由曲線y=2-x²及直線y=2x-1,x=0圍成的在y軸右邊的區(qū)域D及D繞x軸旋轉(zhuǎn)所得的旋轉(zhuǎn)體
樓主的題目敘述不完整.應(yīng)為：
求由曲線y=2-x²及直線y=2x-1,x=0圍成的圖形在y軸右邊的區(qū)域D的面積及D繞x軸旋轉(zhuǎn)所得的旋轉(zhuǎn)體的體積.
解 曲線y=2-x²與直線y=2x-1在y軸右邊的交點(diǎn)為(1,1),所以區(qū)域D的面積
A=∫[(2-x²)-(2x-1)]dx
=∫[3-x²-2x]dx
=[3x-x^3/3-x^2]
=3-1/3-1
=5/3.
D繞x軸旋轉(zhuǎn)所得的旋轉(zhuǎn)體的體積:
Vx=π∫(2-x^2)^2dx-π∫(2x-1)^2dx
=π∫(4-4x^2+x^4)dx-(π/2)∫(2x-1)^2d(2x-1)
=π[4x-(4/3)x^3+x^5/5]-(π/2)(2x-1)^3/3|
=π[4-4/3+1/5]-(π/2)(1/3)
=27π/10.