find volumes of solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y=square root of x
y=0
x=3
2 answers
when the line x = 9 please
The region bounded by
y=√x
y=0
x=3
has its "vertices" at (0,0), (3,0), and (3,√3)
To rotate it about the line x=9, it's best to think of flat discs with holes in them. So, since the volume of a thin disc of thickness dy is πr^2 dy, and we need to subtract the holes, we have
v = ∫[0,√3] π(R^2-r^2) dy
where R = 9-x and r = 6
That makes
v = ∫[0,√3] π((9-y^2)^2-6^2) dy
= π(y^5/5 - 6y^3 + 45y) [0,√3]
= 144/5 √3 π
You can do it with thin shells of thickness dx, as well. The volume of such a shell is 2πrh dx, so
v = ∫[0,3] 2πrh dx
where r = 9-x h = y
v = ∫[0,3] 2π(9-x)√x dx
= 4/5 π (15-x) x^(3/2) [0,3]
= 144/5 √3 π
y=√x
y=0
x=3
has its "vertices" at (0,0), (3,0), and (3,√3)
To rotate it about the line x=9, it's best to think of flat discs with holes in them. So, since the volume of a thin disc of thickness dy is πr^2 dy, and we need to subtract the holes, we have
v = ∫[0,√3] π(R^2-r^2) dy
where R = 9-x and r = 6
That makes
v = ∫[0,√3] π((9-y^2)^2-6^2) dy
= π(y^5/5 - 6y^3 + 45y) [0,√3]
= 144/5 √3 π
You can do it with thin shells of thickness dx, as well. The volume of such a shell is 2πrh dx, so
v = ∫[0,3] 2πrh dx
where r = 9-x h = y
v = ∫[0,3] 2π(9-x)√x dx
= 4/5 π (15-x) x^(3/2) [0,3]
= 144/5 √3 π
1 answer