2=4. We can take this equation into z=x 2+y 2 and get z = 4.

Therefore, the surface z=x 2+y 2 is cut by the cylindrical surface x 2+y 2=4 and is a cylindrical surface with a height of 4

How to find the area.

Surface area. When intercepting a surface, we need to find the intersection equation of the two surfaces so that they are on the same plane. Here, the intersection equation for surface z=x 2+y 2 is x 2+y 2 and the intersection equation for cylinder x 2+y 2=4 is x 2+y 2=4.

From this, we can see that x 2+y 2 = x 2+y 2 = 4 is the intersection equation of two surfaces, which are in the same plane. Bringing this equation into z=x 2+y 2, we get z = 4. This is the surface that is intercepted, and this surface is a cylindrical surface with a height of 4.

Because the cylindrical surface is formed by the extension of the lead in a specific direction by the same circle, we can know that in this case, the surface obtained after the interception is the cylindrical surface.

The surface area is generally solved using the surface division method. The most commonly used of these is the Lagrangian integral method. The Lagrangian integral method solves the surface area based on a set of parametric equations for the surface.

Assuming that the surface equation is z=f(x,y), then the surface area can be expressed by the following equation: s = f(x,y) ds, where ds is the element (unit area) on the surface. This method requires parameterization of the surface, then divides the surface into many elements, finds the area of each element, and finally sums the surface area object width talk.

In addition, there are Euler's formulas and Gaussian formulas that can be used to solve surface areas, but these methods require a high level of mathematical knowledge and expertise. For the specific surface area solution, Qiaoran needs to be analyzed according to the specific surface equation.

z=2-x -y, then z x=-2x, z y=-2yds= late dress[1+( z x) +z y) ]dxdy (1+4x +4y )dxdy

Therefore, the surface area sought = ds (d represents the projection garden where the surface area of the code stove is in the xy plane: x hall god + y = 1).

√1+4x²+4y²)dxdy

d 1+4r )rdr (for polar transformation)2 1+4r )rdr

/4)∫√1+4r²)d(1+4r²)π5√5-1)/6.,3,