求由函数y=sinx,y=cosx,x轴上的线段【0,π/2】所围图形绕X轴旋转所成的旋转体体积？求详细计算过程! 求由函数y=sinx,y=cosx,x轴上的线段【0,π/2...
\u6c42\u7531\u51fd\u6570y=sinx,y=cosx,x\u8f74\u4e0a\u7684\u7ebf\u6bb5\u30100,\u03c0/2\u3011\u6240\u56f4\u56fe\u5f62\u7ed5X\u8f74\u65cb\u8f6c\u6240\u6210\u7684\u65cb\u8f6c\u4f53\u4f53\u79ef\uff1f\u89e3\uff1a V=[0\uff0c\u03c0/4)\u03c0\u222bsin²xdx+[\u03c0/4\uff0c\u03c0/2]\u03c0\u222bcos²xdx
=[0\uff0c\u03c0/4](\u03c0/2)\u222b(1-cos2x)dx+[\u03c0/4\uff0c\u03c0/2](\u03c0/2)\u222b(1+cosx)dx
=[0\uff0c\u03c0/4][(\u03c0/2)\u222bdx-(\u03c0/4)\u222b(cos2xd(2x)]+[\u03c0/4\uff0c\u03c0/2][(\u03c0/2)\u222bdx+(\u03c0/4)\u222bcos2xd(2x)]
=[(\u03c0/2)x-(\u03c0/4)sin2x]\ufe31[0\uff0c\u03c0/4]+[(\u03c0/2)x+(\u03c0/4)sin2x]\ufe31[\u03c0/4\uff0c\u03c0/2]
=\u03c0²/8-(\u03c0/4)+[\u03c0²/4-\u03c0²/8-\u03c0/4]
=\u03c0²/4-\u03c0/2
=(\u03c0/4)(\u03c0-2)
❤\u60a8\u7684\u95ee\u9898\u5df2\u7ecf\u88ab\u89e3\u7b54~~(>^\u03c9^<)\u55b5
\u5982\u679c\u91c7\u7eb3\u7684\u8bdd\uff0c\u6211\u662f\u5f88\u5f00\u5fc3\u7684\u54df(\uff5e o \uff5e)~zZ
V = (0..\u03c0/2) \u222b \u03c0 sin²x dx+ (\u03c0/4..\u03c0/2) \u222b \u03c0 cos²x dx
= (0..\u03c0/4) \u222b \u03c0/2 (1-cos2x)dx + (\u03c0/4..\u03c0/2) \u222b \u03c0/2 (1+cos2x) dx
= \u03c0/2 (x-0.5sin2x)|(0..\u03c0/4) + \u03c0/2 (x+0.5sin2x)|(\u03c0/4..\u03c0/2)
= \u03c0/2 * (\u03c0/4-0.5) + \u03c0/2 * (\u03c0/2-\u03c0/4-0.5)
= \u03c0²/4 - \u03c0/2
\u2248 0.8966
y2 = cosx
y1 and y2 相交于（pi/4,sqrt(2)/2)
体积= pi y1^2 dx (0->pi/4 积分）+ pi y2^2 dx (pi/4->pi/2 积分）
= pi sin^2 x dx (0->pi/4) + pi cos^2 x dx (pi/4 ->pi/2)
= pi /2 (1-cos2x) dx + pi / 2 (1+cos2x) dx
= pi/2 * pi/2 - pi/4 sin2x (0->pi/4) + pi/4 sin2x(pi/4->pi/2)
= pi^2 / 4 - pi/2
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