Solution:
Solve a system
y=x2
y=4x-x2
We obtain x=0,x=2
Then in a new coordinates y'=y-6 we have two functions:y'1=x2-6, y'2=4x-x2-6
V="\\pi" "\\int" 20((y'1)2-(y'2)2)dx=...=
π"\\int" 20(8x3-40x2+48x)dx="\\pi" (2x4-40x3/3+24x2)|20="\\pi" (32-320/3+96)=64"\\pi"/3
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