no need for calculus.
your solid is a fulcrum formed by the following:
You have a cone on its side, (the x-axis is the axis of the cone)
height = 3, radius = 9
the other :
height = 5, radius = 15
V = (1/3)π(15^2)(5) - (1/3)π(9^2)(3)
= (1/3)π (1125 - 243)
= 882π/3
= 294π cubic units
The region bounded by y=3x, y=0, x=3, and x=5 is rotated about the x-axis. Find the volume of the resulting solid.
2 answers
However, just so you get to use your calculus to analyze the frustrum, just think of it as a stack of thin discs of thickness dx:
v = ∫πr^2 dx
where r=y=3x
v = ∫[3,5] π(3x)^2 dx = 294π
Or, as a bunch of nested shells, it gets a bit more complicated, since the interior is a solid cylinder of radius 9 and height 2. The volume of the portion bounded by y=3x is
∫[9,15] 2πrh dy
where r=y and h=(5-x)=(5-y/3)
∫[9,15] 2πy(5-y/3) dy = 132π
add that to the cylinder (162π) and you get 294π
v = ∫πr^2 dx
where r=y=3x
v = ∫[3,5] π(3x)^2 dx = 294π
Or, as a bunch of nested shells, it gets a bit more complicated, since the interior is a solid cylinder of radius 9 and height 2. The volume of the portion bounded by y=3x is
∫[9,15] 2πrh dy
where r=y and h=(5-x)=(5-y/3)
∫[9,15] 2πy(5-y/3) dy = 132π
add that to the cylinder (162π) and you get 294π