dV=2PI y(dA)
where dA=dy*(4-y^2)) from y=0 to 2
dV=2PI(4y-y^3)dy from y= 0 to 2
V= 2PI ( 2y^2-y^4/4)
V=2PI (8-4)=8PI=your answer
Given the area is in the first quadrant bounded by y²=x, the line x=4 and the 0X
What is the volume generated when this area is revolved about the 0X?
the answer is 25.13 but I don't know how. help me please
2 answers
or using discs
V π∫ y^2 dx from x= 0 to 4
= π∫x dx from x = 0 to 4
= π[ (1/2)x^2] from 0 to 4
= π(16/2 - 0)
= 8π
V π∫ y^2 dx from x= 0 to 4
= π∫x dx from x = 0 to 4
= π[ (1/2)x^2] from 0 to 4
= π(16/2 - 0)
= 8π