数学微积分中的积分 求旋转的体积 只要答案即可

题目

数学微积分中的积分 求旋转的体积 只要答案即可
Find the volume of the solid obtained by rotating the region bounded by y=x^2,y=0,x=5,and  about
the  y-axis.求绕y轴旋转后的体积.
Find the volume of the solid formed by rotating the region enclosed by
y=e^x +5,y=0,x=0,x=0.5, about the y-axis.求绕y轴旋转后的体积.
A ball of radius 15 has a round hole of radius 3 drilled through its center.
Find the volume of the resulting solid.
Find the volume of the solid obtained by rotating the region bounded by the
given curves about the specified axis.y=0,y=(cos6x),x=0,x=pi/12, about the axis y=-6.
有一个答案算一个,多多益善.

答案

it's easier to use y-axis as variable.  The range is 0 to 5 for x,0 to 25 for y; y = x^2,x = √y
V = ∫₀²⁵π[5² - (√y)²]dy = π(25y - y²/2)|₀²⁵ = 625π/2


Just consider the first quadrant in a plane.  A circle of raduius 15 is expressed as y = √(15² - x²); the hole can be considered as line y = 3; 3 = √(15² - x²),x² = 216
they inercept at (√216,3)
The volume is double the result from rotating the region about the x-axis
V = 2∫π[(15² - x²) - 3²]dx  0 to √216
= 2π(216x - x³/3)   0 to √216
= (1064√54)π/3


x = 0,y = 1; x = π/12,y = cos(π/2) = 0
The inner radius of the solid is r = 0 - (-6) = 6; the outer radius of the solid is R = cos(6x) - (-6) = 6 + cos(6x)
V = ∫π(((6 + cos(6x))² - 6²)dx   0 to π/12
= π∫[12cos(6x) + cos²(6x)]dx = π∫[12cos(6x) + cos²(6x)]dx
= π∫[12cos(6x) + 1/2 + (1/2)cos(12x)]dx
= π[x/2 + 2sin(6x) + (1/24)cos(12x)]    0 to π/12
= π(π/12 + 2 - 1/24) - π(0 + 0 + 1/24)
= π(2 + π/12)