Who can clearly tell me how to calculate double integral Just in contact with two-dimensional random variables in probability theory, it's difficult to be solved by double integral. How can this problem be solved For example, ∫ e ^ - (x + 2Y) x, Y > 0 Is the upper and lower limits x ~ 0,0 Should I separate ∫∫ e ^ - (x + 2Y) into ∫ e ^ - x ∫ e ^ - 2Y and then calculate the integral separately, or should I integrate the e ^ - (x + 2Y) integral again, but if so, there are two variables in it. How should I calculate the integral The upper limit is x and Y. who will write me a specific process The original problem is f (x, y) = AE ^ - (x + 2Y), x, Y > 0, finding constant A. I just want to know the algorithm of this type of problem

Who can clearly tell me how to calculate double integral Just in contact with two-dimensional random variables in probability theory, it's difficult to be solved by double integral. How can this problem be solved For example, ∫ e ^ - (x + 2Y) x, Y > 0 Is the upper and lower limits x ~ 0,0 Should I separate ∫∫ e ^ - (x + 2Y) into ∫ e ^ - x ∫ e ^ - 2Y and then calculate the integral separately, or should I integrate the e ^ - (x + 2Y) integral again, but if so, there are two variables in it. How should I calculate the integral The upper limit is x and Y. who will write me a specific process The original problem is f (x, y) = AE ^ - (x + 2Y), x, Y > 0, finding constant A. I just want to know the algorithm of this type of problem

Divide the double product into quadratic integral, that is, take one of the variables as a constant, such as y, and then integrate only one variable to get an integrand function containing only y, and then integrate with y. you can find a Book of advanced mathematics to read

The cross section method of triple integral is two before one
To find ∫ ∫ 3zdv, the integral region is Ω, which is Z = 1-x & # 178; - & # 188; Y & # 178; (0 ≤ Z ≤ 1), I know that it can be changed into 3 ∫ ZDZ ∫ DXDY, but the section related to Z will not be solved. At the same time, the general triple integral like vertebral body or spherical domain uses the "two" section algorithm in the first two then one method

First of all, the section must be a plane figure that you are familiar with, and the area is easy to calculate
In fact, the section is very simple: it is the surface equation of the side, but Z is treated as a constant when making the section
So the cross section equation is: X & # 178; + Y & # 178 / 4 = 1-z, which is an ellipse, a = √ (1-z), B = 2 √ (1-z)
The area of ellipse is: π AB = 2 π (1-z)
So the original formula = 3 ∫ [0 → 1] Z DZ ∫ ∫ DXDY
=6π∫[0→1] z(1-z) dz
=6π[(1/2)z²-(1/3)z³] |[0→1]
=π
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Surface area and volume formula of sphere?

Square of surface area 4 * π * radius
Cube of Volume 4 / 3 * π * radius

What are the volume formula and surface area formula of a sphere?

Volume: v = 4 / 3 π R ^ 3
Surface area: S = 4 π R ^ 2

Derivation of volume formula and surface area formula of sphere

The process of deriving the formula for calculating the volume and surface area of a sphere is as follows:
Assuming that the radius of the sphere is equal to the radius of the bottom of the cylinder, both are R, then the height of the cylinder is 2R, or D. then the letters and symbols are used to express the calculation formula of the volume and surface area of the cylinder, and then the volume and surface area of the sphere are obtained by multiplying them respectively
V cylinder = π R2 × 2R
=πr2×（r+r）
=πr3×2
V-ball = π R3 × 2 ×
= πr3
S Cylinder = π R2 × 2 + π D × D
=πdr+πdd
=(r+d) πd
=3r×2πr
=6πr2
S-sphere = 6 π R2 ×
=4πr2
In this way, the formulas for calculating the volume and surface area of a sphere can be obtained

Derivation formula of sphere surface area
I think so
Think of the top half of the sphere as a fan-shaped circle, and connect the two sides of the central angle
Then s = π D / π ^ 2R × 1 / 16 π ^ 3D ^ 2
Then 2S = sphere surface area = π D / π ^ 2R × 1 / 16 π ^ 3D ^ 2 × 2 = π ^ 2R ^ 2
D is the diameter of the ball, R is the radius, π is the circumference, and ^ 2 is the square
How do you end up with π ^ 2R ^ 2 instead of 4 π R ^ 2
Thank you, 3Q!

To be sure, you are good at thinking, but not comprehensive
If you are a junior or senior high school student, you don't need to deduce, because your knowledge reserve is not enough. If you are a college student, you can use the integral method to deduce

The volume and area formula of sphere

Hello!
The volume formula of sphere: V sphere = 4 / 3 π R ^ 3
The formula of sphere area: s sphere = 4 π R ^ 2
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Derivation of sphere area formula

The upper half of a sphere with radius R is cut into N equal height parts, and each part is regarded as a cylinder, where the radius is equal to the radius of the bottom circle, then the side product of the k-th cylinder from bottom to top s (k) = 2 π R (k) * h, where h = R / N R (k) = radical [R ^ - (KH) ^] s (k) = radical [R ^ - (KR / N) ^] * 2 π R

Triple integral calculation of cone x ^ 2 + y ^ 2

XX + YY = Z is an elliptic paraboloid, not a cone
So XX + YY

Using triple integral to calculate the volume enclosed by z = √ (5-x ^ 2-y ^ 2) and x ^ 2 + y ^ 2 = 4Z

Z = √ (5-x ^ 2-y ^ 2) and x ^ 2 + y ^ 2 = 4Z,
We get x ^ 2 + y ^ 2 = 4, which is the projection of the intersection line on the xoy plane
V =∫∫∫ dv =∫dt∫rdr∫dz
= π∫r[√(5-r^2)-r^2/4]dr
= -π∫√(5-r^2)d(5-r^2) - π∫r^3/4]dr
= π(2/3)[(5-r^2)^(3/2)] - π[r^4/16]
=(2π/3)(5√5-1)-π = 5π(2√5-1)/3.