Vx=π∫abf2(x)dxSo, we will have
Vx=V1+V2=π∫02x2dx+π∫24(4−(x−2)2)2dx∫02x2dx=3x3∣02=38
∫24(4−x2+4x−4)dx=∫24(−x2+4x)dx=(−3x3+42x2)∣24=−364+2⋅16+38−2⋅4=−356+24
From here
Vx=(38−356+24)π=(24−348)π=(24−16)π=8π sq.units.
Answer. 8π sq.units.
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