Find the surface area and volume of the figure below 1. A vertical cylinder is 6cm wide and 12cm high 2. A cube is 20 cm high, 15 cm long and 10 cm wide 3. The cylinder is 14 cm wide and 5 cm high Find the surface area and volume of each

1. Surface area:
6 / 2 = 3 (CM)
3.14 * 2 * 3 * 12 = 226.08 (cm2)
3.14 * 3 * 3 = 28.26 (cm2)
226.08 + 28.26 = 254.34 (cm2)
Volume:
28.26 * 12 = 339.12 (cm3)
2. Surface area:
(20 * 15 + 20 * 10 + 15 * 10) * 2 = (300 + 200 + 150) * 2 = 650 * 2 = 1300 (cm2)
Volume:
20 * 15 * 10 = 3000 (cm3)
3. Surface area:
14 * 3.14 * 5 = 219.8 (cm2)
14 / 2 = 7 (CM)
3.14 * 7 * 7 = 153.89 (cm2)
219.8 + 153.89 = 373.69 (cm2)
Volume:
153.89 * 5 = 769.45 (cm3)

Find the surface area and volume of this figure. (unit: cm)

10 × 10 × 6 = 600 (square centimeter), 10 × 10 × 10-4 × 4, = 1000-64, = 936 (cubic centimeter). A: the surface area of this figure is 600 square centimeter, and the volume is 936 cubic centimeter

Find the solid surrounded by the rotating paraboloid x ^ 2 + y ^ 2 = AZ and the cone z = 2A radical (x ^ 2 + y ^ 2) (a > 0)
The volume I calculated is 5A ^ 3 pies / 6, which is better for the process

Find the volume of the solid surrounded by the cone z = √ x ^ 2 + y ^ 2 and the hemisphere z = √ 1-x ^ 2-y ^ 2

There are two methods: one is to use integral, the other is to use solid angle, one is to use integral, the other is to use spherical coordinates, let the included angle between radius R and Z axis be φ, and the included angle between R projection on xoy plane and X axis be θ, then the integral region is: 0 ≤ R ≤ 1,0 ≤ φ ≤ π / 4,0 ≤ θ ≤ 2 π, the solid volume enclosed by two curved surfaces is v = ∫ DV = ∫∫∫ dxdydz = ∫∫∫ R & M

A plane parallel to 6x + y + 6Z + 5 = 0 and the volume of the tetrahedron enclosed with the coordinate axis is 1, the equation of the plane is obtained

Let the plane 6x + y + 6Z + B = 0, then use B to express the intercept of XYZ, calculate the volume as 1, and get that there should be more than one B.B

Through the plane of point (1,3,2), make the volume of tetrahedron besieged with coordinate plane minimum, find the plane equation and minimum body!

Let the plane equation be ax + by + CZ = 1, the intercept equation be x (1 / a) + y (1 / b) + Z (1 / C) = 1, the passing point equation be a + 3B + 2C = 1, the cone volume formula v = (1 / 3) (1 / 2) (1 / ABC) = (1 / 6) / ABC Let f = (1 / 6) / ABC - λ (a + 3B + 2c-1) be f'a = (- 1 / 6) / A ^ 2BC - λ = 0f'b = (- 1 / 6) / AB ^ 2c-3 λ = 0f

In the plane passing through point P (2,1,13), find a plane to minimize the volume of the solid in the first hexagram

Let the plane equation passing through point P (2,1,13) be a (X-2) + B (Y-1) + C (z-13) = 0, that is, ax + by + CZ = 2A + B + 13C is transformed into intercept equation & nbsp; x2a + B + C3a + y2a + B + C3b + z2a + B + C3c = 1

In the first hexagram limit, the tangent plane of x ^ 2 + y ^ 2 + Z ^ 2 = 3 is made to minimize the volume of the tetrahedron surrounded by the plane and three coordinate planes, and the coordinates of the tangent point are obtained
The minimum volume of the tetrahedron is obtained

Let the tangent coordinates be (a, B, c), then the equation of the tangent plane is ax + by + CZ = 1. The axial intercept of the tangent plane is 1 / A, 1 / B, 1 / C respectively, and the tetrahedral volume v = 1 / (6abc). (ABC) & # 178; ≤ (A & # 178; + B & # 178; + C & # 178;) & # 179 / / 27 = 1 / 27V = 1 / (6abc) ≥ 6 √ 27 = 18 √ 3. When a = b = C = √ 3, V is the minimum

Find the volume of the solid surrounded by the surface AZ = a ^ 2-x ^ 2-y ^ 2 and the plane x + y + Z = a (a > 0) and three coordinate planes
For example, after thinking for a long time, I really can't figure out how to form a three-dimensional enclosure, and MATLAB is not very proficient. Ask friends for help, how to solve this problem and how to form a three-dimensional enclosure?

The key is the surface AZ = a ^ 2-x ^ 2-y ^ 2. First, consider the graph obtained by using the plane section surface parallel to xoy, where Z is a constant, so the section x ^ 2 + y ^ 2 = a ^ 2-az is a circle. Then consider the projection of the surface on the YOZ plane, where x = 0, so the projection is y ^ 2 = a ^ 2-az, which is a parabola with the opening downward, By synthesizing two points, we can imagine that the surface is an inverted paraboloid (note that although the intersection of the surface and the X, y, z axes is a, but it is not a sphere! If we investigate the curvature at the intersection, we will find that they are different). Now the volume is equal to the volume of the paraboloid in the first trigram minus the volume of the cone, and the volume of the cone is a ^ 3 / 6, Using the method of "two before one", the volume = (1 / 4) ∫ DZ ∫ DXDY, the integral of Z is limited from 0 to a, and ∫ ∫ DXDY is equal to the area of the paraboloid cut by the plane parallel to xoy (note that the result contains z). From X ^ 2 + y ^ 2 = a ^ 2-az, we know that ∫ DXDY = π (a ^ 2-az), the integral of this part of the volume = π a ^ 3 / 8, so the volume = (π / 8-1 / 6) a ^ 3

Make the tangent plane of the ellipsoid x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 + Z ^ 2 / C ^ 2 = 1 in the first hexagram to minimize the volume of the tetrahedron enclosed by the tangent plane and three coordinate planes, and find the coordinates of the tangent point. I want to ask why we set u = lnx0 + lny0 + lnz0 when we do Lagrange multiplication? We should not directly bring in his volume formula v = ABC / (6x0y0z0)?

Because the volume is the largest, as long as the three intercept x0, Y0, Z0 of tangent plane satisfy: x0y0z0 is the largest
For the convenience of calculation, take logarithm, ln (x0y0z0) = lnx0 + lny0 + lnz0

Find the surface area and volume of the figure below 1. A vertical cylinder is 6cm wide and 12cm high 2. A cube is 20 cm high, 15 cm long and 10 cm wide 3. The cylinder is 14 cm wide and 5 cm high Find the surface area and volume of each