Given that the radius of the ball is a, the radius of the bottom surface of the inner cylinder of the ball is R and the height is h, what are the values of R and h, the volume of the inner cylinder is the largest?

Given that the radius of the ball is a, the radius of the bottom surface of the inner cylinder of the ball is R and the height is h, what are the values of R and h, the volume of the inner cylinder is the largest?

R = 4th power under root sign 2A ^ 2 / 3
A ^ 2-r ^ 2 under H = 2 radical

Ask the next simple calculus question~ y = Ln(sec(2^(1/x))) The answer to calculus is not: 1/sec(2^(1/x)) X sec(2^(1/x))tan(2^(1/x)) X 2^(1/x)Ln1/x I'm based on the formula: D / DX (LN U) = 1 / u Du / DX The question is, how do you take calculus like 2 ^ (1 / x) or x ^ x? Mathematics has been neglected for some time, but now it's a little rusty

From the derivation formula: the derivative of a ^ x is a ^ x * LNX, so the derivative of 2 ^ (1 / x) is 2 ^ (1 / x) * (- 1 / x ^ 2) (this is the derivation rule of composite function)
Note: This is not calculus, but differentiation, but derivation

Find the integral to the 10th power of (3 + 2x)

∫(3+2x)＾10dx
=∫1/2*(3+2x)＾10d（3+2x）
=1/2*1/11(3+2x)＾11+C
=[(3+2x)＾11]/22+C

2 questions Find the area between two intersections of two curves: 1) y = - 2x ^ 2 + 4x and y = 2x ^ 2.2) y = 2x + 6 and Y ^ 2 = 16x + 132

1）2/3
2)250/3

Two calculus problems ①∫sin ² 3X dx ②∫(sec(1/X)tan(1/X))/X dx No one paid any attention to me... TWT

∫sin2 3xdx
=1/2∫(1-cos6x)dx
=1/2x-1/2 × 1/6∫cos6xd6x
=1/2x-1/12sin6x+c
Did you copy the second question correctly? Why is it so cumbersome? Please check it, and then I'll write it down. It's really hard to type

Two calculus problems 1. Find the general solution of the differential equation y '= (x + y) ^ 2 2. Find the power series Σ [n = 1 to + ∞] (x ^ (3n-1)) / (3n-2)! The sum function of If the answer is good, there will be additional points

1、 Let t = y + X, y '= t' - 1, t '= T ^ 2 + 1, DT / (t ²+ 1) = DX, arctant = x + C, arctan (y + x) = x + C, y = - x + Tan (x + C). Second, Let f (x) = ∑ [n = 1 to + ∞] (x ^ (3n-1)) / (3n-2)= G (0) = 0, G (0) = 0, G (0) = 0, G (0) = 1, G (0) = 1, G (0) = 0, G (0) = 0, G (0) = 0, G (0) = 0, G (0) = 0. G = (e ^ (- x (2)) * cos (x * √3 / 2 (x * √3 / 2)) GG (x), then G '' '' '' (0, G (0) = g, G (0) = g (0), G (0) = 0, G (0) = 0, G (0) = 0, G (0) = 0, G (0) = 0 (0) = 0, G (0) = 0) (g = (e ^ (- X / 2)) (e ^ (- X / 2)) * cos (e ^ (- x (x (x ^ (- X / 2)))) + (- x (e ^ (- x (- X / 2 / 2 / 2)) * cos (x ^ (x * √ (- 3 / 2)) * 3 / 2)) *)) * cos (x^ (- X / 2)) * sin (x * √ 3 / 2)) * (x / 3)