Higher mathematics problem: how to calculate the surface area of the ball by the integral of the circumference? Sphere surface area s = 4 * Pai * r * r How to use it Circumference C = 2 * Pai * r Integral? Is it S=2∏＊∫0 R∫0 ∏/2 r* sinθ dr * dθ =4∏R2 ? (for format problem, superscript and subscript are inseparable, ∫ is the integral sign, ∫ 0 R means the product from 0 to R)

Higher mathematics problem: how to calculate the surface area of the ball by the integral of the circumference? Sphere surface area s = 4 * Pai * r * r How to use it Circumference C = 2 * Pai * r Integral? Is it S=2∏＊∫0 R∫0 ∏/2 r* sinθ dr * dθ =4∏R2 ? (for format problem, superscript and subscript are inseparable, ∫ is the integral sign, ∫ 0 R means the product from 0 to R)

Integral formula has its mathematical meaning. What do you mean by that? It should be written for the purpose of getting the form of 4 Π R2. The correct integral should be written as follows: S = ∫ 2 Π R * DS (where DS is not the area but the arc length, it is equal to R * D θ) = ∫ (0 - Π) 2 Π R * sin θ * r * D θ = 4 Π R

How is the surface area formula derived?
How does cosr in the surface area formula come out? How does it have something to do with the partial reciprocal

The surface R (x, y) = (x, y, f (x, y)) takes (x, y) as a parameter, and its two natural tangent vectors are
rx = (1, 0, fx)
ry = (0, 1, fy)
Where Rx is the partial derivative of R to x, the other symbols are similar
Since the vectors n = (- FX, - fy, 1) and Rx, ry are perpendicular, n is the normal vector of the surface at P = R (x, y), that is, the normal vector of the tangent plane P passing through P
Let k = (0, 0, 1) be the positive direction of the z-axis unit, that is, the normal vector of the XY plane, so that the angle between P and XY plane is equal to the angle between N and K, and its cosine is equal to / | n | K | = 1 / (sqrt (FX ^ 2 + FY ^ 2 + 1)
Where ⁃ sqrt is the square root

Given the square of X - 3x + 1 = 0, find the square of X + 1 / X and the square of X / X to the fourth power + the square of X + 1

x^2-3x+1=0
x^2+1=3x
x+1/x=3
(x^4+x^2+1)/x^2
=x^+1+1/x^2
=(x+1)^2-1
=9-1
=8
So x ^ 2 / (x ^ 4 + x ^ 2 + 1) = 1 / 8

The power a of 2 = 3, the power B of 2 = 5, the power B of 3 = 13, find the power 2Ab of 2

2^2ab
=(2^a)^2b
=3^2b
=(3^b)2
=13^2
=169

(3 / 5 of 2ab-3 / 7) / 2Ab

(3 / 5 of 2ab-3 / 7) / 2Ab
=b²/5-3/14;
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(5 / 3 a squared B - 1 / 3 a squared b) times (- 0.2ab)

(5a^2b/3-a^3b^2/3-1)*(-0.2ab)
=5a^2b/3*(-0.2ab)-a^3b^2/3*(-0.2ab)+(0.2ab)
=-a^3b^2/3+a^4b^3/15+ab/5

Given that the square of a is equal to B + 5 and the square of B is equal to a + 5, find the cubic power of a-2ab + B

a³-2ab+b³=a³-ab+b³-ab=a(a²-b)+b(b²-a)=5a+5b=5(a+b)=5*（-1）=-5

Find the area of the figure surrounded by the curve r = asin θ

This is a double leaf rose. Because of the symmetry, just multiply the first quadrant by 4
A=1/2∫r^2dθ=1/2∫a^2sin^2θdθ 0→π/2
=a^2/4∫(1-cos2θ)dθ 0→π/2
=a^2/4(θ-1/2sin2θ) 0→π/2
=a^2/8 * π/2=a^2π/16
Area of enclosed figure = a ^ 2 π / 16 * 4 = a ^ 2 π / 4

Find the common area of the figure surrounded by the curves ρ = a (1 + cos θ) and ρ = a (a > 0)

ρ = a (1 + cos θ) is an epicycloid, and ρ = a is a circle. The left part of the common area enclosed by them needs to be integrated. The right part is the circle of ρ = a on the positive half axis of X, which is π a ^ 2 / 2. The left part is s = 1 / 2 ∫ (π / 2 product to 3 π / 2) ρ ^ 2D θ = 1 / 2A ^ 2 (θ + 1 / 2sin2 θ) | π / 2 to 3 π / 2 = 1 / 2A ^

Find the area of the common part of the figure enclosed by the curve P = a sin β, P = a (COS β + sin β) (a > 0)

Draw a picture. The intersection of two circles, one is the origin, the other is on the Y axis
Area s = ∫ (0 to π / 2) 1 / 2 × [asin β] ^ 2 D β + ∫ (π / 2 to 3 π / 4) 1 / 2 × [a (COS β + sin β)] ^ 2 D β = π a ^ 2 / 4-A ^ 2 / 4