The surface area of a courtyard is 50 square centimeter, and the bottom area is 15 square centimeter. How many square centimeter is the surface area of a large cylinder made up of two such cylinders? Formula

The surface area of a courtyard is 50 square centimeter, and the bottom area is 15 square centimeter. How many square centimeter is the surface area of a large cylinder made up of two such cylinders? Formula

S column = 2 * 50 - 2 * 15 = 70 square centimeter

Triple integral ∫∫∫∫ ZDV, where Ω is a curved area fraction, z = √ (2-x & # 178; - Y & # 178;) and z = x & # 178; + Y & # 178;)
∫∫∫ ZDV, where Ω is the curved area, z = √ (2-x & # 178; - Y & # 178;) and z = x & # 178; + Y & # 178;
I worked out 0

Because the paraboloid z = x & # 178; + Y & # 178; is open upward, the lowest point is (0,0,0), and z = √ (2 - X & # 178; - Y & # 178;) is the upper hemisphere, the vertex is (0,0, √ 2), so √ (2 - X & # 178; - Y & # 178;) ≥ X & # 178; + Y & # 178; √ (2 - R & # 178;) ≥ R & # 178; = = = > 0 ≤

Calculate the triple integral ∫ ∫ (x ^ 2 + y ^ 2 + Z ^ 2) DV, where Ω is a closed region bounded by z = x ^ 2 + y ^ 2 + Z ^ 2

z = x² + y² + z²
x² + y² + z² - z + 1/4 = 1/4
x² + y² + (z - 1/2)² = (1/2)²
{ x = rsinφcosθ
{ y = rsinφsinθ
{ z = rcosφ
Ω：r² = rcosφ → r = cosφ
∫∫∫ (x² + y² + z²) dV
= ∫∫∫ r² * r²sinφ dV = ∫∫∫ r⁴sinφ dV
= ∫(0→2π) ∫(0→π/2) ∫(0→cosφ) r⁴sinφ drdφdθ
= 2π ∫(0→π/2) sinφ * (1/5)r⁵：(0→cosφ) dφ
= 2π/5 ∫(0→π/2) cos⁵φsinφ dφ
= - 2π/5 ∫(0→π/2) cos⁵φ d(cosφ)
= - 2π/5 * (1/6)cosφ：[0→π/2]
= - π/15 * (0 - 1)
= π/15

∫∫∫∫ (2Z + 1 + 2x) DV, the integral region is the outer side of X & # 178; + Y & # 178; + Z & # 178; = 1,

The integral region is symmetric with respect to xoy and YOZ planes. 2Z is an odd function with respect to Z and 2x is an odd function with respect to X
So 2Z + 2x doesn't need to be integrated, and the result is 0, so the integrand is only 1
Original formula = ∫∫∫ 1 DV
The integrand function is 1, the integral result is the volume of the region, and the volume of the sphere is 4 π / 3
The result is 4 π / 3

Calculate triple integral I = ∫∫∫ Ω (x ^ 2 + y ^ 2 + Z ^ 2) DV, where Ω: x ^ 2 + y ^ 2 + Z ^ 2 = a ^ 2
Seek specific results

The original formula = ∫ D θ∫ D φ∫ R & # 178; * r & # 178; sin φ DR (for spherical coordinate transformation)
=2π∫sinφdφ∫r^4dr
=2π[cos(0)-cos(π)]*a^5/5
=4πa^5/5.

Calculate triple integral ∫ ∫ ∫ Ω (x ^ 2 + y ^ 2) DV, Ω = {(x, y, z) | (x ^ 2 + y ^ 2) / 2 ≤ Z ≤ 2}

The original formula = ∫ D θ∫ RDR ∫ R ^ 2dz (for polar coordinate transformation)
=2π∫r^3(2-r^2/2)dr
=2π∫(2r^3-r^5/2)dr
=2π(2^4/2-2^6/12)
=2π(8-16/3)
=16π/3.

The DV integral domain of triple integral ∫ ∫ (x / A + Y / B + Z / C) is calculated to be the region bounded by three coordinate planes and plane X / A + Y / B + Z / C = 1 (a, B, C > 0)

It can be divided into ∫ ∫ (x / a) DV + ∫ ∫ (Y / b) DV + ∫ ∫ (Z / C) DV, and then it can be used again
∫∫∫(x/a)dV = ∫(x/a)dx∫∫dydz = abc/24
So I = ABC / 8

The triple integral ∫ ∫ (Ω) XY ^ 2Z ^ 3Dv Ω is the space region surrounded by the saddle surface z = xy and the plane y = x, x = 1, z = 0
1/364

∫∫∫(Ω)xy²z³dV
=∫[0→1]xdx∫[0→x]y²dy∫[0→xy] z³dz
=(1/4)∫[0→1]xdx∫[0→x] y²z^4 |[0→xy]dy
=(1/4)∫[0→1]xdx∫[0→x] x^4y^6 dy
=(1/28)∫[0→1] x^5y^7 |[0→x] dx
=(1/28)∫[0→1] x^12 dx
=(1/364)x^13 |[0→1]
=1/364

There is a semi cylinder as shown in the figure. It is known that its bottom diameter is 20 cm, its height is 8 cm, and its surface area is ()

Surface area = tangential area + bottom area + 1 / 2 side area
Cut area = bottom diameter × height = 20 × 8 = 160 square centimeter
Bottom area = (20 △ 2) &# 178; × 3.14 = 314cm2
Side area = bottom perimeter × height = 20 × 3.14 × 8 = 502.4 square centimeter
Surface area: 160 + 314 + 502.4 △ 2 = 725.2 square centimeter

A cylinder with a bottom diameter of 10 cm and a height of 20 cm is cut into the same two sections by a knife, and the surface area of one section (as shown in the figure below) is 473 square centimeters
Finding the area of cross section

The area of cross section is 3.14 × 5 & # 178; = 78.5 (square centimeter)