'Find the volume of the solid generated by revolving the region bounded by x=16-y^2, x-axis, and y-axis around the x-axis.'
I know in theory how to do it, but I'm a little confused by what constitutes the bounded region because it's a sideways parabola, but if the x-axis is a bound, does that mean only half of it counts?
So far I have that Volume equals pi times the integral from 0 to 16, but then getting the integrand conuses me.
Thamks!
2 answers
confuses* my apologies
If you are using a limit of 16, then you must be integrating on x, so using discs of thickness dx,
v = ∫[0,16] πr^2 dx
where r = y = √(16-x)
v = ∫[0,16] π(16-x) dx = 128π
Or, you could use shells of thickness dy, where
v = ∫[0,4] 2πrh dy
where r=y and h=x = 16-y^2
v = ∫[0,4] 2πy(16-y^2) dy = 128π
v = ∫[0,16] πr^2 dx
where r = y = √(16-x)
v = ∫[0,16] π(16-x) dx = 128π
Or, you could use shells of thickness dy, where
v = ∫[0,4] 2πrh dy
where r=y and h=x = 16-y^2
v = ∫[0,4] 2πy(16-y^2) dy = 128π
5 answers