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 rotatin

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数学微积分中的积分求旋转的体积只要答案即可Findthevolumeofthesolidobtainedbyrotatingtheregionboundedbyy=x^2,y=0,x=5,andabo

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 rotatin
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.

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 rotatin
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)

1:题错了吧,这片区域不是封闭的,没法求体积的
2:13/4π-π√e
3:864√6
4:π²/24+2π

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 rotatin 求定积分,旋转体积, 大学数学微积分,这两道用换元积分法怎么求,定积分的题 数学一道积分,什么时候用定积分的几何意义求,什么时候用微积分 大学数学微积分,定积分的题 数学微积分中的dx是对e^x3进行定积分的意思? 一道定积分题目.关于求面积的.由y=4/(x^2),y=1,x=1围成的图形.a:绕着X轴旋转,求体积.b:绕着y轴旋转,求体积.c:绕着x=1旋转,求体积.d:绕着x=2旋转,求体积.只要用定积分表示就好.是求体积,说错了。 数学微积分求体积题目如图 用微积分方法解立体图形的体积,求此图形曲线绕X轴旋转所形成的图形的体积,要求用微积分方法 [微积分] 求 1/cosx 的积分 微积分方法解图形体积求此图形围绕x轴旋转之后得到新图形的体积,要求用微积分方法做 微积分:求定积分. 我的数学不好 微积分中的定理怎么也看不懂 就是高等数学的微积分定理一阿打不出来 郁闷就是变上限积分函数 后的定理一 数学上的积分法是什么 是微积分吗 微积分:积分的计算 定积分 求体积请问定积分求体积,是不是 绕X轴旋转就对X积分,绕Y轴旋转就对Y积分.急, 曲线y=sinx(0≤x≤π)绕y轴旋转一周得到几何体的体积是.记得是Y轴,别写出绕X轴的,好的马上给分我们现在学到用微积分求体积。我做的是这样,y=πfo~1[(arcsiny)^2]dy,我用了分部积分,可还是不会 再求解一道微积分求体积的题~求由y=x^2 和y=16 围成的区域绕y=16 旋转之后的体积;该区域绕y=17旋转之后的体积.